Area of Rectangles MP6: Attend to precision. MP7: Look for & - - PDF document

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Slide 1 / 152 Slide 2 / 152 Geometry Area of Figures 2015-10-27 www.njctl.org Slide 3 / 152 Slide 4 / 152 Table of Contents Throughout this unit, the Standards for Mathematical Practice are used. Click on the topic to go to that section

SLIDE 1

Slide 1 / 152 Slide 2 / 152

Geometry

Area of Figures

2015-10-27 www.njctl.org

Slide 3 / 152

Click on the topic to go to that section

Area of Rectangles Area of Triangles Area of Parallelograms Area of Regular Polygons Area of Circles & Sectors Law of Sines PARCC Sample Questions Area of Other Quadrilaterals Area & Perimeter of Figures in the Coordinate Plane

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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.

Slide 4 (Answer) / 152

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

Slide 5 / 152

Area of Rectangles

Return to Table of Contents

SLIDE 2

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The area of a rectangle is defined to be the number of squares of area "1" that can fit within it. In the below drawing, 6 unit squares fit within the below rectangle of height 2 and based 3.

Area of a Rectangle

3 2

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In general, the area of a rectangle is equal to its base times its

height. This can also be referred to as its length times its width.

Arectangle = length x width (lw) = base x height (bh)

Area of a Rectangle

b h

Slide 8 / 152 Area of a Rectangle

Sometimes, the dimensions will not be given, so you will need to calculate them before calculating the area. Since a rectangle is a quadrilateral with 4 right angles, 2 right triangles can be formed when drawing one of its diagonals. Therefore, Pythagorean Theorem can be a helpful formula. Another helpful formula is the perimeter formula for a rectangle: P = 2l + 2w There might also be questions asking you about the population density

f a town, state, or country. It is the ratio that represents the number of

people living per square mile and can be found by dividing the total population by the total area.

Slide 9 / 152 Example

The diagonal of a rectangle is 34 feet and its length is 14 feet more than its width. Find the length, width, and area of the rectangle. 34 ft x x + 14 We know that the width is unknown, so let's call it "x". Therefore, the length will be "x + 14". Using the Pythagorean Theorem &

ur Algebra skills, we can solve for x.

x2 + (x + 14)2 = 342 x2 + x2 + 28x + 196 = 1,156 2x2 + 28x - 980 = 0 2(x2 + 14x - 480) = 0 2(x + 30)(x - 16) = 0 x + 30 = 0 or x - 16 = 0 x = -30 or x = 16 Since x is the length of a side, it has to be positive. Therefore,

ur final answer is

x = 16 ft = width length = 30 ft Area = 480 ft2

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Slide 9 (Answer) / 152 Example

The diagonal of a rectangle is 34 feet and its length is 14 feet more than its width. Find the length, width, and area of the rectangle. 34 ft x x + 14 We know that the width is unknown, so let's call it "x". Therefore, the length will be "x + 14". Using the Pythagorean Theorem &

ur Algebra skills, we can solve for x.

x2 + (x + 14)2 = 342 x2 + x2 + 28x + 196 = 1,156 2x2 + 28x - 980 = 0 2(x2 + 14x - 480) = 0 2(x + 30)(x - 16) = 0 x + 30 = 0 or x - 16 = 0 x = -30 or x = 16 Since x is the length of a side, it has to be positive. Therefore,

ur final answer is

x = 16 ft = width length = 30 ft Area = 480 ft2

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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)

What labels (or units of measurement) should you use? (MP6) Does anyone have the same answer, but a different way to explain it? (MP7) What concepts that we have learned before were useful in solving this? (MP8)

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1 What is the area of a rectangle that has a length of 8.4 cm and a width of 3.7 cm?

SLIDE 3

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1 What is the area of a rectangle that has a length of 8.4 cm and a width of 3.7 cm?

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Answer

A = 31.08 cm2

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2 Televisions, are advertised using the length of the

diagonal. For example, a 26" TV could have a length of

24" and a width of 10", as shown below. What is the area of an 80" TV if the length is 69.3"?

26" 24" 10"

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2 Televisions, are advertised using the length of the

diagonal. For example, a 26" TV could have a length of

24" and a width of 10", as shown below. What is the area of an 80" TV if the length is 69.3"?

26" 24" 10"

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Answer

w = 39.97 in A = 2,769.92 in2

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3 The population density is the amount of people living per square mile. If the town of Geometryville is a rectangular town that has a length of 24 miles and a width of 13 miles, and its population is 2,500 people, what is the population density of the town?

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3 The population density is the amount of people living per square mile. If the town of Geometryville is a rectangular town that has a length of 24 miles and a width of 13 miles, and its population is 2,500 people, what is the population density of the town?

