Intro to 3-Dimensional Solids Return to Table of Contents Slide 6 - - PDF document

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Geometry

3D Geometry

2015-10-28 www.njctl.org

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Table of Contents Intro to 3-D Solids Views & Drawings of 3-D Solids Surface Area of a Prism Surface Area of a Cylinder Surface Area of a Pyramid Surface Area of a Cone

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Volume of a Prism Volume of a Cylinder Volume of a Pyramid Volume of a Cone Surface Area & Volume of Spheres Cavaleri's Principle Similar Solids PARCC Sample Questions

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Throughout this unit, the Standards for Mathematical Practice are used. MP1: Making sense of problems & persevere in solving them. MP2: Reason abstractly & quantitatively. MP3: Construct viable arguments and critique the reasoning of

• thers.

MP4: Model with mathematics. MP5: Use appropriate tools strategically. MP6: Attend to precision. MP7: Look for & make use of structure. MP8: Look for & express regularity in repeated reasoning. Additional questions are included on the slides using the "Math Practice" Pull-tabs (e.g. a blank one is shown to the right on this slide) with a reference to the standards used. If questions already exist on a slide, then the specific MPs that the questions address are listed in the Pull-tab.

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Throughout this unit, the Standards for Mathematical Practice are used. MP1: Making sense of problems & persevere in solving them. MP2: Reason abstractly & quantitatively. MP3: Construct viable arguments and critique the reasoning of

• thers.

MP4: Model with mathematics. MP5: Use appropriate tools strategically. MP6: Attend to precision. MP7: Look for & make use of structure. MP8: Look for & express regularity in repeated reasoning. Additional questions are included on the slides using the "Math Practice" Pull-tabs (e.g. a blank one is shown to the right on this slide) with a reference to the standards used. If questions already exist on a slide, then the specific MPs that the questions address are listed in the Pull-tab.

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Math Practice

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Intro to 3-Dimensional Solids

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2-dimensional drawings use only the x and y axes

X Y Length width

Y X Length w i d t h Y X Length w i d t h

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Y X Z height height Y X

3-dimensional drawings include the x, y and z-axis. The z-axis is the third dimension. The third dimension is the height of the figure

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Y X Z height height Y X

x

Y

Intro to 3-D Solids Slide 8 / 311

Y X Z height

Y X

X Y

r

Intro to 3-D Solids Slide 9 / 311

To give a figure more of a 3-dimensional look, lines that are not visible from the angle the figure is being viewed are drawn as dashed line segments. These are called hidden lines.

Y X Z height height

Intro to 3-D Solids Slide 10 / 311

A Polyhedron (pl. Polyhedra) is a solid that is bounded by polygons, called faces. An edge is the line segment formed by the intersection of 2 faces. A vertex is a point where 3 or more edges meet Face Edge Vertex

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The 3-Dimensional Figures discussed in this unit are: Pyramids Cylinders Prisms

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The 3-Dimensional Figures discussed in this unit are:

. C

Cones: Spheres:

Intro to 3-D Solids Slide 13 / 311 Right Vs. Oblique

Right Right In Right Prisms & Cylinders, the bases are aligned directly above one another. The edges are perpendicular with both bases.

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Right Vs. Oblique

In Oblique Prisms & Cylinders, the bases are not aligned directly above one another. The edges are not perpendicular with the bases.

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Right Oblique Right Oblique In Right Pyramids & Cones, the vertex is aligned directly above the center of the base. In Oblique Pyramids & Cones, the vertex is not aligned directly above the center of the base.

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Prisms have 2 congruent polygonal bases. The sides of a base are called base edges. The segments connecting corresponding vertices are lateral edges. A B C X Y Z In this diagram: There are 6 vertices: A, B, C, X, Y, & Z There are 2 bases: ABC & XYZ. There are 6 base edges: AB, BC, AC, XY, YZ, & XZ. There are 3 lateral edges: AX, BY, & CZ. This prism has a total of 9 edges.

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The polygons that make up the surface of the figure are called faces. The bases are a type of face and are parallel and congruent to each other. The lateral edges are the sides of the lateral faces. A B C X Y Z In this diagram: There are 2 bases: ABC & XYZ. There are 3 lateral faces: AXBY, BYCZ, & CZAX. This prism has a total of 5 faces.

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1

A AB B DE C FS D CP E FA F CD G NP H BC I DQ

A B C D E F M N P Q R S Choose all of the base edges.

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1

A AB B DE C FS D CP E FA F CD G NP H BC I DQ

A B C D E F M N P Q R S Choose all of the base edges.

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Answer

A, B, E, F, G, H Slide 19 (Answer) / 311

2 A AB B CD C ER D BN E DQ F QR G MS H AM I CP Choose all of the lateral edges. A B C D E F M N P Q R S

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2 A AB B CD C ER D BN E DQ F QR G MS H AM I CP Choose all of the lateral edges. A B C D E F M N P Q R S

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Answer

C, D, E, H, I

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3 Chooses all of the bases. A AFSM B FERS C EDQR D ABCDEF E CDQP F BCPN G MNPQRS H ABNM A B C D E F M N P Q R S

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3 Chooses all of the bases. A AFSM B FERS C EDQR D ABCDEF E CDQP F BCPN G MNPQRS H ABNM A B C D E F M N P Q R S

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Answer

D, G

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4 Chooses all of the lateral faces. A AFSM B FERS C EDQR D ABCDEF E CDQP F BCPN G MNPQRS H ABNM A B C D E F M N P Q R S

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4 Chooses all of the lateral faces. A AFSM B FERS C EDQR D ABCDEF E CDQP F BCPN G MNPQRS H ABNM A B C D E F M N P Q R S

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Answer

A, B, C, E, F, H

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5 Chooses all of the faces. A AFSM B FERS C EDQR D ABCDEF E CDQP F BCPN G MNPQRS H ABNM A B C D E F M N P Q R S

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5 Chooses all of the faces. A AFSM B FERS C EDQR D ABCDEF E CDQP F BCPN G MNPQRS H ABNM A B C D E F M N P Q R S

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Answer

All of the choices are faces

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A pyramid has 1 base with vertices and the lateral edges go to a single vertex. A M N P R S Q This pyramid has: 6 lateral edges, 6 base edges, 12 edges (total) 7 vertices

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A pyramid has faces that are polygons: 1 base and triangles that are the lateral faces.

A M N P Q R S

This pyramid has: 6 lateral faces, 1 base, 7 faces (total)

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6 Choose all of the base edges. A VN B KN C VL D LM E VM F VK K L M N V G KL H NM

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6 Choose all of the base edges. A VN B KN C VL D LM E VM F VK K L M N V G KL H NM

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Answer

B, D, G, H

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7 Choose all of the lateral edges. A VN B KN C VL D LM E VM F VK G KL H NM K L M N V

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7 Choose all of the lateral edges. A VN B KN C VL D LM E VM F VK G KL H NM K L M N V

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Answer

A, C, E, F

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8 How many edges does the pyramid have? K L M N V

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8 How many edges does the pyramid have? K L M N V

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Answer

8

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9 Choose all of the lateral faces. A KNV B NMV C KLMN D VML E KLV K L M N V

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9 Choose all of the lateral faces. A KNV B NMV C KLMN D VML E KLV K L M N V

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Answer

A, B, D, E

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10 Choose all of the bases. A KNV B NMV C KLMN D VML E KLV K L M N V

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10 Choose all of the bases. A KNV B NMV C KLMN D VML E KLV K L M N V

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Answer

C

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11 How many faces does the pyramid have? K L M N V

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11 How many faces does the pyramid have? K L M N V

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Answer

5

Slide 31 (Answer) / 311 . . A B

A cylinder has 2 bases which are congruent circles. The lateral face is a rectangle wrapped around the circles. A & B are the bases

• f the

cylinder.

Intro to 3-D Solids

A cylinder can also be formed by rotating a rectangle about an axis. Click for sample animation

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A cone, like a pyramid, has one base which is a circle.

. N V

N is the base of the cone. V is the vertex of the cone.

Intro to 3-D Solids

A cone can also be formed by rotating a right triangle about

• ne of its legs.

Click for sample animation

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A sphere is a 3-dimensional circle in that every point on the sphere is the same distance from the center.

. C

Similar to a circle, a sphere is named by its center

• point. Sphere C is the solid shown above.

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12 Which solids have 2 bases? A Prism B Pyramid C Cylinder D Cone E Sphere

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12 Which solids have 2 bases? A Prism B Pyramid C Cylinder D Cone E Sphere

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Answer

A & C

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13 Which solid has one vertex? A Prism B Pyramid C Cylinder D Cone E Sphere

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13 Which solid has one vertex? A Prism B Pyramid C Cylinder D Cone E Sphere

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Answer

D

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14 Which solid has more base edges than lateral edges? A Prism B Pyramid C Cylinder D Cone E Sphere

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14 Which solid has more base edges than lateral edges? A Prism B Pyramid C Cylinder D Cone E Sphere

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Answer

A

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15 Which solid(s) have no vertices? A Prism B Pyramid C Cylinder D Cone E Sphere

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15 Which solid(s) have no vertices? A Prism B Pyramid C Cylinder D Cone E Sphere

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Answer

C & E

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16 Which solid is formed when rotating an isosceles triangle about its altitude? A a prism B a cylinder C a pyramid D a cone E a sphere

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16 Which solid is formed when rotating an isosceles triangle about its altitude? A a prism B a cylinder C a pyramid D a cone E a sphere

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Answer

D

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Euler's Theorem states that the number of faces (F), vertices (V), and edges (E) satisfy the formula F + V = E + 2 A B C X Y Z A M N P Q R S F = 5 V = 6 E = 9 5 + 6 = 9 + 2 11 = 11 F = 7 V = 7 E = 12 7 + 7 = 12 + 2 14 = 14

Intro to 3-D Solids Slide 40 / 311

Example: A solid has 12 faces, 2 decagons and 10 trapezoids. How many vertices does the solid have? V + F = E + 2 V + 12 = 30 + 2 V + 12 = 32 V = 20 On their own, the 2 decagons & 10 trapezoids have 2(10) + 10(4) = 60 edges. In a 3-D solid, each side is shared by 2 polygons. Therefore, the number of edges in the solid is 60/2 = 30.

Intro to 3-D Solids

click click click click

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Example: A solid has 12 faces, 2 decagons and 10 trapezoids. How many vertices does the solid have? V + F = E + 2 V + 12 = 30 + 2 V + 12 = 32 V = 20 On their own, the 2 decagons & 10 trapezoids have 2(10) + 10(4) = 60 edges. In a 3-D solid, each side is shared by 2 polygons. Therefore, the number of edges in the solid is 60/2 = 30.

Intro to 3-D Solids

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Math Practice Questioning to help address MP standards: What information do you have? (MP1) What is this problem asking? (MP1) What strategies are you going to use? (MP1) How can you represent the problem with numbers and symbols? (MP2) Create an equation to represent the problem. (MP2) Would it help to draw a picture? (MP4 & MP5) Does your answer seem reasonable? Why or why not? (MP3) Can you find a shortcut to solve the problem? How would your shortcut make the problem easier? (MP8)

• Patterns of a prism: V = 2(# of sides of the

base), E = 3(# of sides of the base) & F = (# of sides of the base) + 2

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Example: A solid has 9 faces, 1 octagon and 8 triangles. How many vertices does the solid have? V + F = E + 2 V + 9 = 16 + 2 V + 9 = 18 V = 9 What information do you have? 9 faces & the 2 types of faces

Intro to 3-D Solids

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What is the problem asking? Create an equation to represent the problem. How are the number of edges in the 2-D faces, related to the number of edges in the polyhedron? Write a number sentence to describe this situation. (1(8) + 8(3))/2 (8 + 24)/2 32/2 16 edges

click click click click click

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Example: A solid has 9 faces, 1 octagon and 8 triangles. How many vertices does the solid have? V + F = E + 2 V + 9 = 16 + 2 V + 9 = 18 V = 9 What information do you have? 9 faces & the 2 types of faces

Intro to 3-D Solids

click click click click

What is the problem asking? Create an equation to represent the problem. How are the number of edges in the 2-D faces, related to the number of edges in the polyhedron? Write a number sentence to describe this situation. (1(8) + 8(3))/2 (8 + 24)/2 32/2 16 edges

click click click click click [This object is a pull tab]

Math Practice Questions on this slide address MP standards 1st Question: MP1 2nd set of questions: MP7 & MP4 3rd set of questions: MP1 & MP2 Additional Questions to help address MP standards: Would it help to draw a picture? (MP4 & MP5) Does your answer seem reasonable? Why or why not? (MP3) Can you find a shortcut to solve the problem? How would your shortcut make the problem easier? (MP8)

• Patterns of a pyramid: V = (# of sides of the

base) + 1, E = 2(# of sides of the base) & F = (# of sides of the base) + 1

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17 A solid has 10 faces, one of them being a nonagon and 9 triangles. How many vertices does it have? A 8 B 9 C 10 D 18

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17 A solid has 10 faces, one of them being a nonagon and 9 triangles. How many vertices does it have? A 8 B 9 C 10 D 18

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Answer

C

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18 A solid has 12 faces, all of them being pentagons. How many vertices does it have? A 30 B 20 C 15 D 10

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18 A solid has 12 faces, all of them being pentagons. How many vertices does it have? A 30 B 20 C 15 D 10

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Answer

B

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19 A solid has 8 faces, all of them being triangles. How many vertices does it have? A 24 B 12 C 8 D 6

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19 A solid has 8 faces, all of them being triangles. How many vertices does it have? A 24 B 12 C 8 D 6

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Answer

D

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A cross-section is the locus of points of the intersection of a plane and a 3-D solid.

Cross-Section Intro to 3-D Solids Slide 46 / 311

Think about it as if the plane were a knife and you were cutting the shape, what would the cut look like?

Cross-Section

Circle Ellipse Parabola (with the inner section shaded)

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Cross-sections of a surface are a 2-dimensional figure. Cross-sections of a solid are a 2-dimensional figure and its interior. The top can be removed to see the cross section. (Try it out)

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20 What is the locus of points (cross-section) of a cube and a plane perpendicular to the base and parallel to the non-intersecting sides? A square B rectangle C trapezoid D hexagon E rhombus F parallelogram G triangle H circle

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20 What is the locus of points (cross-section) of a cube and a plane perpendicular to the base and parallel to the non-intersecting sides? A square B rectangle C trapezoid D hexagon E rhombus F parallelogram G triangle H circle

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Answer

A

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21 If the length of each edge of the cube is 12 inches, what would be the area of the cross-section of the cube and a plane perpendicular to the base and parallel to the non- intersecting sides? A 72 sq inches B 144 sq inches C 187.06 sq inches D 203.65 sq inches 12 in.

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21 If the length of each edge of the cube is 12 inches, what would be the area of the cross-section of the cube and a plane perpendicular to the base and parallel to the non- intersecting sides? A 72 sq inches B 144 sq inches C 187.06 sq inches D 203.65 sq inches 12 in.

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Answer

B

12 in. 12 in.

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22 What is the locus of points of a cube and a plane that contains the diagonal of the base and is perpendicular to the base? A square B rectangle C trapezoid D hexagon E rhombus F parallelogram G triangle H circle

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22 What is the locus of points of a cube and a plane that contains the diagonal of the base and is perpendicular to the base? A square B rectangle C trapezoid D hexagon E rhombus F parallelogram G triangle H circle

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Answer

B

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23 If the length of each edge of the cube is 12 inches, what would be the area of the cross-section of the cube and a plane that contains the diagonal of the base and is perpendicular to the base? A 72 sq inches B 144 sq inches C 187.06 sq inches D 203.65 sq inches 12 in.

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23 If the length of each edge of the cube is 12 inches, what would be the area of the cross-section of the cube and a plane that contains the diagonal of the base and is perpendicular to the base? A 72 sq inches B 144 sq inches C 187.06 sq inches D 203.65 sq inches 12 in.

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Answer

D

12√2 in. 12 in.

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24 What is the locus of points of a cube and a plane that contains the diagonal of the base but does not intersect the opposite base? A square B rectangle C trapezoid D hexagon E rhombus F parallelogram G triangle H circle

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24 What is the locus of points of a cube and a plane that contains the diagonal of the base but does not intersect the opposite base? A square B rectangle C trapezoid D hexagon E rhombus F parallelogram G triangle H circle

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Answer

G

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25 What is the locus of points of a cube and a plane that intersects all of the faces? A square B rectangle C trapezoid D hexagon E rhombus F parallelogram G triangle H circle

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25 What is the locus of points of a cube and a plane that intersects all of the faces? A square B rectangle C trapezoid D hexagon E rhombus F parallelogram G triangle H circle

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Answer

D

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Views & Drawings of 3-D Solids

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Isometric drawings are drawings that look 3-D & are created on a grid

• f dots using 3 axes that intersect to form 120° & 60° angles.

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Example: Create an Isometric drawing of a cube.

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Example: Create an Isometric drawing of a cube.

Views & Drawings

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Answer

OR Slide 57 (Answer) / 311

An Orthographic projection is a 2-D drawing that shows the different viewpoints of an object, usually from the front, top & side. Each drawing depends on your position relative to the figure. Front Side Top (from front)

Views & Drawings Slide 58 / 311

Consider these three people viewing a pyramid:

Views & Drawings Slide 59 / 311

Consider these three people viewing a pyramid: The orange person is standing in front of a face, so their view is a triangle.

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Consider these three people viewing a pyramid: The green person is standing in front of a lateral edge, so from their view they can see 2 faces.

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Consider these three people viewing a pyramid: The purple person is flying over and can see the four lateral faces.