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Answer

8 people per square mile

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4 The diagonal of a rectangle is 10 feet and its width is 2 feet less than its length. What is the length of the rectangle? A 4 feet B 6 feet C 8 feet D 9 feet

SLIDE 4

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4 The diagonal of a rectangle is 10 feet and its width is 2 feet less than its length. What is the length of the rectangle? A 4 feet B 6 feet C 8 feet D 9 feet

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Answer

C Slide 14 / 152

5 The diagonal of a rectangle is 10 feet and its width is 2 feet less than its length. What is the width of the rectangle? A 4 feet B 6 feet C 8 feet D 9 feet

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5 The diagonal of a rectangle is 10 feet and its width is 2 feet less than its length. What is the width of the rectangle? A 4 feet B 6 feet C 8 feet D 9 feet

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Answer

B Slide 15 / 152

6 The diagonal of a rectangle is 10 feet and its width is 2 feet less than its length. What is the area of the rectangle? A 80 ft2 B 60 ft2 C 48 ft2 D 24 ft2

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6 The diagonal of a rectangle is 10 feet and its width is 2 feet less than its length. What is the area of the rectangle? A 80 ft2 B 60 ft2 C 48 ft2 D 24 ft2

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Answer

C Slide 16 / 152

Area of Triangles

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SLIDE 5

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Clearly, the area of each of the below right triangles is equal to half the area of the rectangle they comprise. Since the area of the rectangle is bh, the area of each right triangle is 1/2 bh.

Area of a Right Triangle

b h AΔ = 1/2 bh

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Now, let's find the area formula for an arbitrary scalene triangle ABC with base "b" and height "h." Keep in mind that the height is always measured perpendicular to the base that is opposite the vertex.

Area of Any Triangle

b h A B C

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Let's draw right triangle ACD such that the new triangle plus the

riginal triangle form a larger right triangle ABD.

ΔADC is a right Δ with base "b' " and height "h". ΔADB is a right Δ with base "b' + b" and height "h".

Area of Any Triangle

b h b' A B C D

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AreaΔADC = 1/2 b'h AreaΔADB = 1/2 (b' + b) h AreaΔABC = Area

ΔADB - Area ΔADC

AreaΔABC = 1/2 (b' + b) h - 1/2 b'h AreaΔABC = 1/2 b'h + 1/2bh - 1/2b'h AreaΔABC = 1/2 bh

Area of Any Triangle

b h A B C D b'

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AreaΔADC = 1/2 b'h AreaΔADB = 1/2 (b' + b) h AreaΔABC = Area

ΔADB - Area ΔADC

AreaΔABC = 1/2 (b' + b) h - 1/2 b'h AreaΔABC = 1/2 b'h + 1/2bh - 1/2b'h AreaΔABC = 1/2 bh

Area of Any Triangle

b h A B C D b'

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Math Practice Questioning to help address MP standards: How can we prove that the area of a triangle is equal to 1/2 bh for all triangles? (MP3) 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)

What concepts that you have learned about before were useful in writing this algebraic proof? (MP8)

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So, the area of any triangle, not just right triangles, is given by : AreaΔ = 1/2 bh

Area of Any Triangle

b h A B C

SLIDE 6

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While the formula is the same for any triangle: AreaΔ = 1/2 bh For triangles that are not right triangles, the height is usually not directly given. Instead, you may be given the lengths of one or more sides and the measures of one or more angles.

Area of Any Triangle

4 A B C 5 8 For instance, in this case how would you find the height...and then the area? 40°

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While the formula is the same for any triangle: AreaΔ = 1/2 bh For triangles that are not right triangles, the height is usually not directly given. Instead, you may be given the lengths of one or more sides and the measures of one or more angles.

Area of Any Triangle

4 A B C 5 8 For instance, in this case how would you find the height...and then the area? 40°

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Math Practice

Question to the left of the triangle addresses MP1, MP3, and MP7.

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Draw an altitude from the vertex to the base, creating a right triangle with the same height as our original triangle. ΔABD has the same height as our original triangle ΔABC. And both triangles share a side, AB, and an angle, ∠B. Which trig function would allow us to find the value of "h"?

Area of Any Triangle

h 4 A B C 5 8 D 40°

Slide 23 (Answer) / 152

Draw an altitude from the vertex to the base, creating a right triangle with the same height as our original triangle. ΔABD has the same height as our original triangle ΔABC. And both triangles share a side, AB, and an angle, ∠B. Which trig function would allow us to find the value of "h"?

Area of Any Triangle

h 4 A B C 5 8 D 40°

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Math Practice

Question above the triangle addresses MP7.

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h 4 A B C 5 8 D 40° sin θ = opposite / hypotenuse = opp / hyp

pp = (hyp)(sin θ)

h = (8)(sin 40°) h = (8)(0.64) h = 5.1 units

Area of Any Triangle

Now that you have the base and the height of ΔABC, how would you find the area?

click click click click click

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h 4 A B C 5 8 D 40° sin θ = opposite / hypotenuse = opp / hyp

pp = (hyp)(sin θ)

h = (8)(sin 40°) h = (8)(0.64) h = 5.1 units

Area of Any Triangle

Now that you have the base and the height of ΔABC, how would you find the area?

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

Math Practice Question to the left of the triangle addresses MP1. Questioning to help address MP standards: 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)

What concepts that you have learned about before were useful? (MP8)

SLIDE 7

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A = 1/2 bh A = 1/2 (4)(8)(sin 40°) A = 1/2 (4)(8)(0.64) A = 10.2 units

Area of Any Triangle

h 4 A B C 5 8 40° D

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A = 1/2 bh A = 1/2 (4)(8)(sin 40°) A = 1/2 (4)(8)(0.64) A = 10.2 units

Area of Any Triangle

h 4 A B C 5 8 40° D Examining our solution, we can see that the area of the triangle is given by half the product of two sides multiplied by the sine of the angle between them. In this case, the sides were of length 4 and 8 and the angle is 40°.

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AΔ = 1/2 ac sinB

Area of Any Triangle

a A B C b c Replacing the numbers by labeling the angles with upper case letters and the sides opposite them with matching lower case letters yields this formula. But, there's nothing special about those sides and that angle, so it will also be true that: AΔ = 1/2 ab sinC Or AΔ = 1/2 bc sinA The area of a triangle is equal to the product of any two sides and the sine

f the included angle.

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Find the area of ΔABC. In this case, the altitude you draw will be within the triangle.

Example

4 A B C 9 8 50°

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Find the area of ΔABC. In this case, the altitude you draw will be within the triangle.

Example

4 A B C 9 8 50°

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Answer 4 A B C 9 8

50°

h sin 50° = h 4 h = 4sin50° A = 1/2(9)(4sin50°) A = 13.79 units2

Questioning to help address MP standards: How could you start this problem? (MP1) How can you represent the problem with numbers and symbols? (MP2) Create an equation to represent the

problem. (MP2)

What do you know about trigonometry that could apply to this situation? (MP7)

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7 Find the area of ΔDEF. Round your answer to the nearest hundredth.

34° 8 in. 12 in. D E F

SLIDE 8

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7 Find the area of ΔDEF. Round your answer to the nearest hundredth.

34° 8 in. 12 in. D E F

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Answer

A = 26.84 in2

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8 Find the area of ΔGHI. Round your answer to the nearest hundredth.

28° 16 in. 9 in. G H I

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8 Find the area of ΔGHI. Round your answer to the nearest hundredth.

28° 16 in. 9 in. G H I

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Answer

A = 33.80 in2

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9 Find the area of ΔJKL. Round your answer to the nearest hundredth.

32° 6 in. 15 in. J K L

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9 Find the area of ΔJKL. Round your answer to the nearest hundredth.

32° 6 in. 15 in. J K L

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Answer

A = 23.85 in2

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10 Find the area of ΔPQR. Round your answer to the nearest hundredth.

52° 9 in. 14 in. P Q R

SLIDE 9

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10 Find the area of ΔPQR. Round your answer to the nearest hundredth.

52° 9 in. 14 in. P Q R

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Answer

A = 49.64 in2

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Law of Sines

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We just learned that we can find the area of any triangle with any of these three formulas: AΔ = 1/2 ac sinB AΔ = 1/2 ab sinC AΔ = 1/2 bc sinA Since the area of a triangle will be the same regardless of which formula we use, these three formulas must be equal. Setting them equal to

ne another will give us

a general relationship between the sides and the angles of a triangle.

Law of Sines

a A B C b c

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AΔ = 1/2 ac sinB = 1/2 ab sinC = 1/2 bc sinA Let's look at one pair at a time and simplify. 1/2 ac sinB = 1/2 ab sinC AND 1/2 ab sinC = 1/2 bc sinA ac sinB = ab sinC ab sinC = bc sinA c sinB = b sinC a sinC = c sinA c / sinC = b / sinB a / sinA = c / sinC

Law of Sines

= = a b c sinA sinB sinC

click click click click click click click

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AΔ = 1/2 ac sinB = 1/2 ab sinC = 1/2 bc sinA Let's look at one pair at a time and simplify. 1/2 ac sinB = 1/2 ab sinC AND 1/2 ab sinC = 1/2 bc sinA ac sinB = ab sinC ab sinC = bc sinA c sinB = b sinC a sinC = c sinA c / sinC = b / sinB a / sinA = c / sinC

Law of Sines

= = a b c sinA sinB sinC

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

Math Practice

Questioning to help address MP standards: What concepts that we have learned before are useful in simplify each algebraic equation? (MP8) What strategies are you going to use to get the same letters on the same side of the equation? (MP1)

Slide 36 / 152 Law of Sines

This relationship between the sides and angles of a triangle will be true for all triangles. a A B C b c = = a b c sinA sinB sinC = = a b c sinA sinB sinC OR

SLIDE 10

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Find the missing segment lengths and angle measures in ΔABC if m∠B = 28°, m∠C = 103°, and b = 26 cm.

Example

A B C 28° 103° 26 cm c a Since we have a triangle, we know that the sum of the interior angles is 180°. Therefore, we can find the m∠A using the Triangle Sum Theorem. m∠A = 180 - 103 - 28 = 49°

click click

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Find the missing segment lengths and angle measures in ΔABC if m∠B = 28°, m∠C = 103°, and b = 26 cm.

Example

A B C 28° 103° 26 cm c a Since we have a triangle, we know that the sum of the interior angles is 180°. Therefore, we can find the m∠A using the Triangle Sum Theorem. m∠A = 180 - 103 - 28 = 49°

click click [This object is a pull tab]

Math Practice

Filling in the blanks on this slide accomplish MP1. Alternative questions: What information do you have? (MP1) What strategies are you going to use? (MP1) Can you do this mentally? (MP1 & MP5)

Slide 38 / 152

Find the missing segment lengths and angle measures in ΔABC if m∠B = 28°, m∠C = 103°, and b = 26 cm.