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26 Given the surface shown, what would be the view from point A? A a Rectangle B a Square C a Circle D a Pentagon E a Triangle F a Parallelogram G a Hexagon H a Trapezoid

A

(front)

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26 Given the surface shown, what would be the view from point A? A a Rectangle B a Square C a Circle D a Pentagon E a Triangle F a Parallelogram G a Hexagon H a Trapezoid

A

(front)

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Answer

E

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27 Given the surface shown, what would be the view from point A? A a Rectangle B a Square C a Circle D a Pentagon E a Triangle F a Parallelogram G a Hexagon H a Trapezoid

A (top) Slide 64 / 311

27 Given the surface shown, what would be the view from point A? A a Rectangle B a Square C a Circle D a Pentagon E a Triangle F a Parallelogram G a Hexagon H a Trapezoid

A (top)

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Answer

A

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28 Given the surface shown, what would be the view from point A? A a Rectangle B a Square C a Circle D a Pentagon E a Triangle F a Parallelogram G a Hexagon H a Trapezoid

A (top)

right square prism

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28 Given the surface shown, what would be the view from point A? A a Rectangle B a Square C a Circle D a Pentagon E a Triangle F a Parallelogram G a Hexagon H a Trapezoid

A (top)

right square prism

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Answer

B

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29 Given the surface shown, what would be the view from point A? A a Rectangle B a Square C a Circle D a Pentagon E a Triangle F a Parallelogram G a Hexagon H a trapezoid

A

(front) right square prism

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29 Given the surface shown, what would be the view from point A? A a Rectangle B a Square C a Circle D a Pentagon E a Triangle F a Parallelogram G a Hexagon H a trapezoid

A

(front) right square prism

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Answer

A

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30 Given the surface shown, what would be the view from point A? A a Rectangle B a Square C a Circle D a Pentagon E a Triangle F a Parallelogram G a Hexagon H a Trapezoid

A

(front)

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30 Given the surface shown, what would be the view from point A? A a Rectangle B a Square C a Circle D a Pentagon E a Triangle F a Parallelogram G a Hexagon H a Trapezoid

A

(front)

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Answer

E

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31 Given the surface shown, what would be the view from point A? A a Rectangle B a Square C a Circle D a Pentagon E a Triangle F a Parallelogram G a Hexagon H a Trapezoid

A

(above)

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31 Given the surface shown, what would be the view from point A? A a Rectangle B a Square C a Circle D a Pentagon E a Triangle F a Parallelogram G a Hexagon H a Trapezoid

A

(above)

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Answer

C

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32 Given the surface shown, what would be the view from point A? A a Rectangle B a Square C a Circle D a Pentagon E a Triangle F a Parallelogram G a Hexagon H a Trapezoid

A (above) Slide 69 / 311

32 Given the surface shown, what would be the view from point A? A a Rectangle B a Square C a Circle D a Pentagon E a Triangle F a Parallelogram G a Hexagon H a Trapezoid

A (above)

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Answer

C

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33 Given the surface shown, what would be the view from point A? A a Rectangle B a Square C a Circle D a Pentagon E a Triangle F a Parallelogram G a Hexagon H a Trapezoid

A

(front)

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33 Given the surface shown, what would be the view from point A? A a Rectangle B a Square C a Circle D a Pentagon E a Triangle F a Parallelogram G a Hexagon H a Trapezoid

A

(front)

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Answer

A

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34 Given the surface shown, what would be the view from point A? A a Rectangle B a Square C a Circle D a Pentagon E a Triangle F a Parallelogram G a Hexagon H a Trapezoid

A

sphere

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34 Given the surface shown, what would be the view from point A? A a Rectangle B a Square C a Circle D a Pentagon E a Triangle F a Parallelogram G a Hexagon H a Trapezoid

A

sphere

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Answer

C

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(Looking down from above) What would the view be like from each position?

Views & Drawings Slide 72 / 311 A

What would the view be like from each position? From A, how many columns of blocks are visible?

• 3 columns

How tall is each column?

• first one is 4 high
• second & third columns

are each 2 blocks high

Click to reveal Click to reveal

Views & Drawings Slide 73 / 311

A

What would the view be like from each position? From A, how many columns of blocks are visible?

• 3 columns

How tall is each column?

• first one is 4 high
• second & third columns

are each 2 blocks high

Click to reveal Click to reveal

Views & Drawings

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Math Practice Questions on this slide address MP standards 1st Question: MP1 2nd Question: MP1 Additional Questions to help address MP standards: Does your answer seem reasonable? Why or why not? (MP3) Why does your final drawing make sense? (MP4)

Slide 73 (Answer) / 311 B

What would the view be like from each position? From B, how many columns of blocks are visible?

• 2 columns

How tall is each column?

• left one is 3 high
• right one is 4 high

Click to reveal Click to reveal

Views & Drawings Slide 74 / 311 B

What would the view be like from each position? From B, how many columns of blocks are visible?

• 2 columns

How tall is each column?

• left one is 3 high
• right one is 4 high

Click to reveal Click to reveal

Views & Drawings

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Math Practice Questions on this slide address MP standards 1st Question: MP1 2nd Question: MP1 Additional Questions to help address MP standards: Does your answer seem reasonable? Why or why not? (MP3) Why does your final drawing make sense? (MP4)

Slide 74 (Answer) / 311

C

(Looking down from above) What would the view be like from each position? From C, how many columns of blocks are visible?

• 3 columns

How tall is each column?

• all of them are 2 blocks

high

Click to reveal Click to reveal

Views & Drawings Slide 75 / 311 C

(Looking down from above) What would the view be like from each position? From C, how many columns of blocks are visible?

• 3 columns

How tall is each column?

• all of them are 2 blocks

high

Click to reveal Click to reveal

Views & Drawings

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Math Practice Questions on this slide address MP standards 1st Question: MP1 2nd Question: MP1 Additional Questions to help address MP standards: Does your answer seem reasonable? Why or why not? (MP3) Why does your final drawing make sense? (MP4)

Slide 75 (Answer) / 311

Front Side Above Draw the 3 views.

Side View Top View Front View Move for Answer

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Front Side Above Draw the 3 views.

Side View Top View Front View Move for Answer

Views & Drawings

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Math Practice Questions to help address MP standards: From each viewpoint, how many columns of blocks are visible? (MP1) For this viewpoint, how tall is each column? (MP1) Does your answer seem reasonable? Why or why not? (MP3) Why does your final drawing make sense? (MP4)

Slide 76 (Answer) / 311

Front Side Above Draw the 3 views. Above Front Side

Views & Drawings

Move for Answer

Slide 77 / 311

Front Side Above Draw the 3 views. Above Front Side

Views & Drawings

Move for Answer

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Math Practice Questions to help address MP standards: From each viewpoint, how many columns of blocks are visible? (MP1) For this viewpoint, how tall is each column? (MP1) Does your answer seem reasonable? Why or why not? (MP3) Why does your final drawing make sense? (MP4)

Slide 77 (Answer) / 311

Here are 3 views of a solid, draw a 3-dimensional representation. Top Front Side L R F

Views & Drawings

Move for Answer

Slide 78 / 311

Here are 3 views of a solid, draw a 3-dimensional representation. Top Front Side L R F

Views & Drawings

Move for Answer

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What strategies are you going to use? (MP1) Can you guess & check? (MP5)

• Referring to drawing a "guess diagram" for the

solid & "checking" to see if it works Does your answer seem reasonable? Why or why not? (MP3) Why does your final drawing make sense? (MP4)

Slide 78 (Answer) / 311

Here are 3 views of a solid, draw a 3-dimensional representation. Top F L R Side Front

Views & Drawings

Move for Answer

Slide 79 / 311

Here are 3 views of a solid, draw a 3-dimensional representation. Top F L R Side Front

Views & Drawings

Move for Answer

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What strategies are you going to use? (MP1) Can you guess & check? (MP5)

• Referring to drawing a "guess diagram" for the

solid & "checking" to see if it works Does your answer seem reasonable? Why or why not? (MP3) Why does your final drawing make sense? (MP4)

Slide 79 (Answer) / 311

Surface Area of a Prism

Return to Table of Contents

Slide 80 / 311

A Net is a 2-dimensional shape that folds into a 3-dimensional figure. The Net shows all of the faces of the surface.

Net

6 6 4 6 4 12 4 Shown is the net of a right rectangular prism. 12 6 4

Slide 81 / 311

A Net is a 2-dimensional shape that folds into a 3-dimensional figure. The Net shows all of the faces of the surface.

Net

6 6 4 6 4 12 4 Shown is the net of a right rectangular prism. 12 6 4

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Teacher's Note

The PARCC Reference sheet for the HS level does NOT contain any formulas to calculate the Surface Area (reference sheet is linked to this pull tab - just click on it to download it). Encourage students to either memorize future formulas,

• r draw the net each time so that

they can break down the solid into smaller 2-D shapes.

Slide 81 (Answer) / 311

The net shown is a right triangular prism. The lateral faces are

• rectangles. The bases are on opposite sides of the rectangles,

although they do not need to be on the same rectangle.

Net Slide 82 / 311

The nets shown are for the same right triangular prism.

Net Slide 83 / 311

Nets of oblique prisms have parallelograms as lateral faces.

Nets Slide 84 / 311 Rectangular Prisms

cube w w w H H H ℓ ℓ ℓ

Slide 85 / 311 Rectangular Prisms

cube w w w H H H ℓ ℓ ℓ

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Teacher's Note

To avoid confusion with the "heights" when calculating the surface area of a triangular prism, the height of the prism has been assigned "H". The triangular height will be "h" starting on slide #98

Slide 85 (Answer) / 311

Base Base height Base height Base A prism has 2 bases. The base of a rectangular prism is a rectangle. The height of the prism is the length between the two bases.

Rectangular Prisms Slide 86 / 311

The Surface Area of a figure is the total amount of area that is needed to cover the entire figure (e.g. the amount of wrapping paper required to wrap a gift). Area Area Area Area Area Area

Top Area Side Area Front Area Bottom Area Back Area Side Area

The Surface Area of a figure is the sum of the areas of each side of the figure.

Rectangular Prisms Slide 87 / 311 Finding the Surface Area of a Rectangular Prism

H w ℓ Area of the Top = ℓ x w Area of the Bottom = ℓ x w Area of the Front = ℓ x H Area of the Back = ℓ x H Area of Left Side = w x H Area of Right Side = w x H The Surface Area is the sum of all the areas S.A. = ℓw + ℓw + ℓH + ℓH + wH + wH S.A. = 2 ℓw + 2 ℓH + 2wH

Slide 88 / 311

Finding the Surface Area of a Rectangular Prism

H w ℓ Area of the Top = ℓ x w Area of the Bottom = ℓ x w Area of the Front = ℓ x H Area of the Back = ℓ x H Area of Left Side = w x H Area of Right Side = w x H The Surface Area is the sum of all the areas S.A. = ℓw + ℓw + ℓH + ℓH + wH + wH S.A. = 2 ℓw + 2 ℓH + 2wH

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Teacher's Note

To avoid confusion with the "heights" when calculating the surface area of a triangular prism, the height of the prism has been assigned "H". The triangular height will be "h" starting on slide #98

Slide 88 (Answer) / 311

Example: Find the surface area of the prism 7 4 3 Area of Top & Bottom Area of Right & Left Area of Front & Back A = 7(4) = 28u2 A = 3(4) = 12 u2 A = 3(7) = 21 u2

Click Click Click

Total Surface Area = 2(28) + 2(12) + 2(21) = 56 + 24 + 42 = 122 units2

Click Click

Finding the Surface Area of a Rectangular Prism Slide 89 / 311

Example: Find the surface area of the prism 7 4 3 Area of Top & Bottom Area of Right & Left Area of Front & Back A = 7(4) = 28u2 A = 3(4) = 12 u2 A = 3(7) = 21 u2

Click Click Click

Total Surface Area = 2(28) + 2(12) + 2(21) = 56 + 24 + 42 = 122 units2

Click Click

Finding the Surface Area of a Rectangular Prism

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Can you do this mentally? (MP5) What labels could you use? (MP6)

Slide 89 (Answer) / 311

35 What is the total surface area, in square units? 4 5 9

Slide 90 / 311

35 What is the total surface area, in square units? 4 5 9

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Answer

SA = 2(4)(5) + 2(4)(9) + 2(5)(9) SA = 40 + 72 + 90 SA = 202 sq. units

Slide 90 (Answer) / 311

36 What is the total surface area, in square units? 8 8 8

Slide 91 / 311

36 What is the total surface area, in square units? 8 8 8

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Answer

SA = 6(8)(8) SA = 6(64) SA = 384 sq. units

Slide 91 (Answer) / 311

37 Troy wants to build a cube out of straws. The cube is to have a total surface area of 96 in2, what is the total length of the straws, in inches?

Slide 92 / 311

37 Troy wants to build a cube out of straws. The cube is to have a total surface area of 96 in2, what is the total length of the straws, in inches?

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Answer

96 = 6(x)(x) 96 = 6x2 16 = x2 x = 4 in.

Slide 92 (Answer) / 311

S.A. = 2B + PH The Surface Area is the sum of the areas of the 2 Bases plus the Lateral Area (Perimeter of the base, P, times the height of the prism, H) The Lateral Area is the area of the Lateral Surface. The Lateral Surface is the part that wraps around the middle of the figure (in between the two bases).

Another Way of Looking at Surface Area

Lateral Surface B a s e B a s e

Base Base

Slide 93 / 311

S.A. = 2B + PH The Surface Area is the sum of the areas of the 2 Bases plus the Lateral Area (Perimeter of the base, P, times the height of the prism, H) The Lateral Area is the area of the Lateral Surface. The Lateral Surface is the part that wraps around the middle of the figure (in between the two bases).

Another Way of Looking at Surface Area

Lateral Surface B a s e B a s e

Base Base

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Teacher's Note

To avoid confusion with the "heights" when calculating the surface area of a triangular prism, the height of the prism has been assigned "H". The triangular height will be "h" starting on slide #98

Slide 93 (Answer) / 311

Base Base w H ℓ Another formula for Surface Area of a right prism: S.A. = 2B + PH B = Area of the base B = ℓw P = Perimeter of the base P = 2 ℓ + 2w H = Height of the prism S.A. = 2B + PH S.A. = 2 ℓw + (2 ℓ +2w)H S.A. = 2 ℓw + 2 ℓH + 2wH

Rectangular Prisms Slide 94 / 311

Base Base w H ℓ In the surface area formula, 2B is the sum of the area of the 2 bases. What does PH represent? The area of lateral faces or Lateral Area

Click

Rectangular Prisms

Another formula for Surface Area of a right prism: S.A. = 2B + PH B = Area of the base B = ℓw P = Perimeter of the base P = 2 ℓ + 2w H = Height of the prism

Slide 95 / 311

38 If the base of the prism is 12 by 6, what is the lateral area, in sq ft? 12 ft 6 ft 4 ft

Slide 96 / 311

38 If the base of the prism is 12 by 6, what is the lateral area, in sq ft? 12 ft 6 ft 4 ft

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Answer

LA = PH P = 2(12) + 2(6) = 36 ft LA = 36(4) LA = 144 sq. ft

Slide 96 (Answer) / 311

39 The surface area of the rectangular prism is : A 24 sq ft B 144 sq ft C 288 sq ft D 48 sq ft E 72 sq ft 12 ft 6 ft 4 ft

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39 The surface area of the rectangular prism is : A 24 sq ft B 144 sq ft C 288 sq ft D 48 sq ft E 72 sq ft 12 ft 6 ft 4 ft

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Answer

C SA = 2B + PH SA = 2(12)(6) + 36(4) SA = 144 + 144 sq ft

Slide 97 (Answer) / 311

40 If 7 by 6 is base of the prism, what is the lateral area, in sq units? 7 9 6

Slide 98 / 311

40 If 7 by 6 is base of the prism, what is the lateral area, in sq units? 7 9 6

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Answer

LA = PH P = 2(6) + 2(7) = 26 units LA = 26(9) LA = 234 sq. units

Slide 98 (Answer) / 311

41 What is the total square units of the surface area? 7 9 6

Slide 99 / 311

41 What is the total square units of the surface area? 7 9 6

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Answer

SA = 2B + PH SA = 2(6)(7) + 26(9) SA = 84 + 234 SA = 318 units2

Slide 99 (Answer) / 311

42 Find the value of y, if the lateral area is 144 sq units, and y by 6 is the base. y 6 8

Slide 100 / 311

42 Find the value of y, if the lateral area is 144 sq units, and y by 6 is the base. y 6 8

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Answer

LA = PH P = 2(6) + 2y = 12 + 2y 144 = (12 + 2y)8 144 = 96 + 16y 48 = 16y 3 = y

Slide 100 (Answer) / 311

43 What is the value of the missing variable if the surface area is 350 sq. ft. A 7 ft B 8.3 ft C 12 ft D 15 ft x ft 5 ft 10 ft

Slide 101 / 311

43 What is the value of the missing variable if the surface area is 350 sq. ft. A 7 ft B 8.3 ft C 12 ft D 15 ft x ft 5 ft 10 ft

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Answer

B SA = 2B + PH P = 2(5) + 2x = 10 + 2x B = 5x 350 = 2(5x) + (10 + 2x)10 350 = 10x + 100 + 20x 350 = 30x + 100 250 = 30x x = 8.3 ft

Slide 101 (Answer) / 311

44 Sharon was invited to Maria's birthday party. For a present, she purchased an iHome (a clock radio for an iPod or iPhone) which is contained in a box that measures 7 inches in length, 5 inches in width, and 4 inches in height. How much wrapping paper does Sharon need to wrap Maria's present?

Slide 102 / 311

44 Sharon was invited to Maria's birthday party. For a present, she purchased an iHome (a clock radio for an iPod or iPhone) which is contained in a box that measures 7 inches in length, 5 inches in width, and 4 inches in height. How much wrapping paper does Sharon need to wrap Maria's present?

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Answer

SA = 2(5)(7) + 2(5)(4) + 2(7)(4) = 70 + 40 + 56 SA = 166 sq. inches

Slide 102 (Answer) / 311

Other Prisms Slide 103 / 311

base base height base base height base base height base base height A Prism has 2 Bases The Base of a Prism matches the first word in the name of the prism. e.g. the Base of a Triangular Prism is a Triangle The Height of the Prism is the length between the two bases

Other Prisms Slide 104 / 311

The Surface Area of a figure is the total amount of Area that is needed to cover the entire figure (e.g. the amount of wrapping paper required to wrap a gift). The Surface Area of a figure is the sum of the areas of each side of the figure Area Area Area Area Area Area Area

Area

Area Area

Other Prisms

Triangular Prism Net of the Triangular Prism

Slide 105 / 311

Finding the Surface Area of a Right Prism

Surface Area: S.A. = 2B + PH B = Area of the triangular base = ½bh P = Perimeter of the triangular base = a + b + c H = Height of the prism Lateral Area = PH = (a + b + c)H The Lateral Area is the area of the Lateral Surface, the rectangular area that wraps around the prism between the triangular bases. base base Prism's height a b c H P = a + b + c a c b c a Lateral Surface H h b B = ½ bh Note: The formula above will work for any right prism.

Slide 106 / 311

Example: Find the lateral area and surface area of the right triangular prism. 10 6 11 Since it has a base that is a right triangle, we need to find the base of the triangle using Pythagorean Theorem. 62 + b2 = 102 36 + b2 = 100 b2 = 64 b = 8 units Next, calculate the perimeter of your base. P = 6 + 8 + 10 = 24 units Use this to find the Lateral Area LA = PH = 24(11) = 264 units

2

Other Prisms Slide 107 / 311

Example: Find the lateral area and surface area of the right triangular prism. 10 6 11 Since it has a base that is a right triangle, we need to find the base of the triangle using Pythagorean Theorem. 62 + b2 = 102 36 + b2 = 100 b2 = 64 b = 8 units Next, calculate the perimeter of your base. P = 6 + 8 + 10 = 24 units Use this to find the Lateral Area LA = PH = 24(11) = 264 units

2

Other Prisms

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Can you do this mentally? (MP5)

• Referring to the Pythagorean Triple

Can you find a shortcut to solve the problem? How would the shortcut make the problem easier? (MP8)

• Referring to the Pythagorean Triple

What labels could you use? (MP6)

Slide 107 (Answer) / 311

10 6 11 Example: Find the lateral area and surface area of the right triangular prism. Then, calculate the area of your base, B B = (1/2)(8)(6) = 24 units2 Finally, calculate your Surface Area. SA = 2B + PH SA = 2(24) + (24)(11) SA = 48 + 264 = 312 units2

Other Prisms Slide 108 / 311

10 6 11 Example: Find the lateral area and surface area of the right triangular prism. Then, calculate the area of your base, B B = (1/2)(8)(6) = 24 units2 Finally, calculate your Surface Area. SA = 2B + PH SA = 2(24) + (24)(11) SA = 48 + 264 = 312 units2

Other Prisms

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Can you do this mentally? (MP5) What labels could you use? (MP6)

Slide 108 (Answer) / 311

Example: Find the lateral area and surface area of the triangular prism. 9 9 9 12 Since it has a base that is an equilateral triangle, we need to find the height of the triangle using Pythagorean Theorem or the 30-60-90 Triangle Theorem. 4.52 + b2 = 92 20.25 + b2 = 81 b2 = 60.75 b = 4.5√3 units = 7.79 units Next, calculate the perimeter of your base. P = 9 + 9 + 9 = 27 units Use this to find the Lateral Area LA = PH = 27(12) = 324 units2

Other Prisms Slide 109 / 311

Example: Find the lateral area and surface area of the triangular prism. 9 9 9 12 Since it has a base that is an equilateral triangle, we need to find the height of the triangle using Pythagorean Theorem or the 30-60-90 Triangle Theorem. 4.52 + b2 = 92 20.25 + b2 = 81 b2 = 60.75 b = 4.5√3 units = 7.79 units Next, calculate the perimeter of your base. P = 9 + 9 + 9 = 27 units Use this to find the Lateral Area LA = PH = 27(12) = 324 units2

Other Prisms

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Can you do this mentally? (MP5)

• Referring to the 30-60-90 triangle

Can you find a shortcut to solve the problem? How would the shortcut make the problem easier? (MP8)

• Referring to the 30-60-90 triangle

What labels could you use? (MP6)

Slide 109 (Answer) / 311

Example: Find the lateral area and surface area of the triangular prism. Then, calculate the area of your base, B B = (1/2)(9)(4.5√3) = 20.25√3 units2 = 35.07 units2 Finally, calculate your Surface Area. SA = 2B + PH SA = 2(35.07) + (27)(12) SA = 70.14 + 324 = 394.14 units2

Other Prisms

9 9 9 12

Slide 110 / 311

Example: Find the lateral area and surface area of the triangular prism. Then, calculate the area of your base, B B = (1/2)(9)(4.5√3) = 20.25√3 units2 = 35.07 units2 Finally, calculate your Surface Area. SA = 2B + PH SA = 2(35.07) + (27)(12) SA = 70.14 + 324 = 394.14 units2

Other Prisms

9 9 9 12

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Can you do this mentally? (MP5) What labels could you use? (MP6)

Slide 110 (Answer) / 311

45 The height of the triangular prism below is 11 ft, the height of the base is 3 ft, and the triangular base is an isosceles triangle. Find the surface area. A 88 sq ft B 132 sq ft C 198 sq ft D 222 sq ft

3 ft

5 ft 11 ft

Slide 111 / 311

45 The height of the triangular prism below is 11 ft, the height of the base is 3 ft, and the triangular base is an isosceles triangle. Find the surface area. A 88 sq ft B 132 sq ft C 198 sq ft D 222 sq ft

3 ft

5 ft 11 ft

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Answer

D (1/2) of the base in the triangle is 4...3-4-5 Pyth. Triple, so the base of the triangle is 8. P = 5 + 5 + 8 = 18 B = (1/2)(8)(3) = 12 SA = 2(12) + 18(11) SA = 24 + 198 SA = 222 sq ft

Slide 111 (Answer) / 311

46 The height of the triangular prism below is 3, and the triangular base is an equilateral triangle. Find the surface area. A 64 sq ft B 127.43 sq ft C 72 sq ft D 55.43 sq ft 8 ft 3 ft

Slide 112 / 311

46 The height of the triangular prism below is 3, and the triangular base is an equilateral triangle. Find the surface area. A 64 sq ft B 127.43 sq ft C 72 sq ft D 55.43 sq ft 8 ft 3 ft

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Answer

B The height of the triangular base is 4√3 = 6.93...30-60-90 triangle. P = 8(3) = 24 B = (1/2)(8)(4√3) = 16√3 SA = 2(16√3) + 24(3) SA = 32√3 + 72 SA = 127.43 sq ft

4 4 8 h

60o 30o

Slide 112 (Answer) / 311

47 Find the lateral area of the right prism. 5 5 6

Slide 113 / 311

47 Find the lateral area of the right prism. 5 5 6

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Answer

hypotenuse of the right triangle is 5√2...45-45-90 triangle P = 5 + 5 + 5√2 = 10 + 5√2 = = 17.07 LA = (10 + 5√2)6 LA = 60 + 30√2 LA = 102.43 sq units

Slide 113 (Answer) / 311

Finding the Surface Area of a Right Prism

Surface Area : S.A. = 2B + PH B = Area of the regular hexagonal base = ½aP

• a is the apothem of the regular base

P = Perimeter of the base = b + c + d + e + f + g H = Height of the prism = H Lateral Area = PH = (b + c + d + e + f + g)H a B = ½ aP

g c b H e f c d e f b d P = b + c + d + e + f + g

base base Prism's height

Slide 114 / 311

a B = ½ aP

Finding the Surface Area of a Right Prism

P = b + c + d + e + f + g

base base Prism's height

g c b H e f c d e f b d

The Lateral Area is the area of the Lateral Surface, the rectangular area that wraps around the prism between the triangular bases.

Slide 115 / 311

8 in 7 in 30° 4 in. a Example: Find the lateral area and surface area of the regular hexagonal prism. Because the base is a regular polygon, we need to calculate the apothem. To begin, figure out the central angle & top angle in the triangle. = 60° = central angle = 30° = top angle of the triangle. 360 6 60 2

Click Click Click

Other Prisms Slide 116 / 311

8 in 7 in 30° 4 in. a Example: Find the lateral area and surface area of the regular hexagonal prism. Because the base is a regular polygon, we need to calculate the apothem. To begin, figure out the central angle & top angle in the triangle. = 60° = central angle = 30° = top angle of the triangle. 360 6 60 2

Click Click Click

Other Prisms

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Teacher's Note

When calculating the surface area and/or volume of regular polygonal prisms/ pyramids, students could also find the measurement of the 2 congruent base angles using the Triangle Sum Theorem. Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) Can you do this mentally? (MP5) What labels could you use? (MP6)

Slide 116 (Answer) / 311

Example: Find the lateral area and surface area of the regular hexagonal prism. Next find the apothem using trigonometry, or special right triangles (if it applies). tan 30 = atan30 = 4 tan30 tan30 4 a a = 4√3 = 6.93 in.

Click Click

Other Prisms

Click

8 in 7 in

Slide 117 / 311

Example: Find the lateral area and surface area of the regular hexagonal prism. Next find the apothem using trigonometry, or special right triangles (if it applies). tan 30 = atan30 = 4 tan30 tan30 4 a a = 4√3 = 6.93 in.

Click Click

Other Prisms

Click

8 in 7 in

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Can you do this mentally? (MP5) What labels could you use? (MP6) How is right triangle trigonometry used to calculate the segment lengths of regular polygons? (MP7) Can you find a shortcut to solve the problem? How would your shortcut make the problem easier? (MP8)

• Referring to 30-60-90 triangle

Slide 117 (Answer) / 311

B = (1/2)aP = (1/2)(4√3)(48) = 96√3 in2 = 166.28in2 Example: Find the lateral area and surface area of the regular hexagonal prism. Next, calculate the perimeter of your base. P = 8(6) = 48 in Use this to find the Lateral Area LA = PH = 48(7) = 336 in2 Then, calculate the area of your base, B Finally, calculate your Surface Area. SA = 2B + PH SA = 2(166.28) + (48)(7) SA = 332.56 + 336 = 668.56 in2

Click Click Click Click Click Click Click Click Click

Other Prisms

8 in 7 in

Slide 118 / 311

B = (1/2)aP = (1/2)(4√3)(48) = 96√3 in2 = 166.28in2 Example: Find the lateral area and surface area of the regular hexagonal prism. Next, calculate the perimeter of your base. P = 8(6) = 48 in Use this to find the Lateral Area LA = PH = 48(7) = 336 in2 Then, calculate the area of your base, B Finally, calculate your Surface Area. SA = 2B + PH SA = 2(166.28) + (48)(7) SA = 332.56 + 336 = 668.56 in2

Click Click Click Click Click Click Click Click Click

Other Prisms

8 in 7 in

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Can you do this mentally? (MP5) What labels could you use? (MP6)

Slide 118 (Answer) / 311

36° 3 in. a

Example: Find the lateral area and surface area of the right prism. The base is a regular pentagon. 6 ft 10 ft Because the base is a regular polygon, we need to calculate the apothem. To begin, figure out the central angle & top angle in the triangle. = 72° = central angle = 36° = top angle of the triangle. 360 5 72 2

Other Prisms

Click Click Click

Slide 119 / 311

36° 3 in. a

Example: Find the lateral area and surface area of the right prism. The base is a regular pentagon. 6 ft 10 ft Because the base is a regular polygon, we need to calculate the apothem. To begin, figure out the central angle & top angle in the triangle. = 72° = central angle = 36° = top angle of the triangle. 360 5 72 2

Other Prisms

Click Click Click [This object is a pull tab]

Teacher's Note

When calculating the surface area and/or volume of regular polygonal prisms/ pyramids, students could also find the measurement of the 2 congruent base angles using the Triangle Sum Theorem. Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) Can you do this mentally? (MP5) What labels could you use? (MP6)

Slide 119 (Answer) / 311

Example: Find the lateral area and surface area of the right prism. The base is a regular pentagon. Next find the apothem using trigonometry, or special right triangles (if it applies). tan 36 = 3 a atan36 = 3 tan36 tan36 a = 4.13 in. 6 ft 10 ft

Other Prisms

Click Click Click

Slide 120 / 311

Example: Find the lateral area and surface area of the right prism. The base is a regular pentagon. Next find the apothem using trigonometry, or special right triangles (if it applies). tan 36 = 3 a atan36 = 3 tan36 tan36 a = 4.13 in. 6 ft 10 ft

Other Prisms

Click Click Click [This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) How is right triangle trigonometry used to calculate the segment lengths of regular polygons? (MP7) What labels could you use? (MP6)

Slide 120 (Answer) / 311

Example: Find the lateral area and surface area of the right prism. The base is a regular pentagon. Next, calculate the perimeter of your base. P = 5(6) = 30 in Use this to find the Lateral Area LA = PH = 30(10) = 300 in2 Then, calculate the area of your base, B B = (1/2)aP = (1/2)(4.13)(30) = 61.95 in2 Finally, calculate your Surface Area. SA = 2B + PH SA = 2(61.95) + (30)(10) SA = 123.9 + 300 = 423.9 in2 6 ft 10 ft

Other Prisms

Click Click Click Click Click Click Click Click

Slide 121 / 311

Example: Find the lateral area and surface area of the right prism. The base is a regular pentagon. Next, calculate the perimeter of your base. P = 5(6) = 30 in Use this to find the Lateral Area LA = PH = 30(10) = 300 in2 Then, calculate the area of your base, B B = (1/2)aP = (1/2)(4.13)(30) = 61.95 in2 Finally, calculate your Surface Area. SA = 2B + PH SA = 2(61.95) + (30)(10) SA = 123.9 + 300 = 423.9 in2 6 ft 10 ft

Other Prisms

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Can you do this mentally? (MP5) What labels could you use? (MP6)

Slide 121 (Answer) / 311

Example: Find the lateral area and surface area of the right prism. 8 3 7 6 5

Angles are right angles.

First, calculate the perimeter of your base. P = 8 + 7 + 5 + 4 + 3 + 3 P = 30 units Use this to find the Lateral Area LA = PH = 30(6) = 180 units2

Other Prisms

Then, calculate the area of your base, B B = 7(5)+3(3) = 44 units2 Finally, calculate your Surface Area. SA = 2B + PH SA = 2(44) + (30)(6) SA = 88 + 180 = 268 units2

Slide 122 / 311

Example: Find the lateral area and surface area of the right prism. 8 3 7 6 5

Angles are right angles.

First, calculate the perimeter of your base. P = 8 + 7 + 5 + 4 + 3 + 3 P = 30 units Use this to find the Lateral Area LA = PH = 30(6) = 180 units2

Other Prisms

Then, calculate the area of your base, B B = 7(5)+3(3) = 44 units2 Finally, calculate your Surface Area. SA = 2B + PH SA = 2(44) + (30)(6) SA = 88 + 180 = 268 units2

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Can you do this mentally? (MP5) What labels could you use? (MP6)

Slide 122 (Answer) / 311

48 Find the lateral area of the right prism. 8 11 The base is a regular hexagon.

Slide 123 / 311

48 Find the lateral area of the right prism. 8 11 The base is a regular hexagon.

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Answer

P = 6(8) = 48 LA = 48(11) LA = 528 sq units

Slide 123 (Answer) / 311

49 Find the total surface area of the right prism. The base is a regular hexagon. 8 11

Slide 124 / 311

49 Find the total surface area of the right prism. The base is a regular hexagon. 8 11

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Answer

need to start by finding the apothem = 4√3...30-60-90 triangle P = 6(8) = 48 B = (1/2)(4√3)(48) = 96√3 SA = 2(96√3) + 48(11) SA = 192√3 + 528 sq units SA = 860.55 sq units

Slide 124 (Answer) / 311

50 Find the total surface area of the right prism. 4 4 3 2 10 9 All angles are right angles.

Slide 125 / 311

50 Find the total surface area of the right prism. 4 4 3 2 10 9 All angles are right angles.

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Answer

P = 4 + 2 + 2 + 2 + 4 + 3 + 10 + 3 P = 30 B = 4(3) + 4(3) + 2(1) = 26 SA = 2(26) + 30(9) SA = 52 + 270 SA = 322 sq units

Slide 125 (Answer) / 311

y 5 6 51 The right triangular prism has a surface area of 150 sq

• ft. Find the height of the prism.

A 5 ft B 6 ft C 7.81 ft D 6.38 ft

Slide 126 / 311

y 5 6 51 The right triangular prism has a surface area of 150 sq

• ft. Find the height of the prism.

A 5 ft B 6 ft C 7.81 ft D 6.38 ft

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Answer

Need hyp. of the right triangle. 52 + 62 = c2 c = 7.81 B = (1/2)(5)(6) = 15 P = 5 + 6 + 7.81 = 18.81 150 = 2(15) + 18.81(y) 150 = 30 + 18.81y 120 = 18.81y y = 6.38 ft D

Slide 126 (Answer) / 311

Surface Area of a Cylinder

Return to Table of Contents

Slide 127 / 311

height radius base base height radius base base

Cylinders

A Cylinder is a solid w/ 2 circular bases that lie in || planes. Because each base is a circle, it contains a radius. The remaining measurement that connects the 2 bases is the height of the cylinder.

Slide 128 / 311

8

radius

The net of a right cylinder is two circles and a rectangle that forms the lateral surface. 8 x What is the length of x?

• The circumference of the

circle (base)

radius

Click to reveal

Cylinders Slide 129 / 311

Base Base height Base height Lateral Surface Base

Finding the Surface Area of a Right Cylinder

Surface Area : S.A. = 2B + PH B = Area of the circular base = πr2 C = Perimeter of the Circular base (Circumference) = 2πr H = Height of the prism Lateral Area = CH = 2πrH

Slide 130 / 311

Base Base height Base height Lateral Surface Base

Finding the Surface Area of a Right Cylinder

Surface Area : S.A. = 2B + PH B = Area of the circular base = πr2 C = Perimeter of the Circular base (Circumference) = 2πr H = Height of the prism Lateral Area = CH = 2πrH

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Teacher's Note

The PARCC Reference sheet for the HS level does NOT contain any formulas to calculate the Surface Area (reference sheet is linked to this pull tab - click on tab). Encourage students to either memorize the formulas, or draw the net each time so that they can break down the solid into smaller 2- D shapes.

Slide 130 (Answer) / 311

Base Base height Base height Lateral Surface Base

Finding the Surface Area of a Right Cylinder

The Lateral Area is the area of the Lateral Surface, the rectangular area that wraps around the cylinder between the circular bases. Therefore, the Surface Area of a Cylinder can be simplified to the equation below. SA = 2πr2 + 2πrH

Slide 131 / 311

8 r = 4 Example: Find the lateral area and surface area of the right cylinder. LA = 2πrh LA = 2π(4)(8) LA = 64π units2 LA = 201.06 units2 SA = 2πr2 + 2πrh SA = 2π(4)2 + 2π(4)(8) SA = 32π + 64π SA = 96π units2 SA = 301.59 units2

Finding the Surface Area of a Right Cylinder Slide 132 / 311

8 r = 4 Example: Find the lateral area and surface area of the right cylinder. LA = 2πrh LA = 2π(4)(8) LA = 64π units2 LA = 201.06 units2 SA = 2πr2 + 2πrh SA = 2π(4)2 + 2π(4)(8) SA = 32π + 64π SA = 96π units2 SA = 301.59 units2

Finding the Surface Area of a Right Cylinder

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) What labels could you use? (MP6)

Slide 132 (Answer) / 311

34 d = 16 SA = 2πr2 + 2πrh SA = 2π(8)2 + 2π(8)(30) SA = 128π + 480π SA = 608π units2 SA = 1,910.09 units2 Example: Find the lateral area and surface area of the right cylinder. LA = 2πrh LA = 2π(8)(30) LA = 480π units2 LA = 1507.96 units2 162 + h2 = 342 256 + h2 = 1156 h2 = 900 h = 30 Note: 16-30-34 = 2(8-15-17) Pyth. Triple

click click

Cylinders

click click click click click click click click click click click click

Slide 133 / 311

34 d = 16 SA = 2πr2 + 2πrh SA = 2π(8)2 + 2π(8)(30) SA = 128π + 480π SA = 608π units2 SA = 1,910.09 units2 Example: Find the lateral area and surface area of the right cylinder. LA = 2πrh LA = 2π(8)(30) LA = 480π units2 LA = 1507.96 units2 162 + h2 = 342 256 + h2 = 1156 h2 = 900 h = 30 Note: 16-30-34 = 2(8-15-17) Pyth. Triple

click click

Cylinders

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Can you do this mentally? (MP5)

• Referring to the Pythagorean Triple

16-30-34 = 2(8-15-17) What labels could you use? (MP6)

Slide 133 (Answer) / 311

Example: Find the lateral area and surface area

• f the right cylinder when the base

circumference is 16π ft & the height is 10 ft. SA = 2πr2 + 2πrh SA = 2π(8)2 + 2π(8)(10) SA = 128π + 160π SA = 288π ft2 SA = 904.78 ft

2

LA = 2πrh LA = 2π(8)(10) LA = 160π ft2 LA = 502.64 ft

2

C = 2πr 16π = 2πr 2π 2π 8 ft = r

Cylinders

click click click click click click click click click click click click click

Slide 134 / 311

Example: Find the lateral area and surface area

• f the right cylinder when the base

circumference is 16π ft & the height is 10 ft. SA = 2πr2 + 2πrh SA = 2π(8)2 + 2π(8)(10) SA = 128π + 160π SA = 288π ft2 SA = 904.78 ft

2

LA = 2πrh LA = 2π(8)(10) LA = 160π ft2 LA = 502.64 ft

2

C = 2πr 16π = 2πr 2π 2π 8 ft = r

Cylinders

click click click click click click click click click click click click click [This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) Would it help to draw a picture? (MP4 & MP5) What tools do you need? (MP5) Can you do this mentally? (MP5) What labels could you use? (MP6)

Slide 134 (Answer) / 311

h = 12 r = 7 52 Find the lateral area of the right cylinder.

Slide 135 / 311

h = 12 r = 7 52 Find the lateral area of the right cylinder.