Example

A B C 28° 103° 26 cm c a Now that we have the measurements of all

f the angles and one side length given, we

can find the remaining sides using the Law

f Sines.

26 sin28° a sin49° 26sin49° = asin28° a = a = 41.80 cm = 26sin49° sin28° 26 sin28° c sin103° 26sin103° = csin28° c = c = 53.96 cm = 26sin103° sin28°

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Slide 38 (Answer) / 152

Find the missing segment lengths and angle measures in ΔABC if m∠B = 28°, m∠C = 103°, and b = 26 cm.

Example

A B C 28° 103° 26 cm c a Now that we have the measurements of all

f the angles and one side length given, we

can find the remaining sides using the Law

f Sines.

26 sin28° a sin49° 26sin49° = asin28° a = a = 41.80 cm = 26sin49° sin28° 26 sin28° c sin103° 26sin103° = csin28° c = c = 53.96 cm = 26sin103° sin28°

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

Math Practice

Questioning to address the MP standards: How could you start this problem? (MP1) How can you represent the problem with numbers and symbols? (MP2) Create a representation of the problem. (MP2) What labels could you use? (MP6)

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( )

Find the missing segment lengths and angle measures in ΔABC if m∠C = 122°, a = 12 cm and c = 18 cm.

Example

A B C 122° b 18 cm 12 cm In this problem, we need to determine one

f the missing angles first. However, we

can't use the Triangle Sum Theorem yet, since we only know the measurement of 1

angle. Therefore, we need to use the Law
f Sines.

18 sin122° 12 sinA 18sinA = 12sin122° sinA = m∠A = sin-1 m∠A = 34.43° = 12sin122° 18 12sin122° 18

click to reveal

Slide 39 (Answer) / 152

( )

Find the missing segment lengths and angle measures in ΔABC if m∠C = 122°, a = 12 cm and c = 18 cm.

Example

A B C 122° b 18 cm 12 cm In this problem, we need to determine one

f the missing angles first. However, we

can't use the Triangle Sum Theorem yet, since we only know the measurement of 1

angle. Therefore, we need to use the Law
f Sines.

18 sin122° 12 sinA 18sinA = 12sin122° sinA = m∠A = sin-1 m∠A = 34.43° = 12sin122° 18 12sin122° 18

click to reveal [This object is a pull tab]

Math Practice

Questioning to address the MP standards: How could you start this problem? (MP1) How can you represent the problem with numbers and symbols? (MP2) Create a representation of the problem. (MP2) What labels could you use? (MP6)

SLIDE 11

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Find the missing segment lengths and angle measures in ΔABC if m∠C = 122°, a = 12 cm and c = 18 cm.

Example

A B C 122° b 18 cm 12 cm Now, that we know 2 angle measures, we can calculate the measurement of our final missing angle. m∠B = 180 - 122 - 34.43 = 23.57° Finally, we can calculate the length of our final side. 18 sin122° b sin23.57° 18sin23.57° = bsin122° b = b = 8.49 cm = 18sin23.57° sin122°

click click to reveal

Slide 40 (Answer) / 152

Find the missing segment lengths and angle measures in ΔABC if m∠C = 122°, a = 12 cm and c = 18 cm.

Example

A B C 122° b 18 cm 12 cm Now, that we know 2 angle measures, we can calculate the measurement of our final missing angle. m∠B = 180 - 122 - 34.43 = 23.57° Finally, we can calculate the length of our final side. 18 sin122° b sin23.57° 18sin23.57° = bsin122° b = b = 8.49 cm = 18sin23.57° sin122°

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

Math Practice Questioning to address the MP standards (finding the value of b): How could you start this problem? (MP1) How can you represent the problem with numbers and symbols? (MP2) Create a representation of the problem. (MP2) What labels could you use? (MP6) Filling in the blank on this slide accomplishes MP1. Alternative questions: What information do you have? (MP1) What strategies are you going to use? (MP1) Can you do this mentally? (MP1 & MP5)

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11 Find the missing segment lengths and angle measures in ΔABC if m∠A = 70°, m∠B = 64°, and c = 5 yd. What is m∠C? A 36° B 46° C 56° D 136°

A B C 70° b 5 yd a 64°

Slide 41 (Answer) / 152

11 Find the missing segment lengths and angle measures in ΔABC if m∠A = 70°, m∠B = 64°, and c = 5 yd. What is m∠C? A 36° B 46° C 56° D 136°

A B C 70° b 5 yd a 64°

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Answer

B Slide 42 / 152

12 Find the missing segment lengths and angle measures in ΔABC if m∠A = 70°, m∠B = 64°, and c = 5 yd. What is the value of a? Round your answer to the nearest hundredth.

A B C 70° b 5 yd a 64°

Slide 42 (Answer) / 152

12 Find the missing segment lengths and angle measures in ΔABC if m∠A = 70°, m∠B = 64°, and c = 5 yd. What is the value of a? Round your answer to the nearest hundredth.