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Answer

LA = 2π(7)(12) LA = 168π sq units LA = 527.79 sq units

Slide 135 (Answer) / 311

h = 12 r = 7 53 Find the surface area of the right cylinder. Use 3.14 as your value of π & round to two decimal places. A 1200 sq in. B 307.72 sq in. C 835.24 sq in. D 1670.48 sq in.

Slide 136 / 311

h = 12 r = 7 53 Find the surface area of the right cylinder. Use 3.14 as your value of π & round to two decimal places. A 1200 sq in. B 307.72 sq in. C 835.24 sq in. D 1670.48 sq in.

[This object is a pull tab]

Answer

SA = 2π(7)(12) + 2π(7)2 SA = 168π + 98π SA = 266π sq in SA = 835.24 sq in C

Slide 136 (Answer) / 311

54 Find the lateral area of the right cylinder. 13 r = 5

Slide 137 / 311

54 Find the lateral area of the right cylinder. 13 r = 5

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Answer

52 + h2 = 132 25 + h2 = 169 h2 = 144 h = 12 units LA = 2π(5)(12) LA = 120π sq units LA = 376.99 sq units

Slide 137 (Answer) / 311

h = 12 55 Find the lateral area of the right cylinder. Base area is 36π units2

Slide 138 / 311

h = 12 55 Find the lateral area of the right cylinder. Base area is 36π units2

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Answer

LA = 2π(6)(12) LA = 144π sq units LA = 452.39 sq units A = πr2 36π = πr2 π π 36 = r2 6 = r

Slide 138 (Answer) / 311

h = 12 56 Find the surface area of the right cylinder. Base area is 36π units2

Slide 139 / 311

h = 12 56 Find the surface area of the right cylinder. Base area is 36π units2

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Answer

SA = 2π(6)(12) + 2π62 SA = 144π + 72 SA = 216π sq units SA = 678.58 sq units A = πr2 36π = πr2 π π 36 = r2 6 = r

Slide 139 (Answer) / 311

r = 8 in. h 57 The surface area of the right cylinder is 653.12 sq in. Find the height of the cylinder. Use 3.14 as your value

• f π.

A 7 in. B 8 in. C 5 in. D 6 in.

Slide 140 / 311

r = 8 in. h 57 The surface area of the right cylinder is 653.12 sq in. Find the height of the cylinder. Use 3.14 as your value

• f π.

A 7 in. B 8 in. C 5 in. D 6 in.

[This object is a pull tab]

Answer

C

Slide 140 (Answer) / 311

58 A food company packages soup in aluminum cans that have a diameter of 2 1/2 inches and a height of 4 inches. Before shipping the cans off to the stores, they add their company label to the can which does not cover the top and bottom. If the company is shipping 200 cans of soup to one store, how much paper material is required to make the labels?

Slide 141 / 311

58 A food company packages soup in aluminum cans that have a diameter of 2 1/2 inches and a height of 4 inches. Before shipping the cans off to the stores, they add their company label to the can which does not cover the top and bottom. If the company is shipping 200 cans of soup to one store, how much paper material is required to make the labels?

[This object is a pull tab]

Answer

1 can: LA = 2πrh LA = 2π(1.25)(4) LA = 10π sq. inches LA = 31.42 sq. inches 200 cans: LA = 200(10π) LA = 2000π sq. inches LA = 6,283.19 sq. inches

Slide 141 (Answer) / 311

59 Maria's mom baked a cake for her daughter's birthday

• party. The diameter of the cake is 9 inches and the

height is 2 inches. How much base frosting (pink in the picture below) was required to cover the cake?

Slide 142 / 311

59 Maria's mom baked a cake for her daughter's birthday

• party. The diameter of the cake is 9 inches and the

height is 2 inches. How much base frosting (pink in the picture below) was required to cover the cake?

[This object is a pull tab]

Answer

Area of frosting = LA + area of the top = π(4.5)2 + 2π(4.5)(2) = 20.25π + 18π = 38.25π sq. inches = 120.17 sq. inches

Slide 142 (Answer) / 311

Surface Area of a Pyramid

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Slide 143 / 311

A Pyramid is a polyhedron in which the base is a polygon & the lateral faces are triangles with a common vertex. Lateral Edges are the intersection

• f 2 lateral faces

Vertex Lateral Face Lateral Edge Base

Pyramids Slide 144 / 311

Net

This is a right square pyramid. Another name for it is pentahedron. Hedron is a suffix that means face. Why is this a pentahedron?

Slide 145 / 311 Net

This is a right square pyramid. Another name for it is pentahedron. Hedron is a suffix that means face. Why is this a pentahedron?

[This object is a pull tab]

Answer There are 5 polygonal faces that create this solid.

• "Penta" means 5
• "Hedon" means face

Therefore, it's a solid w/ 5 faces

Slide 145 (Answer) / 311

Slant Height

ℓ

The Pyramid has a square base and 4 triangular faces The triangular faces are all isosceles triangles if its a right pyramid. The Height of each triangular face is the Slant Height of the pyramid if it is a regular pyramid (labeled as , or a cursive lower case L).

Surface Area = Sum of the Areas of all the sides ℓ

Height

• f the

Triangle

Slide 146 / 311

Square Base (B) Slant Height ( )

ℓ

Pyramid's Height (h)

Segment Lengths in a Pyramid Slide 147 / 311

Example: Find the value of x. a2 + 122 = 132 a2 + 144 = 169 a2 = 25 a = 5 Note: 5-12-13 Right Triangle Therefore x = 2(5) = 10 x 13 12

Segment Lengths in a Pyramid Slide 148 / 311

Example: Find the value of x. a2 + 122 = 132 a2 + 144 = 169 a2 = 25 a = 5 Note: 5-12-13 Right Triangle Therefore x = 2(5) = 10 x 13 12

Segment Lengths in a Pyramid

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Can you do this mentally? (MP5)

• Referring to the Pythagorean Triple: 5-12-13

Can you find a shortcut to solve this problem? How would your shortcut make the problem easier? (MP8)

• Referring to the Pythagorean Triple

Slide 148 (Answer) / 311

Example: Find the value of x. Base Area of the right square pyramid is 64 u2. x 8 Square Base has an area

• f 64, so

64 = y2 y = 8, so a = 4 of the right triangle. 42 + 82 = x2 16 + 64 = x2 x2 = 80 x = 8.94 units

Segment Lengths in a Pyramid Slide 149 / 311

Example: Find the value of x. Base Area of the right square pyramid is 64 u2. x 8 Square Base has an area

• f 64, so

64 = y2 y = 8, so a = 4 of the right triangle. 42 + 82 = x2 16 + 64 = x2 x2 = 80 x = 8.94 units

Segment Lengths in a Pyramid

[This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5)

Slide 149 (Answer) / 311

Example: Find the length of the slant height.

r

This is a regular hexagonal pyramid. r = 6 lateral edge = 12

Segment Lengths in a Pyramid Slide 150 / 311

First, find the height of the pyramid using Pythagorean Theorem. h 12 6 62 + h2 = 122 36 + h2 = 144 h2 = 108 h = 6√3 = 10.39 Note: 30-60-90 triangle

r Segment Lengths in a Pyramid

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Slide 151 / 311

First, find the height of the pyramid using Pythagorean Theorem. h 12 6 62 + h2 = 122 36 + h2 = 144 h2 = 108 h = 6√3 = 10.39 Note: 30-60-90 triangle

r Segment Lengths in a Pyramid

click click click click click [This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Can you do this mentally? (MP5)

• Referring to the 30-60-90 triangle

Can you find a shortcut to solve this problem? How would your shortcut make the problem easier? (MP8)

• Referring to the 30-60-90 triangle

Slide 151 (Answer) / 311

Second, find the apothem of the hexagonal base. a 6 6 3 3 32 + a2 = 62 9 + a2 = 36 a2 = 27 a = 3√3 = 5.20 Note: 30-60-90 triangle = 60° = central Note: equilateral = 30° = top of the . 360 6 60 2 r

click click click click click click click click

Segment Lengths in a Pyramid Slide 152 / 311

Second, find the apothem of the hexagonal base. a 6 6 3 3 32 + a2 = 62 9 + a2 = 36 a2 = 27 a = 3√3 = 5.20 Note: 30-60-90 triangle = 60° = central Note: equilateral = 30° = top of the . 360 6 60 2 r

click click click click click click click click

Segment Lengths in a Pyramid

[This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Can you do this mentally? (MP5)

• Referring to the 30-60-90 triangle

Can you find a shortcut to solve this problem? How would your shortcut make the problem easier? (MP8)

• Referring to the 30-60-90 triangle

Slide 152 (Answer) / 311

(3√3)2 + (6√3)2 = 2 27 + 108 = 2 2 = 135 = 3√15 = 11.62

ℓ ℓ ℓ ℓ

Last, find the slant height of your pyramid w/ the apothem & height. a = 3√3

ℓ

h = 6√3

r

click click click click Click Click

Segment Lengths in a Pyramid Slide 153 / 311

(3√3)2 + (6√3)2 = 2 27 + 108 = 2 2 = 135 = 3√15 = 11.62

ℓ ℓ ℓ ℓ

Last, find the slant height of your pyramid w/ the apothem & height. a = 3√3

ℓ

h = 6√3

r

click click click click Click Click

Segment Lengths in a Pyramid

[This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5)

Slide 153 (Answer) / 311

60 Find the value of the variable. 16 x 6

Slide 154 / 311

60 Find the value of the variable. 16 x 6

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Answer

10 62 + 82 = x2 36 + 64 = x2 100 = x2 10 = x

Slide 154 (Answer) / 311

61 Find the value of the variable. 12 11 x

Slide 155 / 311

61 Find the value of the variable. 12 11 x

[This object is a pull tab]

Answer

√85 or 9.22 62 + x2 = 11

2

36 + x2 = 121 x2 = 85 x =√85 = 9.22

Slide 155 (Answer) / 311

62 Find the value of the variable. x 6

area of the base is 36 u2

Slide 156 / 311

62 Find the value of the variable. x 6

area of the base is 36 u2

[This object is a pull tab]

Answer

3√5 or 6.71 Note: Area of base = 36, then side length of square is 6. 62 + 32 = x2 36 + 9 = x2 x2 = 45 x =3√5 = 6.71

Slide 156 (Answer) / 311

63 Find the value of the slant height.

r r = 8 lateral edge = 17 Regular Hexagonal Pyramid

Slide 157 / 311

63 Find the value of the slant height.

r r = 8 lateral edge = 17 Regular Hexagonal Pyramid

[This object is a pull tab]

Answer

16.52 82 + h2 = 172 h = 15 apothem in hexagon = 4√3 (4√3)2 + 152 = 2 = 16.52

ℓ ℓ

Slide 157 (Answer) / 311

64 Find the value of the slant height.

a a = 9 lateral edge = 12 Regular Hexagonal Pyramid

Slide 158 / 311

64 Find the value of the slant height.

a a = 9 lateral edge = 12 Regular Hexagonal Pyramid

[This object is a pull tab]

Answer

10.82 apothem in hexagon = 9 9 = x√3, so x = 3√3 & r = 6√3 (6√3)2 + h2 = 122 h = 6 62 + 92 = 2 = 10.82

ℓ ℓ

9

r r

x x

Slide 158 (Answer) / 311

Square Base (B) Slant Height ( ) Pyramid's Height (h)

ℓ

Surface Area = B + ½P and Lateral Area = ½P = Slant Height P = Perimeter of Base B = Area of Base

Surface Area of a Regular Pyramid

ℓ ℓ

ℓ Slide 159 / 311

Square Base (B) Slant Height ( ) Pyramid's Height (h)

ℓ

Surface Area = B + ½P and Lateral Area = ½P = Slant Height P = Perimeter of Base B = Area of Base

Surface Area of a Regular Pyramid

ℓ ℓ

ℓ

[This object is a pull tab]

Teacher's Note

The PARCC Reference sheet for the HS level does NOT contain any formulas to calculate the Surface Area (reference sheet is linked to this pull tab - click on tab). Encourage students to either memorize the formulas, or draw the net each time so that they can break down the solid into smaller 2- D shapes.

Slide 159 (Answer) / 311

Square Base (B) Slant Height ( ) Pyramid's Height (h) Why is the Surface Area SA = B + P ? 1 2 Surface Area is the sum of all of the areas that make up the solid. In

• ur diagram, these are 4 triangles & 1 square.

Asquare = s s = s2 = B A∆ = s 1 2

ℓ ℓ ℓ

Surface Area of a Regular Pyramid Slide 160 / 311

Why is the Surface Area SA = B + P ? 1 2 Since there are 4 ∆s, we can multiply the area of each ∆ by 4. Therefore, our Surface Area for the Pyramid above is SA = s

2 + 4(1/2)s

SA = s

2 + (1/2)(4s)

SA = B +

1/2 P

s

ℓ

Net of Pyramid

ℓℓ ℓ

Surface Area of a Regular Pyramid

ℓ

Slide 161 / 311

ℓ = 7 s = 6

Example: Find the lateral area and the surface area of the pyramid. LA =

1/2 P ℓ

LA =

1/2 (24)(7)

LA = 12(7) LA = 84 units

2

SA = B +

1/2 P ℓ

SA = 6

2 + 1/2 (24)(7)

SA = 36 + 84 SA = 120 units

2

Surface Area of a Regular Pyramid Slide 162 / 311

ℓ = 7 s = 6

Example: Find the lateral area and the surface area of the pyramid. LA =

1/2 P ℓ

LA =

1/2 (24)(7)

LA = 12(7) LA = 84 units

2

SA = B +

1/2 P ℓ

SA = 6

2 + 1/2 (24)(7)

SA = 36 + 84 SA = 120 units

2

Surface Area of a Regular Pyramid

[This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) What labels could you use? (MP6)

Slide 162 (Answer) / 311

Example: Find the lateral area and the surface area of the pyramid. First, calculate the slant height. 32 + 82 = ℓ 2 9 + 64 = ℓ2 73 = ℓ2 ℓ = 8.54 Next, calculate the LA & SA LA = 1/2 P ℓ LA = 1/2 (24)(8.54) LA = 12(8.54) LA = 102.48 units2 SA = B + 1/2 P ℓ SA = 62 + 1/2 (24)(8.54) SA = 36 + 102.48 SA = 138.48 units2

h = 8 s = 6

Surface Area of a Regular Pyramid Slide 163 / 311

Example: Find the lateral area and the surface area of the pyramid. First, calculate the slant height. 32 + 82 = ℓ 2 9 + 64 = ℓ2 73 = ℓ2 ℓ = 8.54 Next, calculate the LA & SA LA = 1/2 P ℓ LA = 1/2 (24)(8.54) LA = 12(8.54) LA = 102.48 units2 SA = B + 1/2 P ℓ SA = 62 + 1/2 (24)(8.54) SA = 36 + 102.48 SA = 138.48 units2

h = 8 s = 6

Surface Area of a Regular Pyramid

[This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) What labels could you use? (MP6)

Slide 163 (Answer) / 311

Example: Find the lateral area and the surface area of the pyramid. 10 8

ℓ

First, calculate the slant height. 82 + ℓ 2 = 102 64 + ℓ 2 = 100 ℓ 2 = 36 ℓ = 6 s = 16 e = 10

click click click

Surface Area of a Regular Pyramid Slide 164 / 311

Example: Find the lateral area and the surface area of the pyramid. 10 8

ℓ

First, calculate the slant height. 82 + ℓ 2 = 102 64 + ℓ 2 = 100 ℓ 2 = 36 ℓ = 6 s = 16 e = 10

click click click

Surface Area of a Regular Pyramid

[This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Can you solve this problem mentally? (MP5)

• Referring to Pythagorean Triple: 6-8-10 =

2(3-4-5) What labels could you use? (MP6) Can you find a shortcut to solve the problem? How would your shortcut make the problem easier? (MP8)

Slide 164 (Answer) / 311

Example: Find the lateral area and the surface area of the pyramid. LA = 1/2 P ℓ LA = 1/2 (64)(6) LA = 32(6) LA = 192 units2 SA = B + 1/2 P ℓ SA = 162 + 1/2 (64)(6) SA = 256 + 192 SA = 448 units2 Next, calculate the LA & SA

click click click click click click click click

s = 16 e = 10

Surface Area of a Regular Pyramid Slide 165 / 311

Example: Find the lateral area and the surface area of the pyramid. LA = 1/2 P ℓ LA = 1/2 (64)(6) LA = 32(6) LA = 192 units2 SA = B + 1/2 P ℓ SA = 162 + 1/2 (64)(6) SA = 256 + 192 SA = 448 units2 Next, calculate the LA & SA

click click click click click click click click

s = 16 e = 10

Surface Area of a Regular Pyramid

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) What labels could you use? (MP6)

Slide 165 (Answer) / 311

Example: Find the lateral area and the surface area of the pyramid.

a

a = 4 lateral edge = 8 Regular Pentagonal Pyramid

Surface Area of a Regular Pyramid Slide 166 / 311

72° 36°36°

4

x r Example: Find the lateral area and the surface area of the pyramid. First, find the radius & side length of the regular pentagon using the apothem & trigonometric ratios tan36 = x = 4tan36 = 2.91 Therefore, s = 2(2.91) = 5.82 = 36° = top

• f the .

360 5 72 2 x 4 4 r cos36 = rcos36 = 4 cos36 cos36 r = 4.94 = 72° = central

Click Click Click Click Click Click Click Click

Click Click Click

Surface Area of a Regular Pyramid Slide 167 / 311

72° 36°36°

4

x r Example: Find the lateral area and the surface area of the pyramid. First, find the radius & side length of the regular pentagon using the apothem & trigonometric ratios tan36 = x = 4tan36 = 2.91 Therefore, s = 2(2.91) = 5.82 = 36° = top

• f the .

360 5 72 2 x 4 4 r cos36 = rcos36 = 4 cos36 cos36 r = 4.94 = 72° = central

Click Click Click Click Click Click Click Click

Click Click Click

Surface Area of a Regular Pyramid

[This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) How is right triangle trigonometry used to calculate the segment lengths in regular polygons? (MP7)

Slide 167 (Answer) / 311

Next, find the slant height of the pyramid using the lateral edge, the value of x from the previous slide & Pythagorean Theorem.

8 2.91

2.912 + ℓ 2 = 82 8.4681 + ℓ 2 = 64 ℓ 2 = 55.5319 ℓ = 7.45

ℓ

click click click click

Surface Area of a Regular Pyramid Slide 168 / 311

Next, find the slant height of the pyramid using the lateral edge, the value of x from the previous slide & Pythagorean Theorem.

8 2.91

2.912 + ℓ 2 = 82 8.4681 + ℓ 2 = 64 ℓ 2 = 55.5319 ℓ = 7.45

ℓ

click click click click

Surface Area of a Regular Pyramid

[This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) How is Pythagorean Theorem used to calculate the segment lengths of pyramids? (MP7)

Slide 168 (Answer) / 311

Last, find the lateral area & surface area of the pyramid. LA =

1/2 P ℓ

LA =

1/2 (29.1)(7.45)

LA = 108.40 units

2

SA = B +

1/2 P ℓ

SA =

1/2 (4)(29.1) + 1/2 (29.1)(7.45)

SA = 58.2 + 108.40 SA = 166.6 units

2 click click click click click click click

Surface Area of a Regular Pyramid Slide 169 / 311

Last, find the lateral area & surface area of the pyramid. LA =

1/2 P ℓ

LA =

1/2 (29.1)(7.45)

LA = 108.40 units

2

SA = B +

1/2 P ℓ

SA =

1/2 (4)(29.1) + 1/2 (29.1)(7.45)

SA = 58.2 + 108.40 SA = 166.6 units

2 click click click click click click click

Surface Area of a Regular Pyramid

[This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) What labels could you use? (MP6)

Slide 169 (Answer) / 311

65 Find the lateral area of the right pyramid.

s = 10

ℓ = 9

Slide 170 / 311

65 Find the lateral area of the right pyramid.

s = 10

ℓ = 9

[This object is a pull tab]

Answer

LA = 1/2 P ℓ LA = 1/2 (40)(9) LA = 180 units2

Slide 170 (Answer) / 311

66 Find the surface area of the right pyramid.

s = 10 ℓ = 9

Slide 171 / 311

66 Find the surface area of the right pyramid.

s = 10 ℓ = 9

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Answer

SA = B + 1/2 P ℓ SA = 102 + 1/2 (40)(9) SA = 100 + 180 SA = 280 units2

Slide 171 (Answer) / 311

67 Find the lateral area of the right pyramid. base e = 10 area = 16

Slide 172 / 311

67 Find the lateral area of the right pyramid. base e = 10 area = 16

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Answer

LA = 1/2 P ℓ LA = 1/2 (16)(9.80) LA = 78.4 units2 B = 16, so s = 4 & small leg of rt ∆ is 2 22 + ℓ2 = 102 ℓ = 9.80

Slide 172 (Answer) / 311

68 Find the surface area of the right pyramid. base e = 10 area = 16

Slide 173 / 311

68 Find the surface area of the right pyramid. base e = 10 area = 16

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Answer

ℓ = 9.80 (see previous slide) SA = B + 1/2 P ℓ SA = 16 + 78.4 SA = 94.4 units2

Slide 173 (Answer) / 311

a

a = 5 h = 12 Regular Octagonal Pyramid 69 Find the lateral area of the right pyramid.

Slide 174 / 311

a

a = 5 h = 12 Regular Octagonal Pyramid 69 Find the lateral area of the right pyramid.

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Answer

P = 8(4.14) = 33.12 LA = 1/2 P ℓ LA = 1/2 (33.12)(13) LA = 215.28 units2 a = 5 & h = 12 52 + 122 = ℓ2 ℓ = 13 units tan22.5 = x 5 x = 2.07 1 side = 4.14

Slide 174 (Answer) / 311

a

a = 5 h = 12 Regular Octagonal Pyramid 70 Find the surface area of the right pyramid.

Slide 175 / 311

a

a = 5 h = 12 Regular Octagonal Pyramid 70 Find the surface area of the right pyramid.

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Answer

SA = B + 1/2 P ℓ SA = 82.8 + 215.28 SA = 298.08 units2 ℓ = 13 units P = 33.12 units B = 1/2aP ℓ B = 1/2(5)(33.12) B = 82.8 units2

Slide 175 (Answer) / 311

Hint: The pyramid is NOT regular. So, B + 1/2 P ℓ doesn't work. Instead, draw a net of the pyramid & find each area.

71 Find the surface area of the right pyramid.

30 12 8

Hint

Slide 176 / 311

Hint: The pyramid is NOT regular. So, B + 1/2 P ℓ doesn't work. Instead, draw a net of the pyramid & find each area.

71 Find the surface area of the right pyramid.

30 12 8

Hint

[This object is a pull tab]

Answer

SA = sum of all areas SA = 30(12) + 2(1/2)(12)(17) + 2(1/2)(30)(10) SA = 360 + 204 + 300 SA = 864 units2 ℓ1 = 17 ℓ2 = 10

Slide 176 (Answer) / 311

Surface Area of a Cone

Return to Table of Contents

Slide 177 / 311

r height S l a n t H e i g h t

ℓ

Lateral Surface Slant Height

ℓ

Base The Base of the cone is a circle The length of the circular portion of the Lateral Surface is the same as the Circumference of the Circlular Base. The Slant Height is the length of the diagonal slant of the cone from the top to the edge of the base. The Height of the cone is the length from the top to the center of the circular base.

Cones Slide 178 / 311

Surface Area = Area of the Base + Lateral Area Lateral Area= ½P ℓ S.A. = B + ½P ℓ ℓ = Slant Height P = Perimeter of Circular Base B = Area of Circular Base Because the base is a circle. P = Circumference = 2πr L.A. = ½(2πr) ℓ

= πr ℓ

S.A. = πr2 + πr ℓ

Finding the Surface Area of a Right Cone

Lateral Surface Slant Height

ℓ

Base

Slide 179 / 311

Surface Area = Area of the Base + Lateral Area Lateral Area= ½P ℓ S.A. = B + ½P ℓ ℓ = Slant Height P = Perimeter of Circular Base B = Area of Circular Base Because the base is a circle. P = Circumference = 2πr L.A. = ½(2πr) ℓ

= πr ℓ

S.A. = πr2 + πr ℓ

Finding the Surface Area of a Right Cone

Lateral Surface Slant Height

ℓ

Base

[This object is a pull tab]

Teacher's Note

The PARCC Reference sheet for the HS level does NOT contain any formulas to calculate the Surface Area (reference sheet is linked to this pull tab - click on tab). Encourage students to either memorize the formulas, or draw the net each time so that they can break down the solid into smaller 2- D shapes.

Slide 179 (Answer) / 311

LA = πr ℓ = π(6)(8) LA = 48π units2 LA = 150.80 units2 SA = πr2 + πr ℓ = π(6)2 + π(6)(8) = 36π + 48π SA = 84π units2 SA = 263.89 units2 Example: Find the lateral area and surface area of the right cone. = 8 r = 6

ℓ

click

Cones

click click click click click click click click

Slide 180 / 311

LA = πr ℓ = π(6)(8) LA = 48π units2 LA = 150.80 units2 SA = πr2 + πr ℓ = π(6)2 + π(6)(8) = 36π + 48π SA = 84π units2 SA = 263.89 units2 Example: Find the lateral area and surface area of the right cone. = 8 r = 6

ℓ

click

Cones

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) What labels could you use? (MP6)

Slide 180 (Answer) / 311

Example: Find the lateral area and surface area of the right cone. h = 8 C = 12π units C = 2πr 12π = 2πr 2π 2π 6 units = r 62 + 82 = ℓ2 36 + 64 = ℓ2 100 = ℓ2 10 units = ℓ

Cones

click click click click click click click click

Slide 181 / 311

Example: Find the lateral area and surface area of the right cone. h = 8 C = 12π units C = 2πr 12π = 2πr 2π 2π 6 units = r 62 + 82 = ℓ2 36 + 64 = ℓ2 100 = ℓ2 10 units = ℓ

Cones

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How could you start this problem? (MP1) What does 12π represent in this problem? (MP2) What other value can it help you calculate? (MP2) How does Pythagorean Theorem related to calculating the slant height? (MP7) How can you represent the problem with symbols and numbers? (MP2) Can you find a shortcut to solve the problem? How would your shortcut make the problem easier? (MP8) What tools do you need? (MP5) What labels could you use? (MP6)

Slide 181 (Answer) / 311

Example: Find the lateral area and surface area of the right cone. h = 8 C = 12π units SA = πr2 + πr ℓ = π(6)2 + π(6)(10) = 36π + 60π SA = 96π units2 SA = 301.59 units2 LA = πr ℓ = π(6)(10) LA = 60π units2 LA = 188.50 units2

Cones

click click click click click click click click click

Slide 182 / 311

Example: Find the lateral area and surface area of the right cone. h = 8 C = 12π units SA = πr2 + πr ℓ = π(6)2 + π(6)(10) = 36π + 60π SA = 96π units2 SA = 301.59 units2 LA = πr ℓ = π(6)(10) LA = 60π units2 LA = 188.50 units2

Cones

click click click click click click click click click [This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How could you start this problem? (MP1) What does 12π represent in this problem? (MP2) What other value can it help you calculate? (MP2) How does Pythagorean Theorem related to calculating the slant height? (MP7) How can you represent the problem with symbols and numbers? (MP2) Can you find a shortcut to solve the problem? How would your shortcut make the problem easier? (MP8) What tools do you need? (MP5) What labels could you use? (MP6)

Slide 182 (Answer) / 311

72 Find the lateral area of the right cone, in square units. r = 4

ℓ = 9 Slide 183 / 311

72 Find the lateral area of the right cone, in square units. r = 4

ℓ = 9

[This object is a pull tab]

Answer

LA = πr ℓ = π(4)(9) LA = 36π units2 LA = 113.10 units

2

Slide 183 (Answer) / 311

r = 4

ℓ = 9

73 Find the surface area of the right cone, in square units.

Slide 184 / 311

r = 4

ℓ = 9

73 Find the surface area of the right cone, in square units.

[This object is a pull tab]

Answer

SA = πr2 + πr ℓ = π(4)2 + π(4)(9) = 16π + 36π SA = 52π units2 SA = 163.36 units2

Slide 184 (Answer) / 311

74 Find the lateral area of the right cone, in square units. h = 9 Base Area = 16π units2

Slide 185 / 311

74 Find the lateral area of the right cone, in square units. h = 9 Base Area = 16π units2

[This object is a pull tab]

Answer

LA = πr ℓ = π(4)(9.85) LA = 39.4π units2 LA = 123.78 units2 B = πr2 16π = πr2 π π 16 = r2 4 = r 42 + 92 = ℓ2 16 + 81 = ℓ2 97 = ℓ2 9.85 = ℓ

Slide 185 (Answer) / 311

75 Find the surface area of the right cone, in square units. h = 9 Base Area = 16π units2

Slide 186 / 311

75 Find the surface area of the right cone, in square units. h = 9 Base Area = 16π units2

[This object is a pull tab]

Answer

SA = πr2 + πr ℓ = π(4)2 + π(4)(9.85) = 16π + 39.4π SA = 55.4π units2 SA = 174.04 units2 B = πr2 16π = πr2 π π 16 = r2 4 = r 42 + 92 = ℓ2 16 + 81 = ℓ2 97 = ℓ2 9.85 = ℓ

Slide 186 (Answer) / 311

76 Find the length of the radius of the right cone if the lateral area is 50π units2?

ℓ = 10 Slide 187 / 311

76 Find the length of the radius of the right cone if the lateral area is 50π units2?

ℓ = 10

[This object is a pull tab]

Answer

LA = πr ℓ 50π = πr(10) 10π 10π r = 5 units

Slide 187 (Answer) / 311

ℓ = 10

77 Find the height of the right cone if the lateral area is 50π units2?

Slide 188 / 311 ℓ = 10

77 Find the height of the right cone if the lateral area is 50π units2?

[This object is a pull tab]

Answer

r = 5 units 52 + h2 = 102 25 + h2 = 100 h2 = 75 h = 5√3 = 8.66

Slide 188 (Answer) / 311

78 Find the slant height of the right cone if the surface area is 45π units2 and the diameter of the base is 6 units?

Slide 189 / 311

78 Find the slant height of the right cone if the surface area is 45π units2 and the diameter of the base is 6 units?

[This object is a pull tab]

Answer

SA = πr2 + πr ℓ 45π = π(3)2 + π(3) ℓ 45π = 9π + 3π ℓ

• 9π -9π

36π = 3π ℓ 3π 3π ℓ = 12 units

Slide 189 (Answer) / 311

79 Find the height of the right cone if the surface area is 45π units2 and the diameter of the base is 6 units?

Slide 190 / 311

79 Find the height of the right cone if the surface area is 45π units2 and the diameter of the base is 6 units?

[This object is a pull tab]

Answer

ℓ = 12 units 32 + h2 = 122 9 + h2 = 144 h2 = 135 h = 3√15 units h = 11.62 units

Slide 190 (Answer) / 311

80 The Department of Transportation keeps 4 piles of road salt for snowy days. Each conical shaped pile is 20 feet high and 30 feet across at the base. During the summer the piles are covered with tarps to prevent erosion. How much tarp is needed to cover the conical shaped piles so that no part of them are exposed?

Slide 191 / 311

80 The Department of Transportation keeps 4 piles of road salt for snowy days. Each conical shaped pile is 20 feet high and 30 feet across at the base. During the summer the piles are covered with tarps to prevent erosion. How much tarp is needed to cover the conical shaped piles so that no part of them are exposed?

[This object is a pull tab]

Answer

LA = πr ℓ = π(15)(25) LA = 375π ft2 LA = 1,178.10 ft2 d = 30 ft r = 15 ft h = 20 ft 152 + 202 = ℓ2 225 + 400 = ℓ2 625 = ℓ2 25 = ℓ 4 piles, so the total area for the tarps is 1500π ft2 = 4,712.39 ft2

Slide 191 (Answer) / 311

Volume of a Prism

Return to Table of Contents

Slide 192 / 311

The volume of a solid is the amount of cubic units that a solid can hold. Where area used square units, volume will use cubic units.

Prisms Slide 193 / 311

Base height Base ℓ w H V = BH Specific Prisms Rectangular Prism: V = ℓwH Cube: V = s3

Finding the Volume of a Prism

Prisms Slide 194 / 311

Base height Base ℓ w H V = BH Specific Prisms Rectangular Prism: V = ℓwH Cube: V = s3

Finding the Volume of a Prism

Prisms

[This object is a pull tab]

Teacher's Note

To avoid confusion with the "heights" when calculating the volume of a triangular prism, the height of the prism has been assigned "H". The triangular height will be "h" starting on slide #190

Slide 194 (Answer) / 311

Does a prism need to be a right prism for the volume formula to work? Think of a ream of paper Stacked nicely it has 500 sheets. If the stack is fanned, it still has 500 sheets. So the volume doesn't change if the prism, stack of paper, is right or oblique. The formula V = BH works for all prisms.

Prisms Slide 195 / 311

Example: Find the volume of the rectangular prism with a length of 2, a width of 6, and a height of 5. V = ℓ w H V = 2(6)(5) V = 60 units3

Prisms Slide 196 / 311

Example: Find the volume of the rectangular prism with a length of 2, a width of 6, and a height of 5. V = ℓ w H V = 2(6)(5) V = 60 units3

Prisms

[This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) Can you do this problem mentally? (MP5) What tools do you need? (MP5) What labels could you use? (MP6)

Slide 196 (Answer) / 311

Example: The volume of a box is 48 ft3. If the height is 4 ft and width is 6 ft, what is the length? V = ℓ w H 48 = ℓ(6)(4) 48 = 24 ℓ 24 24 2 ft = ℓ

Prisms Slide 197 / 311

Example: The volume of a box is 48 ft3. If the height is 4 ft and width is 6 ft, what is the length? V = ℓ w H 48 = ℓ(6)(4) 48 = 24 ℓ 24 24 2 ft = ℓ

Prisms

[This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) Can you do this problem mentally? (MP5) What tools do you need? (MP5) What labels could you use? (MP6)

Slide 197 (Answer) / 311

Example: Find the volume of the prism shown below. 10 6 11 Since it has a base that is a right triangle, we need to find the base

• f the triangle using Pythagorean Theorem.

62 + b2 = 102 36 + b2 = 100 b2 = 64 b = 8 units

Prisms

Next, calculate the area of your base, B B = (1/2)(8)(6) = 24 units2 Finally, calculate your Volume. V = BH V = 24(11) V = 264 units3

Slide 198 / 311

Example: Find the volume of the prism shown below. 10 6 11 Since it has a base that is a right triangle, we need to find the base

• f the triangle using Pythagorean Theorem.

62 + b2 = 102 36 + b2 = 100 b2 = 64 b = 8 units

Prisms

Next, calculate the area of your base, B B = (1/2)(8)(6) = 24 units2 Finally, calculate your Volume. V = BH V = 24(11) V = 264 units3

[This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Can you do this mentally? (MP5)

• Referring to the Pythagorean Triple

Can you find a shortcut to solve the problem? How would the shortcut make the problem easier? (MP8)

• Referring to the Pythagorean Triple

What labels could you use? (MP6)

Slide 198 (Answer) / 311

Example: The volume of a cube is 64 m3, what is area of one face? V = s3 64 = s3 4 m = s Area of one face A = 4(4) A = 16 m

2

Prisms Slide 199 / 311

Example: The volume of a cube is 64 m3, what is area of one face? V = s3 64 = s3 4 m = s Area of one face A = 4(4) A = 16 m

2

Prisms

[This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) Can you do this problem mentally? (MP5) What tools do you need? (MP5) What labels could you use? (MP6)

Slide 199 (Answer) / 311

4 in 7 in 30° 4 in. x in. Because the base is a regular polygon, we need to calculate the side length. To begin, figure out the central angle & top angle in the triangle. = 60° = central angle = 30° = top angle of the triangle. 360 6 60 2

Click Click Click

Example: Find the volume of the prism with a height 7 in. and hexagon base with an apothem of 4 in.

Prisms Slide 200 / 311

4 in 7 in 30° 4 in. x in. Because the base is a regular polygon, we need to calculate the side length. To begin, figure out the central angle & top angle in the triangle. = 60° = central angle = 30° = top angle of the triangle. 360 6 60 2

Click Click Click

Example: Find the volume of the prism with a height 7 in. and hexagon base with an apothem of 4 in.

Prisms

[This object is a pull tab]

Teacher's Note

When calculating the surface area and/or volume of regular polygonal prisms/ pyramids, students could also find the measurement of the 2 congruent base angles using the Triangle Sum Theorem. Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) Can you do this mentally? (MP5) What labels could you use? (MP6)

Slide 200 (Answer) / 311

4 in 7 in Example: Find the volume of the prism with a height 7 in. and hexagon base with an apothem of 4 in. Then, calculate the side length of your base. s = 2(2.31) = 4.62 in Next, find the value of x using trigonometry, or special right triangles (if it applies). tan 30 = 4tan30 = x x = 4√3 = 2.31 in. 3 x 4

Prisms

30° 4 in. x in.

Click Click Click Click

Slide 201 / 311

4 in 7 in Example: Find the volume of the prism with a height 7 in. and hexagon base with an apothem of 4 in. Then, calculate the side length of your base. s = 2(2.31) = 4.62 in Next, find the value of x using trigonometry, or special right triangles (if it applies). tan 30 = 4tan30 = x x = 4√3 = 2.31 in. 3 x 4

Prisms

30° 4 in. x in.