A B C 70° b 5 yd a 64°

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Answer

a = 6.53 yd

SLIDE 12

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13 Find the missing segment lengths and angle measures in ΔABC if m∠A = 70°, m∠B = 64°, and c = 5 yd. What is the value of b? Round your answer to the nearest hundredth.

A B C 70° b 5 yd a 64°

Slide 43 (Answer) / 152

13 Find the missing segment lengths and angle measures in ΔABC if m∠A = 70°, m∠B = 64°, and c = 5 yd. What is the value of b? Round your answer to the nearest hundredth.

A B C 70° b 5 yd a 64°

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Answer

b = 6.25 yd

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14 Find the missing segment lengths and angle measures in ΔABC if m∠C = 111°, b = 3 in., and c = 5 in. What is m∠B? Round your answer to the nearest hundredth.

A B C 111° a 5 in. 3 in.

Slide 44 (Answer) / 152

14 Find the missing segment lengths and angle measures in ΔABC if m∠C = 111°, b = 3 in., and c = 5 in. What is m∠B? Round your answer to the nearest hundredth.

A B C 111° a 5 in. 3 in.

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Answer

m∠B = 34.07°

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15 Find the missing segment lengths and angle measures in ΔABC if m∠C = 111°, b = 3 in., and c = 5 in. What is m∠A? Round your answer to the nearest hundredth. A 124.93° B 44.93° C 34.93° D 24.93°

A B C 111° a 5 in. 3 in.

Slide 45 (Answer) / 152

15 Find the missing segment lengths and angle measures in ΔABC if m∠C = 111°, b = 3 in., and c = 5 in. What is m∠A? Round your answer to the nearest hundredth. A 124.93° B 44.93° C 34.93° D 24.93°

A B C 111° a 5 in. 3 in.

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Answer

C

SLIDE 13

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16 Find the missing segment lengths and angle measures in ΔABC if m∠C = 111°, b = 3 in., and c = 5 in. What is the value of a? Round your answer to the nearest hundredth.

A B C 111° a 5 in. 3 in.

Slide 46 (Answer) / 152

16 Find the missing segment lengths and angle measures in ΔABC if m∠C = 111°, b = 3 in., and c = 5 in. What is the value of a? Round your answer to the nearest hundredth.

A B C 111° a 5 in. 3 in.

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Answer

a = 3.07 in.

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Area of Parallelograms

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Slide 48 / 152

From these results, we can find the area formula for any parallelogram by dividing the parallelogram into two triangles.

Area of a Parallelogram

b h

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We can see below that a parallelogram is comprised of two triangles. Since the area of each triangle is 1/2 bh. The area of a parallelogram is bh.

Area of a Parallelogram

b h Aparallelogram = bh

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As with triangles, the height of a parallelogram is often not given. The height must be found by drawing an altitude and using trigonometry to find the height of the parallelogram.

Area of a Parallelogram

b h Aparallelogram = bh

SLIDE 14

Slide 51 / 152

5 8 110° 20° 5 Find the area of ABCD. This time, use the cosine rather than the sine function, just for

practice. Either can be used depending on the angle you choose.

Example

To use cosine, our 1st step will be to drop down the altitude, or height. Since we know that the entire obtuse angle is 110° and the right angle measures 90°, the remaining acute angle is 20°

Slide 52 / 152

Find the area of ABCD. This time, use the cosine rather than the sine function, just for

practice. Either can be used depending on the angle you choose.

Example

5 8 110° 20° 5 Now, let's find the height of

ur parallelogram.

cos20° = h 5 h = 5cos20° h = 4.70 units And its area A = 8(4.70) A = 37.60 units2

click click

Slide 52 (Answer) / 152

Find the area of ABCD. This time, use the cosine rather than the sine function, just for

practice. Either can be used depending on the angle you choose.

Example

5 8 110° 20° 5 Now, let's find the height of

ur parallelogram.

cos20° = h 5 h = 5cos20° h = 4.70 units And its area A = 8(4.70) A = 37.60 units2

click click [This object is a pull tab]

Math Practice

Questioning to address the MP standards: How could you start this problem? (MP1) How can you represent the problem with numbers and symbols? (MP2) Create a representation of the problem. (MP2) What labels could you use? (MP6)

Slide 53 / 152

17 A diagonal parking space creates a parallelogram. Its length is 20.5 feet, its distance along the curb is 9 feet 4 inches, and the acute angle that is made with the curb is 75°. Find the height of the parking space. A 2.4 ft B 9 ft C 9.7 ft D 34.8 ft

20.5 ft 9 ft 4 in. 9 ft 4 in. 75° 75°

Slide 53 (Answer) / 152

17 A diagonal parking space creates a parallelogram. Its length is 20.5 feet, its distance along the curb is 9 feet 4 inches, and the acute angle that is made with the curb is 75°. Find the height of the parking space. A 2.4 ft B 9 ft C 9.7 ft D 34.8 ft

20.5 ft 9 ft 4 in. 9 ft 4 in. 75° 75°

[This object is a pull tab]