Click Click Click Click [This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Can you do this mentally? (MP5) What labels could you use? (MP6) How is right triangle trigonometry used to calculate the segment lengths of regular polygons? (MP7) Can you find a shortcut to solve the problem? How would your shortcut make the problem easier? (MP8)

• Referring to 30-60-90 triangle

Slide 201 (Answer) / 311

4 in 7 in Example: Find the volume of the prism with a height 7 in. and hexagon base with an apothem of 4 in. Next, use your value of s to find the Perimeter of your base P = 6(4.62) = 27.72 in

Prisms

Click

Then, calculate the area of your base, B B = (1/2)aP = (1/2)(2.31)(27.72) = 32.02 in2 Finally, calculate your Volume. V = Bh V = 32.02(7) V = 224.14 in3

Click Click Click Click Click Click

Slide 202 / 311

4 in 7 in Example: Find the volume of the prism with a height 7 in. and hexagon base with an apothem of 4 in. Next, use your value of s to find the Perimeter of your base P = 6(4.62) = 27.72 in

Prisms

Click

Then, calculate the area of your base, B B = (1/2)aP = (1/2)(2.31)(27.72) = 32.02 in2 Finally, calculate your Volume. V = Bh V = 32.02(7) V = 224.14 in3

Click Click Click Click Click Click [This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Can you do this mentally? (MP5) What labels could you use? (MP6)

Slide 202 (Answer) / 311

81 What is the volume of a rectangular prism with edges of 4, 5, and 7?

Slide 203 / 311

81 What is the volume of a rectangular prism with edges of 4, 5, and 7?

[This object is a pull tab]

Answer

V = 4(5)(7) V = 140 units3

Slide 203 (Answer) / 311

82 What is the volume of a cube with edges of 5 units?

Slide 204 / 311

82 What is the volume of a cube with edges of 5 units?

[This object is a pull tab]

Answer

V = 53 V = 125 units3

Slide 204 (Answer) / 311

83 If the volume of a rectangular prism is 64 u3 and has height 8 and width 4, what is the length?

Slide 205 / 311

83 If the volume of a rectangular prism is 64 u3 and has height 8 and width 4, what is the length?

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Answer

64 = ℓ(8)(4) 64 = 32 ℓ 32 32 ℓ = 2 units

Slide 205 (Answer) / 311

84 If a cube has volume 27 u3, what is the cubes surface area?

Slide 206 / 311

84 If a cube has volume 27 u3, what is the cubes surface area?

[This object is a pull tab]

Answer

27 = s3 3 = s SA = 6(3)2 SA = 54 units2

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85 Find the volume of the prism. 15 12 20

Slide 207 / 311

85 Find the volume of the prism. 15 12 20

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Answer

a2 + 122 = 152 a = 9 units B = (1/2)(12)(9) = 54 units2 V = 54(20) V = 1080 units3

Slide 207 (Answer) / 311

86 Find the volume of the prism. 7 2 6 6 6

Slide 208 / 311

86 Find the volume of the prism. 7 2 6 6 6

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Answer

22 + H2 = 72 H = 3√5 H = 6.71 units B = (1/2)(6)(3√3) B = 9√3 units2 B = 15.59 units2 V = (9√3)(3√5) V = 27√15 units3 V = 104.6 units3

Slide 208 (Answer) / 311

87 Find the volume of the prism. 8 11 The base is a regular hexagon.

Slide 209 / 311

87 Find the volume of the prism. 8 11 The base is a regular hexagon.

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Answer need to start by finding the apothem = 4√3...30-60-90 triangle P = 6(8) = 48 B = (1/2)(4√3)(48) = 96√3 V = (96√3)(11) V = 1056√3 units3 V = 1829.05 units3

Slide 209 (Answer) / 311

88 A high school has a pool that is 25 yards in length, 60 feet in width, and contains the depth dimensions shown in the figure below. If one cubic yard is about 201.974 gallons, how much water is required to fill the pool? Shallow end Deep end 3 ft 9 ft 2 yds 4 yds 19 yds

Slide 210 / 311

88 A high school has a pool that is 25 yards in length, 60 feet in width, and contains the depth dimensions shown in the figure below. If one cubic yard is about 201.974 gallons, how much water is required to fill the pool? Shallow end Deep end 3 ft 9 ft 2 yds 4 yds 19 yds

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Answer

V = 1(2)(20) + 4(3)(20) + 1/2(19)(1 + 3)(60) V = 40 + 240 + 2,280 V = 2,560 cubic yards Water: 2,560(201.974) = 517,053.44 gallons of water

Slide 210 (Answer) / 311

Volume of a Cylinder

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Slide 211 / 311

base base height r r

Finding the Volume of a Cylinder

V = Bh V = πr2h

Slide 212 / 311

Example: Find the volume of the cylinder with a radius of 4 and a height of 11. V = π(4)2 (11) V = 176π units3 V = 552.92 units3

Cylinders Slide 213 / 311

Example: Find the volume of the cylinder with a radius of 4 and a height of 11. V = π(4)2 (11) V = 176π units3 V = 552.92 units3

Cylinders

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) Can you do this problem mentally? (MP5) What tools do you need? (MP5) What labels could you use? (MP6)

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Example: The surface area of a cylinder is 96π units2, and its radius is 4 units. What is the volume? V = π(4)2 (8) V = 128π units3 V = 402.12 units3 SA = 2πr2 + 2πrh 96π = 2π(4)2 + 2π(4)h 96π = 32π + 8πh

• 32π -32π

64π = 8πh 8π 8π h = 8 units

Cylinders Slide 214 / 311

Example: The surface area of a cylinder is 96π units2, and its radius is 4 units. What is the volume? V = π(4)2 (8) V = 128π units3 V = 402.12 units3 SA = 2πr2 + 2πrh 96π = 2π(4)2 + 2π(4)h 96π = 32π + 8πh

• 32π -32π

64π = 8πh 8π 8π h = 8 units

Cylinders

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) Can you do this problem mentally? (MP5) What tools do you need? (MP5) What labels could you use? (MP6)

Slide 214 (Answer) / 311

89 Find the volume of the cylinder with radius 6 and height 8.

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89 Find the volume of the cylinder with radius 6 and height 8.

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Answer

V = π(6)2 (8) V = 288π units3 V = 904.78 units3

Slide 215 (Answer) / 311

90 Find the volume of the cylinder with a circumference of 18π units and a height of 6.

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90 Find the volume of the cylinder with a circumference of 18π units and a height of 6.

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Answer

V = π(9)2 (6) V = 486π units3 V = 1,526.81 units3 C = 2πr 18π = 2πr 2π 2π r = 9 units

Slide 216 (Answer) / 311

r = 8 h

91 Find the volume of the cylinder with a surface area of 653.12 u2 & a radius of 8 units. Use 3.14 as your value of π.

Slide 217 / 311

r = 8 h

91 Find the volume of the cylinder with a surface area of 653.12 u2 & a radius of 8 units. Use 3.14 as your value of π.

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Answer

V = π(8)2 (5) V = 1004.8 units3 SA = 2π82 + 2π(8)h 653.12 = 401.92 + 50.24h 251.2 = 50.24h h = 5 units

Slide 217 (Answer) / 311

92 The volume of a cylinder is 108π u3, and the height is 12 units. What is the surface area?

Slide 218 / 311

92 The volume of a cylinder is 108π u3, and the height is 12 units. What is the surface area?

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Answer

108π = πr2 (12) 9 = r2 3 = r SA = 2π32 + 2π(3)(12) SA = 18π + 72π SA = 90π units2 SA = 282.74 units2

Slide 218 (Answer) / 311

93 The height of a cylinder doubles, what happens to the volume? A Doubles B Quadruples C Depends on the cylinder D Cannot be determined

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93 The height of a cylinder doubles, what happens to the volume? A Doubles B Quadruples C Depends on the cylinder D Cannot be determined

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Answer

A

Slide 219 (Answer) / 311

94 The radius of a cylinder doubles, what happens to the volume? A Doubles B Quadruples C Depends on the cylinder D Cannot be determined

Slide 220 / 311

94 The radius of a cylinder doubles, what happens to the volume? A Doubles B Quadruples C Depends on the cylinder D Cannot be determined

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Answer

B

Slide 220 (Answer) / 311

24" 4" 3" 95 A 3" hole is drilled through a solid cylinder with a diameter of 4" forming a tube. What is the volume of the tube?

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24" 4" 3" 95 A 3" hole is drilled through a solid cylinder with a diameter of 4" forming a tube. What is the volume of the tube?

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Answer

Vouter = π(4)2 (24) Vouter = 384π in3 Vouter = 1,206.37 in3 Vtube = 384π - 216π Vtube = 168π in3 Vtube = 527.79 in3 Vinner = π(3)2 (24) Vinner = 216π in3 Vinner = 678.58 in3

Slide 221 (Answer) / 311

Volume of a Pyramid

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Slide 222 / 311 Finding the Volume of a Pyramid

V = 1/3 Bh

Square Base (B) Slant Height ( ) Pyramid's Height (h)

ℓ

Slide 223 / 311

Example: Find the volume of the pyramid. 5 4 6 V = 1/3 Bh B = 5(4) = 20 h = 6 units V = 1/3 (20)(6) V = 40 units3

Volume of Pyramids Slide 224 / 311

Example: Find the volume of the pyramid. 5 4 6 V = 1/3 Bh B = 5(4) = 20 h = 6 units V = 1/3 (20)(6) V = 40 units3

Volume of Pyramids

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) Can you do this problem mentally? (MP5) What tools do you need? (MP5) What labels could you use? (MP6)

Slide 224 (Answer) / 311

Example: Find the volume of the pyramid. 8 8 5 8 8 5 4 h

click for extra diagram

Volume of Pyramids Slide 225 / 311

Example: Find the volume of the pyramid. 8 8 5 8 8 5 4 h

click for extra diagram

Volume of Pyramids

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Answer h2 + 42 = 52 h2 + 16 = 25 h2 = 9 h = 3 units Note: 3-4-5 Pythagorean Triple V = 1/3 Bh B = 8(8) = 64 V = 1/3 (64)(3) V = 64 units3

Slide 225 (Answer) / 311

96 Find the volume of the pyramid. 7 6 5

Slide 226 / 311

96 Find the volume of the pyramid. 7 6 5

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Answer

V = 1/3 Bh B = 7(6) = 42 V = 1/3 (42)(5) V = 70 units3

Slide 226 (Answer) / 311

97 Find the volume of the pyramid. 6 6 8

Slide 227 / 311

97 Find the volume of the pyramid. 6 6 8

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Answer

V = 1/3 Bh B = 6(6) = 36 V = 1/3 (36)(7.42) V = 89.04 units3 h2 + 32 = 82 h2 + 9 = 64 h2 = 55 h = 7.42 units

Slide 227 (Answer) / 311

98 Find the volume of the pyramid. 12 12 10

Slide 228 / 311

98 Find the volume of the pyramid. 12 12 10

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Answer

V = 1/3 Bh B = 12(12) = 144 V = 1/3 (144)(5.29) V = 253.92 units3 2 + 62 = 102 = 8 units ℓ ℓ h2 + 62 = 82 h = 5.29 units

10

6

ℓ

8 6 h

Slide 228 (Answer) / 311

Example: Find the volume of the pyramid. a a = 4 lateral edge = 8 Regular Pentagonal Pyramid First, find the side length of the regular pentagon using the apothem & trigonometric ratios. = 72° = central = 36° = top angle of the . 72 2 360 5 tan36 = x = 4tan36 = 2.91 Therefore, s = 2(2.91) = 5.82 x 4

Volume of Pyramids

Click Click Click Click Click

Slide 229 / 311

Example: Find the volume of the pyramid. a a = 4 lateral edge = 8 Regular Pentagonal Pyramid First, find the side length of the regular pentagon using the apothem & trigonometric ratios. = 72° = central = 36° = top angle of the . 72 2 360 5 tan36 = x = 4tan36 = 2.91 Therefore, s = 2(2.91) = 5.82 x 4

Volume of Pyramids

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) How is right triangle trigonometry used to calculate the segment lengths of regular polygons? (MP7)

Slide 229 (Answer) / 311

Example: Find the volume of the pyramid. Next, find the slant height of the pyramid using the lateral edge, the value of x from the previous slide & Pyth. Theorem. 8 2.91

ℓ

2.912 + ℓ 2 = 82 8.4681 + ℓ 2 = 64 ℓ 2 = 55.5319 ℓ = 7.45 Then, use the slant height & apothem w/

• Pyth. Theorem to

find the height. 7 . 4 5 4 h

C l i c k

42 + h2 = 7.452 16 + h2 = 55.5319 h2 = 39.5319 h = 6.29

Volume of Pyramids

Click Click Click Click Click Click Click Click

a a = 4 lateral edge = 8 Regular Pentagonal Pyramid

Click

Slide 230 / 311

Example: Find the volume of the pyramid. Next, find the slant height of the pyramid using the lateral edge, the value of x from the previous slide & Pyth. Theorem. 8 2.91

ℓ

2.912 + ℓ 2 = 82 8.4681 + ℓ 2 = 64 ℓ 2 = 55.5319 ℓ = 7.45 Then, use the slant height & apothem w/

• Pyth. Theorem to

find the height. 7 . 4 5 4 h

C l i c k

42 + h2 = 7.452 16 + h2 = 55.5319 h2 = 39.5319 h = 6.29

Volume of Pyramids

Click Click Click Click Click Click Click Click

a a = 4 lateral edge = 8 Regular Pentagonal Pyramid

Click [This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) How is Pythagorean Theorem used to calculate the segment lengths of pyramids? (MP7)

Slide 230 (Answer) / 311

Example: Find the volume of the pyramid. Last, find the Area of your Base & Volume. B = 1/2 aP B = 1/2 (4)(29.1) B = 58.2 units2 V = 1/3 Bh V = 1/3 (58.2)(6.29) V = 122.03 units3

Volume of Pyramids

Click Click Click Click Click Click

a a = 4 lateral edge = 8 Regular Pentagonal Pyramid

Slide 231 / 311

Example: Find the volume of the pyramid. Last, find the Area of your Base & Volume. B = 1/2 aP B = 1/2 (4)(29.1) B = 58.2 units2 V = 1/3 Bh V = 1/3 (58.2)(6.29) V = 122.03 units3

Volume of Pyramids

Click Click Click Click Click Click

a a = 4 lateral edge = 8 Regular Pentagonal Pyramid

[This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) What labels could you use? (MP6)

Slide 231 (Answer) / 311

99 Find the volume of the right pyramid.

a

a = 5 h = 12 Regular Octagonal Pyramid

Slide 232 / 311

99 Find the volume of the right pyramid.

a

a = 5 h = 12 Regular Octagonal Pyramid

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Answer

P = 8(4.14) = 33.12 B = 1/2 aP B = 1/2 (5)(33.12) B = 215.28 units2 V = 1/3(215.28)(12) V = 861.12 units3 a = 5 & h = 12 tan22.5 = x 5 x = 2.07 1 side = 4.14

Slide 232 (Answer) / 311

100 Find the volume of the right pyramid. 8 11 The base is a regular hexagon.

Slide 233 / 311

100 Find the volume of the right pyramid. 8 11 The base is a regular hexagon.

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Answer

V = 1/3 Bh V = 1/3 (166.32)(11) V = 609.84 units3 r = 8 units a = 4√3 = 6.93 units B = 1/2aP B = 1/2(6.93)(48) B = 166.32 units2

Slide 233 (Answer) / 311

A truncated pyramid is a pyramid with its top cutoff parallel to its base. Find the volume of the truncated pyramid shown. 2 2 6 6 9 3 Vtruncated = Vbig - Vsmall Bbig = 6(6) = 36 hbig = 3 + 9 = 12 Vbig = 1/3 (36)(12) Vbig = 144 units3 Bsmall = 2(2) = 4 hsmall = 3 Vsmall = 1/3 (4)(3) Vsmall = 4 units3 Vtruncated = 144 - 4 Vtruncated = 140 units3

Volume of Pyramids Slide 234 / 311

A truncated pyramid is a pyramid with its top cutoff parallel to its base. Find the volume of the truncated pyramid shown. 2 2 6 6 9 3 Vtruncated = Vbig - Vsmall Bbig = 6(6) = 36 hbig = 3 + 9 = 12 Vbig = 1/3 (36)(12) Vbig = 144 units3 Bsmall = 2(2) = 4 hsmall = 3 Vsmall = 1/3 (4)(3) Vsmall = 4 units3 Vtruncated = 144 - 4 Vtruncated = 140 units3

Volume of Pyramids

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) Can you do this problem mentally? (MP5) What tools do you need? (MP5) What labels could you use? (MP6)

Slide 234 (Answer) / 311

101 Find the volume of the truncated pyramid.

2 2 8 8 12 3

Slide 235 / 311

101 Find the volume of the truncated pyramid.

2 2 8 8 12 3

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Answer

Vtruncated = 316 units3 Vbig = 1/3 (64)(15) Vbig = 320 units3 Vsmall = 1/3 (4)(3) Vsmall = 4 units3

Slide 235 (Answer) / 311

102 The table shows the approximate measurements of the Red Pyramid in Egypt and the Great Pyramid of Cholula in Mexico. Approximately, what is the difference between the volume of the Red Pyramid and the volume of the Great Pyramid of Cholula? A 6,132,867 cubic meters B 4,455,000 cubic meters C 2,777,133 cubic meters D 1,677,867 cubic meters Length Width Height Red Pyramid 220 m 220m 104 m Great Pyramid of Cholula 450 m 450 m 66 m

Answer

Slide 236 / 311

103 Salt water comes in cylindrical containers that measure 10 feet high and have a diameter of 8 feet. Determine the height of the aquarium that should be used in the design. Show that your design will be able to store at least 3 cylindrical containers of water. When you finish, enter your value for h1 into your SMART Responder. The Geometryville Aquarium is building a new tank space for coral reef fish shown in the figure below. The laws say that the dimensions of the tank must have a maximum length of 14 feet, a maximum width of 10 feet and a maximum height of 16 feet. w h1 h2

ℓ Slide 237 / 311

103 Salt water comes in cylindrical containers that measure 10 feet high and have a diameter of 8 feet. Determine the height of the aquarium that should be used in the design. Show that your design will be able to store at least 3 cylindrical containers of water. When you finish, enter your value for h1 into your SMART Responder. The Geometryville Aquarium is building a new tank space for coral reef fish shown in the figure below. The laws say that the dimensions of the tank must have a maximum length of 14 feet, a maximum width of 10 feet and a maximum height of 16 feet. w h1 h2

ℓ

[This object is a pull tab]

Sample Answer: Full Credit Given that the salt water is transported in cylindrical containers, each container holds 502.65 cubic feet of salt water π(4)2(10) = 160π = 502.65 Three containers will hold 480π = 1507.96 cubic feet of salt water. The volume of the new aquarium equals the volume of the prism plus the volume of the pyramid. I used the maximum length of 14 feet and the maximum width of 10 feet. 14(10)h1 + (1/3)(14)(10)h2 I used the maximum total height of 16 feet. Since the volume of a pyramid involves dividing by 3, I made the height of the pyramid much smaller than the height of the prism. 14(10)h1 + (1/3)(14)(10)h2 14(10)(13) + (1/3)(14)(10)(3) = 1,960 cubic feet Using h1 = 13 feet & h2 = 3 feet, the aquarium will have a volume greater than 1,507.96 cubic feet.

• Note: any two heights that have a sum of 16 and create a volume

greater than 1,507.96 are acceptable.