Answer

B Slide 54 / 152

18 A diagonal parking space creates a parallelogram. Its length is 20.5 feet, its distance along the curb is 9 feet 4 inches, and the acute angle that is made with the curb is 75°. Find the area of the parking space. A 184.5 ft2 B 191.3 ft2 C 198.85 ft2 D 324.8 ft2

20.5 ft 9 ft 4 in. 9 ft 4 in. 75° 75°

SLIDE 15

Slide 54 (Answer) / 152

18 A diagonal parking space creates a parallelogram. Its length is 20.5 feet, its distance along the curb is 9 feet 4 inches, and the acute angle that is made with the curb is 75°. Find the area of the parking space. A 184.5 ft2 B 191.3 ft2 C 198.85 ft2 D 324.8 ft2

20.5 ft 9 ft 4 in. 9 ft 4 in. 75° 75°

[This object is a pull tab]

Answer

A Slide 55 / 152

19 A window frame is in the shape of a parallelogram. Its base is 3 feet long, its other side length is 2.5 feet long, and the obtuse angle created between two of the sides is 114°. How much glass would be required to fill the window? Round your answer to the nearest hundredth.

Slide 55 (Answer) / 152

19 A window frame is in the shape of a parallelogram. Its base is 3 feet long, its other side length is 2.5 feet long, and the obtuse angle created between two of the sides is 114°. How much glass would be required to fill the window? Round your answer to the nearest hundredth.

[This object is a pull tab]

Answer

6.85 ft2

Slide 56 / 152

20 Mrs. Polygon is creating a quilt for her grandson. The quilt is created by stitching together parallelograms that are different colors. Each parallelogram is 4 inches long its other side is 3 inches long, and the acute angle created between the two sides is 32°. What is the area

f each parallelogram used to make the quilt? Round

your answer to the nearest hundredth.

Slide 56 (Answer) / 152

20 Mrs. Polygon is creating a quilt for her grandson. The quilt is created by stitching together parallelograms that are different colors. Each parallelogram is 4 inches long its other side is 3 inches long, and the acute angle created between the two sides is 32°. What is the area

f each parallelogram used to make the quilt? Round

your answer to the nearest hundredth.

[This object is a pull tab]

Answer

6.36 in2

Slide 57 / 152

21 If the quilt requires 5 parallelograms horizontally and 10 parallelograms vertically, how much material will Mrs. Polygon need to make the quilt?

SLIDE 16

Slide 57 (Answer) / 152

21 If the quilt requires 5 parallelograms horizontally and 10 parallelograms vertically, how much material will Mrs. Polygon need to make the quilt?

[This object is a pull tab]

Answer

318 in2

Slide 58 / 152

Area of Regular Polygons

Return to Table of Contents

Slide 59 / 152

A regular polygon is a polygon that has all its sides and angles congruent: it is both equilateral & equiangular.

Area of Regular Polygons Slide 60 / 152

r C

Circumscribing a Regular Polygon

A polygon is circumscribed by drawing around it the smallest circle

n which lie all the vertices of the polygon.

That circle is called the circumcircle of the polygon.

Slide 61 / 152

r C The radius of a regular polygon is the radius of the circumscribed circle, shown here as "r." This is called the circumradius.

Circumcenter and Circumradius

The center of a regular polygon is the center of its circumscribed circle, shown here as "C." This is called the circumcenter of the polygon.

Slide 62 / 152 Central Angle of a Regular Polygon

r C r The degrees of the central angle can be found using the formula Central angle = where n is the number of sides in the regular polygon. 360 n A central angle of a regular polygon is an angle with one vertex at the circumcenter and two vertices on the circumcircle. The sides of the central angle are radii of the circle.

SLIDE 17

Slide 63 / 152

r C r a

The Apothem a Regular Polygon

The apothem of a Regular Polygon is the shortest distance from the center of the polygon to one of its sides. An apothem is perpendicular to a side of the polygon. An apothem is also the altitude (height) of the isosceles triangle formed by the sides of a central angle and the side of the polygon that is opposite the central angle.

Slide 64 / 152

A polygon of n sides is comprised of n triangles. Each triangle has a height equal to the apothem, "a." The base of each triangle is equal to the perimeter, P, divided by the number of sides, n: b = P/n.

Area of Regular Polygons

What is the area of a triangle whose base is P/n and whose height is a? r a

Slide 64 (Answer) / 152

A polygon of n sides is comprised of n triangles. Each triangle has a height equal to the apothem, "a." The base of each triangle is equal to the perimeter, P, divided by the number of sides, n: b = P/n.

Area of Regular Polygons

What is the area of a triangle whose base is P/n and whose height is a? r a

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Math Practice

Question next to the figure addresses MP2 and MP4.

Slide 65 / 152

AΔ = 1/2 bh AΔ = 1/2(P/n)a There are n triangles in a regular polygon, so the area

f the polygon is given by:

A = nA

A = (n)(1/2)(P/n)a A = 1/2Pa

Area of Regular Polygons

r r r r a a a a

click click click click

Slide 65 (Answer) / 152

AΔ = 1/2 bh AΔ = 1/2(P/n)a There are n triangles in a regular polygon, so the area

f the polygon is given by:

A = nA

A = (n)(1/2)(P/n)a A = 1/2Pa

Area of Regular Polygons

r r r r a a a a

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

Math Practice

Questions to address the MP standards: What is this problem asking? (MP1) How could you start this problem? (MP1) How can you represent the problem with numbers and symbols? (MP2) Create a representation of the problem. (MP2) Write an algebraic representation to describe this situation. (MP4) What concepts that we have learned before were useful in solving this? (MP8)

Slide 66 / 152

The formula for the area of a polygon is simple: A = 1/2Pa But, often we are not given both P and a. We have to find one or both of them using trigonometry.

Area of Regular Polygons

r r r r a a a a

SLIDE 18

Slide 67 / 152

Let's find the area of a regular pentagon whose sides have a length of 7.

Example Slide 68 / 152 Example

A = 1/2 Pa For a pentagon, n = 5, so it's perimeter will be 5 times the length of a side. In this case, P = (5)(7) = 35. But, how do we find the apothem? Let's look at one of the five triangles that comprise the pentagon. 7 Let's find the area of a regular pentagon whose sides have a length of 7.

Slide 69 / 152 Example

A = 1/2 Pa For a pentagon, n = 5, so it's perimeter will be 5 times the length of a side. In this case, P = (5)(7) = 35. But, how do we find the apothem? Let's look at one of the five triangles that comprise the pentagon. 7 Let's find the area of a regular pentagon whose sides have a length of 7.

Slide 70 / 152 Example

The central angle is given by: m∠C = 360/n = 360/5 = 72° Since the ∠s must add to 180° and the measures of the base ∠s

f an isosceles triangle are equal,

those base ∠s must = 54°. The apothem, a, is the altitude of this triangle. 7 72° 54° 54° r r 7 C Let's find the area of a regular pentagon whose sides have a length of 7.

Slide 71 / 152

7 36° 54° 54° r r 3.5 a

Example

The two legs of the right triangle shown are a and 3.5. Tangent θ = opposite / adjacent tan θ = opp / adj

pp = adj (tan θ)

a = 3.5 (tan 54°) a = 3.5 (1.38) a = 4.8 units Let's find the area of a regular pentagon whose sides have a length of 7.

Slide 72 / 152 Example

We now know that P = 35 units and a = 4.8 units. A = 1/2 Pa A = 1/2 (35)(4.8) A = 84 units

7 72° 54° 54° r r 7 Let's find the area of a regular pentagon whose sides have a length of 7.

SLIDE 19

Slide 73 / 152

Find the area of a regular octagon whose sides have a length of 8.

Example

8 What is the question asking? To find the area of the octagon How can you represent the problem with symbols and numbers? A = 1/2 Pa How could you start this problem? Is there anything that you can find right away? P = 8(8) = 64 units

click click click

Slide 73 (Answer) / 152

Find the area of a regular octagon whose sides have a length of 8.

Example

8 What is the question asking? To find the area of the octagon How can you represent the problem with symbols and numbers? A = 1/2 Pa How could you start this problem? Is there anything that you can find right away? P = 8(8) = 64 units

click click click [This object is a pull tab]

Math Practice

Questions 1 & 3 address MP1. Question 2 addresses MP2.

Slide 74 / 152

Find the area of a regular octagon whose sides have a length of 8.

Example

8 How do we find the apothem? How could you use a drawing to show your way of thinking? Create an equation to find the apothem. 8

45°

4 4

22.5° 67.5° 67.5° 67.5°

a tan 67.5° = a 4 a = 4 tan 67.5° a = 9.66 units

click click

Slide 74 (Answer) / 152

Find the area of a regular octagon whose sides have a length of 8.

Example

8 How do we find the apothem? How could you use a drawing to show your way of thinking? Create an equation to find the apothem. 8

45°

4 4

22.5° 67.5° 67.5° 67.5°

a tan 67.5° = a 4 a = 4 tan 67.5° a = 9.66 units

click click [This object is a pull tab]

Math Practice

The 1st set of questions address MP5. "Create an equation..." addresses MP2. Additional Q's to address Math Practices: Would it help to draw a picture? (MP4 & MP5) What tools do you need? (MP5) What labels could you use? (MP6)

Slide 75 / 152

Find the area of a regular octagon whose sides have a length of 8.

Example

8 Now that we know all of the needed measurements, find the area of the regular octagon. a = 9.66 units P = 8(8) = 64 units A = 1/2 (9.66)(64) A = 309.12 units2

click

Slide 75 (Answer) / 152

Find the area of a regular octagon whose sides have a length of 8.