Slide 237 (Answer) / 311

Volume of a Cone

Return to Table of Contents

Slide 238 / 311

r height S l a n t H e i g h t

ℓ Finding the Volume of a Cone

V = 1/3 Bh V = 1/3πr2 h

Slide 239 / 311

Example: Find the volume of the cone. 9 7 V = 1/3 πr2 h V = 1/3 π(7)2 (9) V = 147π units3 V = 461.81 units3

Volume of a Cone Slide 240 / 311

Example: Find the volume of the cone. 9 7 V = 1/3 πr2 h V = 1/3 π(7)2 (9) V = 147π units3 V = 461.81 units3

Volume of a Cone

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) Can you do this problem mentally? (MP5) What tools do you need? (MP5) What labels could you use? (MP6)

Slide 240 (Answer) / 311

Example: Find the volume of the cone. 12 4 V = 1/3 πr2 h V = 1/3 π(4)2 (8.94) V = 47.68π units3 V = 149.79 units3 r = 4, so d = 8 With the right triangle, use Pythagorean Theorem to find the height of the pyramid. h2 + 82 = 122 h2 + 64 = 144 h2 = 80, h = √80 = 8.94

Volume of a Cone Slide 241 / 311

Example: Find the volume of the cone. 12 4 V = 1/3 πr2 h V = 1/3 π(4)2 (8.94) V = 47.68π units3 V = 149.79 units3 r = 4, so d = 8 With the right triangle, use Pythagorean Theorem to find the height of the pyramid. h2 + 82 = 122 h2 + 64 = 144 h2 = 80, h = √80 = 8.94

Volume of a Cone

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) What labels could you use? (MP6)

Slide 241 (Answer) / 311

Example: Find the volume of the cone, with lateral area of 15π units2 and a slant height 5 units. LA = πr ℓ 15π = πr(5) 15π = 5πr 5π 5π 3 units = r 1) You know the Lateral area & slant height, so use the Lateral Area formula to calculate the radius.

Volume of a Cone

Click Click Click Click Click

Slide 242 / 311

Example: Find the volume of the cone, with lateral area of 15π units2 and a slant height 5 units. LA = πr ℓ 15π = πr(5) 15π = 5πr 5π 5π 3 units = r 1) You know the Lateral area & slant height, so use the Lateral Area formula to calculate the radius.

Volume of a Cone

Click Click Click Click Click [This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Would it help to draw a picture? (MP4 & MP5) How is the Lateral Area used to calculate the segment lengths of cones? (MP7)

Slide 242 (Answer) / 311

h2 + 32 = 52 h2 + 9 = 25 h2 = 16 h = 4 Note: 3-4-5 Pyth. Triple 2) Next, use the slant height & radius to calculate the height of the cone using Pythagorean Theorem.

Volume of a Cone

Click Click Click Click Click

Slide 243 / 311

h2 + 32 = 52 h2 + 9 = 25 h2 = 16 h = 4 Note: 3-4-5 Pyth. Triple 2) Next, use the slant height & radius to calculate the height of the cone using Pythagorean Theorem.

Volume of a Cone

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Would it help to draw a picture? (MP4 & MP5) How is Pythagorean Theorem used to calculate the segment lengths of cones? (MP7)

Slide 243 (Answer) / 311

V = 1/3 πr2 h V = 1/3 π(3)2 (4) V = 12π units3 V = 37.70 units3 3) Last, calculate the volume of the cone.

Volume of a Cone

Click Click Click Click

Slide 244 / 311

V = 1/3 πr2 h V = 1/3 π(3)2 (4) V = 12π units3 V = 37.70 units3 3) Last, calculate the volume of the cone.

Volume of a Cone

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Can you calculate your answer mentally? (MP5) What labels could you use? (MP6)

Slide 244 (Answer) / 311

104 What is the volume of the cone? 8 d = 10

Slide 245 / 311

104 What is the volume of the cone? 8 d = 10

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Answer

V = 1/3 π(5)2 (8) V = 200πunits3

3

V = 66.6π units3 V = 209.44 units3

Slide 245 (Answer) / 311

105 What is the volume of the cone? r = 4 = 9 ℓ

Slide 246 / 311

105 What is the volume of the cone? r = 4 = 9 ℓ

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Answer

V = 1/3 π(4)2 (8.06) V = 42.99π units3 V = 135.05 units3 42 + h2 = 92 16 + h2 = 81 h2 = 65 h =√65 = 8.06

Slide 246 (Answer) / 311

106 What is the volume of the cone? 10 40°

Slide 247 / 311

106 What is the volume of the cone? 10 40°

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Answer

V = 1/3 π(3.215)2 (7.66) V = 26.39π units3 V = 82.91 units3 sin(40) = d = 10sin(40) d = 6.43, so r = 3.215 d 10

40o h d

10

cos(40) = h = 10cos(40) h = 7.66 h 10

Slide 247 (Answer) / 311

107 What is the volume of the truncated cone? r = 8 r = 4 6 6

Slide 248 / 311

107 What is the volume of the truncated cone? r = 8 r = 4 6 6

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Answer

Vtruncated = 224π units3 Vtruncated = 703.72 units3 Vbig = 1/3 π(8)2 (12) Vbig = 256π units3 Vbig = 804.25 units3 Vsmall = 1/3 π(4)2 (6) Vsmall = 32π units3 Vsmall = 100.53units3

Slide 248 (Answer) / 311

Surface Area & Volume of Spheres

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Slide 249 / 311

Recall the Definition of a Circle The locus of points in a plane that are the same distance from a point called the center of the circle.

X

Y

Every point on the above circle is the same distance from the origin in the x, y plane.

Y X

Spheres Slide 250 / 311

The locus of points in space that are the same distance from a point.

Y X Z

Every point on the sphere above on the left side, is the same distance from the origin in space, the x, y, z plane.

X Y

Y X

Spheres Slide 251 / 311

Y X Z

The Great Circle of a sphere is found at the intersection

• f a plane and a sphere when the plane contains the center
• f the sphere.

Spheres Slide 252 / 311

Y X Z

Great Circles

Each of these planes intersects the sphere, and the plane contains the center of the sphere

Slide 253 / 311

International Date Line

Great Circles

The Earth has 2 Great Circles: Can you name them?

Click to reveal picture

The Equator The Prime Meridian w/ the International Date Line

click click

Slide 254 / 311 Great Circle

The Great Circle separates the Sphere into two equal halves at the center of the sphere.

Slide 255 / 311 Each half is called a Hemisphere Slide 256 / 311

Cross Sections

A Cross Section is found by the intersection of a plane and a solid.

Cross - Section

(Click the top hemisphere to see the cross section.)

Slide 257 / 311

.

small circles great circle The farther the cross section of the sphere is taken from its center the smaller the circle.

Cross Sections Slide 258 / 311

8 2 8 r

Example: Find the radius of the cross section of the sphere that has a radius of 8 if the cross section is 2 from the center. 22 + r2 = 82 4 + r2 = 64 r2 = 60 r = √60 = 2√15 = 7.75

Cross Sections Slide 259 / 311

8 2 8 r

Example: Find the radius of the cross section of the sphere that has a radius of 8 if the cross section is 2 from the center. 22 + r2 = 82 4 + r2 = 64 r2 = 60 r = √60 = 2√15 = 7.75

Cross Sections

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Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) How is Pythagorean Theorem used to calculate the segment lengths of spheres? (MP7)

Slide 259 (Answer) / 311

4 Example: A cross section of a sphere is 4 units from the center of the sphere and has an area of 16π units2. What is area of the great circle? Leave your answer in terms of π. 16π = πr2 r = 4 units in the cross section 42 + 42 = r2 32 = r2 r =√32 = 4√2 = 2.83 = radius of sphere A = π(√32)2 A = 32π units2

Cross Sections Slide 260 / 311

4 Example: A cross section of a sphere is 4 units from the center of the sphere and has an area of 16π units2. What is area of the great circle? Leave your answer in terms of π. 16π = πr2 r = 4 units in the cross section 42 + 42 = r2 32 = r2 r =√32 = 4√2 = 2.83 = radius of sphere A = π(√32)2 A = 32π units2

Cross Sections

[This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) Can you calculate the answer mentally? (MP5) How is Pythagorean Theorem used to calculate the segment lengths of spheres? (MP7)

Slide 260 (Answer) / 311

108 What is the area of the cross section of a sphere that is 6 units from the center of the sphere if the sphere has radius 8 units?

Slide 261 / 311

108 What is the area of the cross section of a sphere that is 6 units from the center of the sphere if the sphere has radius 8 units?

[This object is a pull tab]

Answer

62 + r2 = 82 r of cross section = √28 = 2√7 = 5.29 A = π(√28)2 A = 28π units2 A = 87.96 units2

Slide 261 (Answer) / 311

109 What is the area of the great circle if a cross section that is 3 from the center has a circumference of 10π?

Slide 262 / 311

109 What is the area of the great circle if a cross section that is 3 from the center has a circumference of 10π?

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Answer

C = 2πr 10π = 2πr r = 5 units = radius of cross section circle Sphere radius: 52 + 32 = r2 Sphere radius = √34 = 5.83 A = π(√34)2 A = 34π units2 A = 106.81 units2

Slide 262 (Answer) / 311

110 The circumference of the great circle of a sphere is 12π units and a cross section has a circumference of 8π units. How far is the cross section from the center?

Slide 263 / 311

110 The circumference of the great circle of a sphere is 12π units and a cross section has a circumference of 8π units. How far is the cross section from the center?

[This object is a pull tab]

Answer

C = 2πr 12π = 2πr r = 6 units Distance between great circle & cross section 42 + x2 = 62 x = √20 = 2√5 = 4.47 units C = 2πr 8π = 2πr r = 4 units

Slide 263 (Answer) / 311

r S.A. = 4πr2

Finding the Surface Area of the Sphere

Why is there no formula for lateral area? A sphere doesn't have any bases, so the lateral area is the same as the surface area.

Click to reveal

Slide 264 / 311

r V = πr3 4 3

Finding the Volume of the Sphere Slide 265 / 311

Example: Find the surface area & volume of a sphere with radius of 6 ft. SA = 4π(6)2 SA = 144π units2 SA = 452.39 units2 V = π(6)3 V = 288π units3 V = 904.78 units3 4 3

Finding the Volume of the Sphere Slide 266 / 311

Example: Find the surface area & volume of a sphere with radius of 6 ft. SA = 4π(6)2 SA = 144π units2 SA = 452.39 units2 V = π(6)3 V = 288π units3 V = 904.78 units3 4 3

Finding the Volume of the Sphere

[This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) Can you do this problem mentally? (MP5) What tools do you need? (MP5) What labels could you use? (MP6)

Slide 266 (Answer) / 311

Example: Find the surface area & volume of a sphere that a great circle with area 24π units2? SA = 4π(4.9)2 SA = 96.04π units2 SA = 301.72 units2 V = π(4.9)3 V = 156.87π units3 V = 492.81 units3 4 3 24π = πr2 π π r2 = 24 r = 4.90 units

Finding the Volume of the Sphere Slide 267 / 311

Example: Find the surface area & volume of a sphere that a great circle with area 24π units2? SA = 4π(4.9)2 SA = 96.04π units2 SA = 301.72 units2 V = π(4.9)3 V = 156.87π units3 V = 492.81 units3 4 3 24π = πr2 π π r2 = 24 r = 4.90 units

Finding the Volume of the Sphere

[This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) How is the area of the Great Circle used to calculate the segment lengths of spheres? (MP7) What labels could you use? (MP6)

Slide 267 (Answer) / 311

Example: A cross section of a sphere has area 36π units2 and is 10 units from the center, what is the surface area & volume

• f the sphere?

Radius of Cross Section 36π = πr2 π π r2 = 36 r = 6 units Radius of Sphere 102 + 62 = R2 136 = R2 R = √136 = 11.66 units

Finding the Volume of the Sphere

click click click click click click click

Slide 268 / 311

Example: A cross section of a sphere has area 36π units2 and is 10 units from the center, what is the surface area & volume

• f the sphere?

Radius of Cross Section 36π = πr2 π π r2 = 36 r = 6 units Radius of Sphere 102 + 62 = R2 136 = R2 R = √136 = 11.66 units

Finding the Volume of the Sphere

click click click click click click click [This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) Would it help to draw a picture for this problem? (MP4 & MP5) What tools do you need? (MP5) How is the area of a sphere's cross section used to calculate the segment lengths in the sphere? (MP7) What labels could you use? (MP6)

Slide 268 (Answer) / 311

SA = 4π(√136)2 SA = 544π units2 SA = 1,709.03 units2 V = π(√136)3 V = 2,114.69π units3 V = 6,643.50 units3 4 3 Example: A cross section of a sphere has area 36π units2 and is 10 units from the center, what is the surface area & volume

• f the sphere?

Finding the Volume of the Sphere

click click click click click Click

Slide 269 / 311

SA = 4π(√136)2 SA = 544π units2 SA = 1,709.03 units2 V = π(√136)3 V = 2,114.69π units3 V = 6,643.50 units3 4 3 Example: A cross section of a sphere has area 36π units2 and is 10 units from the center, what is the surface area & volume

• f the sphere?

Finding the Volume of the Sphere

click click click click click Click [This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) Would it help to draw a picture for this problem? (MP4 & MP5) What tools do you need? (MP5) How is the area of a sphere's cross section used to calculate the segment lengths in the sphere? (MP7) What labels could you use? (MP6)

Slide 269 (Answer) / 311

111 Find the surface area of a sphere with radius 10.

Slide 270 / 311

111 Find the surface area of a sphere with radius 10.

[This object is a pull tab]

Answer

SA = 4π(10)2 SA = 400π units2 SA = 1256.64 units2

Slide 270 (Answer) / 311

112 Find the volume of a sphere with radius 10.

Slide 271 / 311

112 Find the volume of a sphere with radius 10.

[This object is a pull tab]

Answer

V = π(10)3 V = 1333.3π units3 V = 4188.79 units3

4 3

Slide 271 (Answer) / 311

113 What is the surface area of a sphere if a cross section 7 units from the center has an area of 50.26 units2?

Slide 272 / 311

113 What is the surface area of a sphere if a cross section 7 units from the center has an area of 50.26 units2?

[This object is a pull tab]

Answer

SA = 4π(√65)2 SA = 260π units2 SA = 816.81 units2 50.26 = πr2 r2 = 16 r = 4 radius of cross section radius of sphere: 42 + 72 = R2 R = √65 = 8.06

Slide 272 (Answer) / 311

114 What is the volume of a sphere if a cross section 7 units from the center has an area of 50.26 units2?

Slide 273 / 311

114 What is the volume of a sphere if a cross section 7 units from the center has an area of 50.26 units2?

[This object is a pull tab]

Answer

V = π(√65)3 V = 698.73π units3 V = 2,195.12 units3

4 3

radius of sphere = √65 (see last problem for how)

Slide 273 (Answer) / 311

115 The volume of a sphere is 24π units3. What is the area

• f a great circle of the sphere?

Slide 274 / 311

115 The volume of a sphere is 24π units3. What is the area

• f a great circle of the sphere?

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Answer

24π = πr3 24 = r3 18 = r3 r = 2.62 units A = π(2.62)2 A = 6.86π units2 A = 21.57 units2

4 3 4 3

Slide 274 (Answer) / 311

116 A recipe calls for half of an orange. Shelly use an

• range that has a diameter of 3 inches. She wraps

the remaining half of orange in plastic wrap. What is the amount of area that Shelly has to cover?

Slide 275 / 311

116 A recipe calls for half of an orange. Shelly use an

• range that has a diameter of 3 inches. She wraps

the remaining half of orange in plastic wrap. What is the amount of area that Shelly has to cover?

[This object is a pull tab]

Answer

SA = 4πr2

1/2 SA = 2πr2

Since she also has to cover the bottom circular region, the total area will be 2πr2 + πr2 = 3πr2 Total Area = 3 π(1.5)2 Total Area = 6.75 π units2 Total Area = 21.21 units

2

Slide 275 (Answer) / 311

Cavalieri's Principle

Return to Table of Contents

Slide 276 / 311 Cavalieri's Principle

If two solids are the same height, and the area of their cross sections are equal, then the two solids will have the same volume.

Slide 277 / 311

14 14 14 Which solid has the greatest volume? 224π 703.72 None: All of the solids have the same volume.

Click

Cavalieri's Principle

2π 8 4π 4 4 224π 703.72 224π 703.72

Click Click Click

Slide 278 / 311

Example: A sphere is submerged in a cylinder. Both solids have a radius of 4. What is the volume of the cylinder not occupied by the sphere? volume of cylinder - volume of sphere

Cavalieri's Principle

π(4)2 (8) - 4/3 π(4)3 128π - 256/3 π

128/3 π units3 Click Click Click Click

Slide 279 / 311

Example: A sphere is submerged in a cylinder. Both solids have a radius of 4. What is the volume of the cylinder not occupied by the sphere? volume of cylinder - volume of sphere

Cavalieri's Principle

π(4)2 (8) - 4/3 π(4)3 128π - 256/3 π

128/3 π units3 Click Click Click Click

[This object is a pull tab]

Math Practice Questions to help address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What tools do you need? (MP5) What do you know about the volumes of solids and area of complex figures that can be applied to this situation? (MP7) What labels could you use? (MP6)

Slide 279 (Answer) / 311

The result shows that the left over volume is equal to what other solid? cone According to Cavalieri, what can be said about the cross section? The cross section of the great circle of the sphere is equal to the circle cross section of the cylinder.

Click

Cavalieri's Principle

Click

Slide 280 / 311

The result shows that the left over volume is equal to what other solid? cone According to Cavalieri, what can be said about the cross section? The cross section of the great circle of the sphere is equal to the circle cross section of the cylinder.

Click

Cavalieri's Principle

Click [This object is a pull tab]

Math Practice Questions on this slide address MP standards: 1st Question: MP2 2nd Question: MP7

Slide 280 (Answer) / 311

Example: What is the radius of a sphere made from the cylinder of modeling clay shown? If you are using clay to model both solids, what measurement is the same? Volume 15 5

Cavalieri's Principle

Click

Slide 281 / 311

Example: What is the radius of a sphere made from the cylinder of modeling clay shown? If you are using clay to model both solids, what measurement is the same? Volume 15 5

Cavalieri's Principle

Click [This object is a pull tab]

Math Practice Question on this slide address MP standards: MP7

Slide 281 (Answer) / 311

Therefore, calculate the volume of the cylinder first.

Cavalieri's Principle

V = π(5)2 (15) V = 375π units3 Then create an equation to represent the problem and solve for r. 375π = 4/3 πr3 375 = 4/3 r3 281.25 = r3 r = 6.55 units

Click Click Click Click Click Click

15 5

Slide 282 / 311

Therefore, calculate the volume of the cylinder first.

Cavalieri's Principle

V = π(5)2 (15) V = 375π units3 Then create an equation to represent the problem and solve for r. 375π = 4/3 πr3 375 = 4/3 r3 281.25 = r3 r = 6.55 units

Click Click Click Click Click Click

15 5

[This object is a pull tab]

Math Practice Questions on this slide address MP standards: "Then create an equation to represent...": MP2 Additional Questions to address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What connections do you see between these two solids? (MP4) What labels could you use? (MP6)

Slide 282 (Answer) / 311

117 These 2 solids have the same volume, find the value of x. 11 r = 6 11 x 9

Slide 283 / 311

117 These 2 solids have the same volume, find the value of x. 11 r = 6 11 x 9

[This object is a pull tab]

Answer

11(9)(x) = π(6)2 (11) 99x = 396π x = 4π units x = 12.57 units

Slide 283 (Answer) / 311

118 These 2 solids have the same volume, find the value of x.

12 x 12 10 8

Slide 284 / 311

118 These 2 solids have the same volume, find the value of x.

12 x 12 10 8

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Answer

12(10)(8) = π(x)2 (12) 960 = 12πx2 x2 =25.46 x = 5.04 units

Slide 284 (Answer) / 311

Prism C B = 20 in2 x Prism D B = 20 in2 y Two prisms each with a base area of 20 square inches are shown. Which statements about prisms C and D are true. Select all that

• apply. (Statements are on the next slide.)

Slide 285 / 311

119 Two prisms each with a base area of 20 square inches are shown. Which statements about prisms C and D are true. Select all that apply. A If x > y, the area of a vertical cross section of prism C is greater than the area of a vertical cross section of prism D. B If x > y, the area of a vertical cross section of prism C is equal to the area of a vertical cross section of prism D. C If x > y, the area of a vertical cross section of prism C is less than the area of a vertical cross section of prism D. D If x = y, the volume of prism C is greater than the volume of prism D, because prism C is a right prism. E If x = y, the volume of prism C is equal to the volume of prism D because the prisms have the same base area. F If x = y, the volume of prism C is less than the volume of prism D because prism D is an oblique prism.

Slide 286 / 311

119 Two prisms each with a base area of 20 square inches are shown. Which statements about prisms C and D are true. Select all that apply. A If x > y, the area of a vertical cross section of prism C is greater than the area of a vertical cross section of prism D. B If x > y, the area of a vertical cross section of prism C is equal to the area of a vertical cross section of prism D. C If x > y, the area of a vertical cross section of prism C is less than the area of a vertical cross section of prism D. D If x = y, the volume of prism C is greater than the volume of prism D, because prism C is a right prism. E If x = y, the volume of prism C is equal to the volume of prism D because the prisms have the same base area. F If x = y, the volume of prism C is less than the volume of prism D because prism D is an oblique prism.

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Answer

A and E

Slide 286 (Answer) / 311

Similar Solids

Return to Table of Contents

Slide 287 / 311

Corresponding sides of similar figures are similar. The prisms shown are similar. Find the values of x and y. 4 x 2 6 9 y 4 6 = 4 6 =

Similar Solids

x 9 36 = 6x 6 6 6 = x 4y = 12 4 4 y = 3 2 y

Click Click Click Click Click Click Click Click

Slide 288 / 311

Corresponding sides of similar figures are similar. The prisms shown are similar. Find the values of x and y. 4 x 2 6 9 y 4 6 = 4 6 =

Similar Solids

x 9 36 = 6x 6 6 6 = x 4y = 12 4 4 y = 3 2 y

Click Click Click Click Click Click Click Click [This object is a pull tab]

Math Practice Additional Questions to address MP standards: What information are you given? (MP1) What is the problem asking? (MP1) How can you represent the problem with symbols and numbers? (MP2) What connections do you see between these two solids? (MP4) What do you know about similar polygons that can be applied to this situation? (MP7) What labels could you use? (MP6) Can you find a shortcut to solve this problem? How would your shortcut make the problem easier? (MP8)

• multiply corresponding side length by 2/3 (if

given in the bigger prism) or 3/2 (if given in the smaller prism)

Slide 288 (Answer) / 311

4 x 2 6 9 y The ratio of similarity, k, is the common value that is multiplied to preimage to get to the image.

• Hint: it's the ratio of image : preimage

If the smaller prism is the preimage, then the value of k is If the larger prism is the preimage, then the value

• f k is

click for the hint

Similar Solids

3/2 2/3

Click Click

Slide 289 / 311

4 x 2 6 9 y The ratio of similarity, k, is the common value that is multiplied to preimage to get to the image.

• Hint: it's the ratio of image : preimage

If the smaller prism is the preimage, then the value of k is If the larger prism is the preimage, then the value

• f k is

click for the hint

Similar Solids

3/2 2/3

Click Click [This object is a pull tab]

Math Practice Questions on this slide address MP standards: Finding the value of k in both cases: MP2 & MP7

Slide 289 (Answer) / 311

120 The pyramid on the left is the preimage and is similar to the image on the right. Find the value

• f x.

8 8 16 h 2 x y 3

Slide 290 / 311

120 The pyramid on the left is the preimage and is similar to the image on the right. Find the value

• f x.

8 8 16 h 2 x y 3

[This object is a pull tab]

Answer

8 8 2 x x = 2 =

Slide 290 (Answer) / 311

121 The pyramid on the left is the preimage and is similar to the image on the right. Find the value

• f y.

8 8 16 h 2 x y 3

Slide 291 / 311

121 The pyramid on the left is the preimage and is similar to the image on the right. Find the value

• f y.

8 8 16 h 2 x y 3

[This object is a pull tab]

Answer

8 16 2 y y = 4 =

Slide 291 (Answer) / 311

122 The pyramid on the left is the preimage and is similar to the image on the right. Find the value

• f h.

8 8 16 h 2 x y 3

Slide 292 / 311

122 The pyramid on the left is the preimage and is similar to the image on the right. Find the value

• f h.

8 8 16 h 2 x y 3

[This object is a pull tab]

Answer

8 h 2 3 h = 12 =

Slide 292 (Answer) / 311

4 6 2 6 9 3 Consider the example of the prisms from earlier. The ratio of similarity from the smaller solid to the larger is 2:3. Calculate the surface area of both solids. How do they compare? SAsmall = 2(6)(2) + 16(4) = 88 units2 SAbig = 2(3)(9) + 24(6) = 198 units2 SA Similarity ratio = 88:198 = 4:9 = 22:32 How do their volumes compare? Vsmall = 2(4)(6) = 48 units3 Vbig = 6(3)(9) = 162 units3 V Similarity ratio = 48:162 = 8:27 = 23:33

Similar Solids

Click Click Click Click Click Click Click Click Click Click Click Click

Slide 293 / 311

4 6 2 6 9 3 Consider the example of the prisms from earlier. The ratio of similarity from the smaller solid to the larger is 2:3. Calculate the surface area of both solids. How do they compare? SAsmall = 2(6)(2) + 16(4) = 88 units2 SAbig = 2(3)(9) + 24(6) = 198 units2 SA Similarity ratio = 88:198 = 4:9 = 22:32 How do their volumes compare? Vsmall = 2(4)(6) = 48 units3 Vbig = 6(3)(9) = 162 units3 V Similarity ratio = 48:162 = 8:27 = 23:33

Similar Solids

Click Click Click Click Click Click Click Click Click Click Click Click [This object is a pull tab]

Math Practice Questions on this slide address MP standards: MP1, MP7 & MP8

Slide 293 (Answer) / 311 Comparing Similar Figures

length in image length in preimage

= k

area in image area in preimage

= k2

volume in image volume in preimage

= k3

Slide 294 / 311

Comparing Similar Figures

length in image length in preimage

= k

area in image area in preimage

= k2

volume in image volume in preimage

= k3

[This object is a pull tab]

Tip

To remember what power of k to use, think

• f the power of the

units used.

Slide 294 (Answer) / 311

How many times bigger is the surface area of the sphere to the right? How many times bigger is the volume of the sphere to the right? r = 3 r = 9 Example: How many times bigger is the radius of the sphere to the right? 3 times bigger 9 times bigger 27 times bigger

Comparing Similar Figures

Click Click Click

SAsmall = 4π(3)2 = 36π units2 SAbig = 4π(9)2 = 324π units2 Vsmall = 4/3 π(3)3 = 36π units3 Vbig = 4/3 π(9)3 = 972π units3

Click Click Click Click Click Click Click Click

Slide 295 / 311

How many times bigger is the surface area of the sphere to the right? How many times bigger is the volume of the sphere to the right? r = 3 r = 9 Example: How many times bigger is the radius of the sphere to the right? 3 times bigger 9 times bigger 27 times bigger

Comparing Similar Figures

Click Click Click

SAsmall = 4π(3)2 = 36π units2 SAbig = 4π(9)2 = 324π units2 Vsmall = 4/3 π(3)3 = 36π units3 Vbig = 4/3 π(9)3 = 972π units3

Click Click Click Click Click Click Click Click [This object is a pull tab]

Math Practice Questions on this slide address MP standards: MP1, MP7 & MP8

Slide 295 (Answer) / 311

123 The scale factor of 2 similar pyramids is 4. If the surface area of the larger one is 64 units2, what is surface area of the smaller one?

Slide 296 / 311

123 The scale factor of 2 similar pyramids is 4. If the surface area of the larger one is 64 units2, what is surface area of the smaller one?

[This object is a pull tab]

Answer

42 64 12 x 16 64 1 x x = 4 units2 = =

Slide 296 (Answer) / 311

124 The scale factor of 2 similar right square pyramids is

• 3. If the area of the base of the larger one is 36 u2

and its height is 12, what is the volume of the smaller one?

Slide 297 / 311

124 The scale factor of 2 similar right square pyramids is

• 3. If the area of the base of the larger one is 36 u2

and its height is 12, what is the volume of the smaller one?

[This object is a pull tab]

Answer

33 144 13 x 27 144 1 x x = 5.3 units3 = = Vbig = 1/3 (36)(12) Vbig = 144 units3

Slide 297 (Answer) / 311

125 An architect builds a scale model of a home using a scale of 2 in to 5 ft. Given the view of the roof of the model, how much roofing material is needed for the house? 12 in 6 in 8 in 5 in 4 in 3 in

Slide 298 / 311

125 An architect builds a scale model of a home using a scale of 2 in to 5 ft. Given the view of the roof of the model, how much roofing material is needed for the house? 12 in 6 in 8 in 5 in 4 in 3 in

[This object is a pull tab]

Answer

22 63 52 x 4 63 25 x = = Amodel roof = 63 in2 x = 393.75 ft2

12in 6in 8in 5in 4in 3in

24 15 24

5 in 3 in

Slide 298 (Answer) / 311

PARCC Sample Questions

The remaining slides in this presentation contain questions from the PARCC Sample Test. After finishing this unit, you should be able to answer these questions. Good Luck! Return to Table

• f Contents

Slide 299 / 311

Question 6/11

Daniel buys a block of clay for an art project. The block is shaped like a cube with edge lengths of 10 inches. Daniel decides to cut the block of clay into two pieces. He places a wire across the diagonal of one face of the cube, as shown in the figure. Then he pulls the wire straight back to create two congruent chunks of clay. PARCC Released Question - PBA - Calculator Section Topic: Intro to 3-D Solids

Slide 300 / 311

126 Part A - Question #1: Daniel wants to keep one chunk of clay for later use. To keep that chunk from drying out, he wants to place a piece of plastic sheeting

• n the surface he exposed when he cut through the cube.

Determine the newly exposed two-dimensional cross section. A Triangle B Parallelogram C Rectangle D Rhombus E Square

Question 6/11

PARCC Released Question - PBA - Calculator Section Topic: Intro to 3-D Solids

Slide 301 / 311

126 Part A - Question #1: Daniel wants to keep one chunk of clay for later use. To keep that chunk from drying out, he wants to place a piece of plastic sheeting

• n the surface he exposed when he cut through the cube.

Determine the newly exposed two-dimensional cross section. A Triangle B Parallelogram C Rectangle D Rhombus E Square

Question 6/11

PARCC Released Question - PBA - Calculator Section Topic: Intro to 3-D Solids

[This object is a pull tab]

Answer

C

Slide 301 (Answer) / 311

127 Part A - Question #2: Daniel wants to keep one chunk of clay for later use. To keep that chunk from drying out, he wants to place a piece of plastic sheeting

• n the surface he exposed when he cut through the cube. Find the

area of this newly exposed two-dimensional cross section. Round your answer to the nearest whole square inch.

Question 6/11

PARCC Released Question - PBA - Calculator Section Topic: Intro to 3-D Solids

Slide 302 / 311

127 Part A - Question #2: Daniel wants to keep one chunk of clay for later use. To keep that chunk from drying out, he wants to place a piece of plastic sheeting

• n the surface he exposed when he cut through the cube. Find the

area of this newly exposed two-dimensional cross section. Round your answer to the nearest whole square inch.

Question 6/11

PARCC Released Question - PBA - Calculator Section Topic: Intro to 3-D Solids

[This object is a pull tab]

Answer

Part A: Full credit

Note to Teachers: make sure that students are showing their work as they complete this question.

Slide 302 (Answer) / 311

128 Part B: Daniel wants to reshape the other chunk of clay to make a set of clay spheres. He wants each sphere to have a diameter of 4

• inches. Find the maximum number of spheres that Daniel can

make from the chunk of clay. Show your work.

Question 6/11

PARCC Released Question - PBA - Calculator Section Topic: Cavaleri's Principle

Slide 303 / 311

128 Part B: Daniel wants to reshape the other chunk of clay to make a set of clay spheres. He wants each sphere to have a diameter of 4

• inches. Find the maximum number of spheres that Daniel can

make from the chunk of clay. Show your work.

Question 6/11

PARCC Released Question - PBA - Calculator Section Topic: Cavaleri's Principle

[This object is a pull tab]

Answer

Part B: Full credit

Note to Teachers: make sure that students are showing their work as they complete this question.

Slide 303 (Answer) / 311

Question 10/11

The Farmer Supply is building a storage building for fertilizer that has a cylindrical base and a cone-shaped top. The county laws say that the storage building must have a maximum width of 8 feet and a maximum height of 14 feet. Topics: Volume of a Prism, Volume of a Cylinder, and Volume of a Cone PARCC Released Question - PBA - Calculator Section

Slide 304 / 311

129 Dump Trucks deliver fertilizer in loads that are 4 feet tall, 6 feet

wide & 12 feet long. Farmer Supply wants to be able to store 2 dump-truck loads of fertilizer. Determine the height of the cylinder, h1, and a height of the cone, h2, that Farmer Supply should use in the design. Show that your design will be able to store at least two dump-truck loads of

• fertilizer. When you finish, enter your value for h1 into your

Responder.

Question 10/11

Topics: Volume of a Prism, Volume of a Cylinder, and Volume of a Cone PARCC Released Question - PBA - Calculator Section - SMART Response Format

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129 Dump Trucks deliver fertilizer in loads that are 4 feet tall, 6 feet

wide & 12 feet long. Farmer Supply wants to be able to store 2 dump-truck loads of fertilizer. Determine the height of the cylinder, h1, and a height of the cone, h2, that Farmer Supply should use in the design. Show that your design will be able to store at least two dump-truck loads of

• fertilizer. When you finish, enter your value for h1 into your

Responder.

Question 10/11

Topics: Volume of a Prism, Volume of a Cylinder, and Volume of a Cone PARCC Released Question - PBA - Calculator Section - SMART Response Format

[This object is a pull tab]

Answer

Full Credit

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130 A rectangle will be rotated 360º about a line which contains the point of intersection of its diagonals and is parallel to a side. What three-dimensional shape will be created as a result of the rotation? A a cube B a rectangular prism C a cylinder D a sphere

Question 4/7

PARCC Released Question - EOY - Non-Calculator Section Topic: Intro to 3-D Solids

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130 A rectangle will be rotated 360º about a line which contains the point of intersection of its diagonals and is parallel to a side. What three-dimensional shape will be created as a result of the rotation? A a cube B a rectangular prism C a cylinder D a sphere

Question 4/7

PARCC Released Question - EOY - Non-Calculator Section Topic: Intro to 3-D Solids

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Answer C a cylinder

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131 The table shows the approximate measurements of the Great

Pyramid of Giza in Egypt and the Pyramid of Kukulcan in Mexico. Approximately, what is the difference between the volume of the Great Pyramid of Giza and the volume of the Pyramid of Kukulcan? A 1,945,000 cubic meters B 2,562,000 cubic meters C 5,835,000 cubic meters D 7,686,000 cubic meters PARCC Released Question - EOY - Calculator Section Topic: Volume of a Pyramid

Question 8/25

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131 The table shows the approximate measurements of the Great

Pyramid of Giza in Egypt and the Pyramid of Kukulcan in Mexico. Approximately, what is the difference between the volume of the Great Pyramid of Giza and the volume of the Pyramid of Kukulcan? A 1,945,000 cubic meters B 2,562,000 cubic meters C 5,835,000 cubic meters D 7,686,000 cubic meters PARCC Released Question - EOY - Calculator Section Topic: Volume of a Pyramid

Question 8/25

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Answer

B

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Question 11/25

Two cylinders each with a height of 50 inches are shown. Topic: Cavaleri's Principle PARCC Released Question - EOY - Calculator Section

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132 Which statements about cylinders P and S are true? Select all that apply. A If x = y, the volume of cylinder P is greater than the volume

• f cylinder S, because cylinder P is a right cylinder.

B If x = y, the volume of cylinder P is equal to the volume of cylinder S, because the cylindres are the same height. C If x = y, the volume of cylinder P is less than the volume of cylinder S, because cylinder S is slanted. D If x < y, the area of a horizontal cross section of cylinder P is greater than the area of a horizontal cross section of cylinder S. E If x < y, the area of a horizontal cross section of cylinder P is equal to the area of a horizontal cross section of cylinder S. F If x < y, the area of a horizontal cross section of cylinder P is less than the area of a hoizontal cross section of cylinder S.

Question 11/25

Topic: Cavaleri's Principle PARCC Released Question - EOY - Calculator Section

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132 Which statements about cylinders P and S are true? Select all that apply. A If x = y, the volume of cylinder P is greater than the volume

• f cylinder S, because cylinder P is a right cylinder.

B If x = y, the volume of cylinder P is equal to the volume of cylinder S, because the cylindres are the same height. C If x = y, the volume of cylinder P is less than the volume of cylinder S, because cylinder S is slanted. D If x < y, the area of a horizontal cross section of cylinder P is greater than the area of a horizontal cross section of cylinder S. E If x < y, the area of a horizontal cross section of cylinder P is equal to the area of a horizontal cross section of cylinder S. F If x < y, the area of a horizontal cross section of cylinder P is less than the area of a hoizontal cross section of cylinder S.

Question 11/25

Topic: Cavaleri's Principle PARCC Released Question - EOY - Calculator Section

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Answer

B and F

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133 Part A The outer surface of the pipe is coated with protective material. How many square feet is the outer surface of the pipe? Give your answer to the nearest integer. A steel pipe in the shape of a right circular cylinder is used for drainage under a road. The length of the pipe is 12 feet and its diameter is 36

• inches. The pipe is open at both ends.

Question 13/25

Topic: Surface Area of a Cylinder PARCC Released Question - EOY - Calculator Section

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133 Part A The outer surface of the pipe is coated with protective material. How many square feet is the outer surface of the pipe? Give your answer to the nearest integer. A steel pipe in the shape of a right circular cylinder is used for drainage under a road. The length of the pipe is 12 feet and its diameter is 36

• inches. The pipe is open at both ends.

Question 13/25

Topic: Surface Area of a Cylinder PARCC Released Question - EOY - Calculator Section

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Answer 113 square feet

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134 Part B A wire screen in the shape of a square is attached at one end of the pipe to allow water to flow through but to keep people from wandering into the pipe. The length of the diagonals of the screen are equal to the diameter of the pipe. The figure represents the placement of the screen at the end of the pipe. A 72 B 102 C 125 D 324 E 648 F 1,018 and the area of the screen is ________ square inches.

Question 13/25

Topic: Surface Area of a Cylinder PARCC Released Question - EOY - Calculator Section The perimeter of the screen is approximately ________ inches, Select from each set of answers to correctly complete the sentence.

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134 Part B A wire screen in the shape of a square is attached at one end of the pipe to allow water to flow through but to keep people from wandering into the pipe. The length of the diagonals of the screen are equal to the diameter of the pipe. The figure represents the placement of the screen at the end of the pipe. A 72 B 102 C 125 D 324 E 648 F 1,018 and the area of the screen is ________ square inches.

Question 13/25

Topic: Surface Area of a Cylinder PARCC Released Question - EOY - Calculator Section The perimeter of the screen is approximately ________ inches, Select from each set of answers to correctly complete the sentence.

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Answer B and E

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