Example

8 Now that we know all of the needed measurements, find the area of the regular octagon. a = 9.66 units P = 8(8) = 64 units A = 1/2 (9.66)(64) A = 309.12 units2

click [This object is a pull tab]

Math Practice

Additional Q's to address Math Practices: What information do you have? (MP1) What do you need to find out? (MP1) What tools do you need? (MP5) What labels could you use? (MP6)

SLIDE 20

Slide 76 / 152

22 Calculate the apothem of the regular polygon shown in the figure below.

16

Slide 76 (Answer) / 152

22 Calculate the apothem of the regular polygon shown in the figure below.

16

[This object is a pull tab]

Answer

m∠C = 72° θ = 54° a = 16 sin(54) a = 12.94 units

Slide 77 / 152

23 Calculate the side length of the regular polygon shown in the figure below.

16

Slide 77 (Answer) / 152

23 Calculate the side length of the regular polygon shown in the figure below.

16

[This object is a pull tab]

Answer

m∠C = 72° θ = 54° b = 16 cos(54) b = 9.40 units s = 18.80 units

Slide 78 / 152

24 Calculate the perimeter & area of the regular polygon shown in the figure below.

16

Slide 78 (Answer) / 152

24 Calculate the perimeter & area of the regular polygon shown in the figure below.

16

[This object is a pull tab]

Answer

a = 12.94 units s = 18.80 units P = 5(18.80) P = 94 units A = 1/2 (12.94)(94) A = 608.18 units2

SLIDE 21

Slide 79 / 152

25 Calculate the side length of the regular polygon shown in the figure below.

15

Slide 79 (Answer) / 152

25 Calculate the side length of the regular polygon shown in the figure below.

15

[This object is a pull tab]

Answer

m∠C = 18° θ = 81° b = 15/tan(81) b = 2.38 units s = 4.76 units

Slide 80 / 152

26 Calculate the perimeter & area of the regular polygon shown in the figure below.

15

Slide 80 (Answer) / 152

26 Calculate the perimeter & area of the regular polygon shown in the figure below.

15

[This object is a pull tab]

Answer

s = 4.76 units P = 10(4.76) P = 47.6 units A = 1/2 (15)(47.6) A = 357 units

Slide 81 / 152

27 Calculate the apothem of the regular polygon shown in the figure below.

7

Slide 81 (Answer) / 152

27 Calculate the apothem of the regular polygon shown in the figure below.

7

[This object is a pull tab]

Answer

m∠C = 60° θ = 60° a = 3.5tan(60) a = 6.06 units Note: 30-60-90 ∆, so a = 3.5√3 = 6.06

SLIDE 22

Slide 82 / 152

28 Calculate the perimeter of the regular polygon shown in the figure below.

7

Slide 82 (Answer) / 152

28 Calculate the perimeter of the regular polygon shown in the figure below.

7

[This object is a pull tab]

Answer

P = 6(7) P = 42 units

Slide 83 / 152

29 Calculate the area of the regular polygon shown below.

7

Slide 83 (Answer) / 152

29 Calculate the area of the regular polygon shown below.

7

[This object is a pull tab]

Answer

a = 6.06 = 3.5√3 P = 6(7) P = 42 units A = 1/2 (6.06)(42) A = 127.26 units2 A = 73.5√3 units2

Slide 84 / 152

Area of Circles and Sectors

Return to Table of Contents

Slide 85 / 152

Interestingly, the formula, A = 1/2Pa, also leads to the formula for the area of a circle. If you let n go to infinity, the regular polygon approaches the shape of a circle. The apothem, a, approaches the radius of the circle, r. And, the perimeter of the polygon approaches the circumference of the circle: 2πr. Then, A = 1/2Pa approaches A = 1/2(2πr)(r) A = πr2

Area of a Circle

r a

SLIDE 23

Slide 86 / 152

30 Find the area of a circle that has a radius of 8 in. A 4π in2 B 8π in2 C 16π in2 D 64π in2

Slide 86 (Answer) / 152

30 Find the area of a circle that has a radius of 8 in. A 4π in2 B 8π in2 C 16π in2 D 64π in2

[This object is a pull tab]

Answer

D A = 64 π in2

Slide 87 / 152

31 Find the area of a circle that has a diameter of 17 in. A 8.5π in2 B 17π in2 C 72.25π in2 D 289π in2

Slide 87 (Answer) / 152

31 Find the area of a circle that has a diameter of 17 in. A 8.5π in2 B 17π in2 C 72.25π in2 D 289π in2

[This object is a pull tab]

Answer

C A = 72.25 π in2

Slide 88 / 152

32 Find the area of a circle that has a circumference of 18π in. A 324π in2 B 81π in2 C 18π in2 D 9π in2

Slide 88 (Answer) / 152

32 Find the area of a circle that has a circumference of 18π in. A 324π in2 B 81π in2 C 18π in2 D 9π in2

[This object is a pull tab]

Answer

B A = 81 π in2

SLIDE 24

Slide 89 / 152

A sector of a circle is the portion of the circle enclosed by two radii and the arc that connects them.

Minor Sector Major Sector

A B C

Sectors of Circles Slide 90 / 152

33 Which arc borders the minor sector? A Arc AB B Arc AC C Arc ADB A B C D

Slide 90 (Answer) / 152

33 Which arc borders the minor sector? A Arc AB B Arc AC C Arc ADB A B C D

[This object is a pull tab]

Answer

A Slide 91 / 152

34 Which arc borders the major sector? A Arc AB B Arc AC C Arc ADB A B C D

Slide 91 (Answer) / 152

34 Which arc borders the major sector? A Arc AB B Arc AC C Arc ADB A B C D

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Answer

C Slide 92 / 152 The Area of a Sector

A sector is part of a circle. The area of a complete circle is given by Acircle = πr2 A B C D If the central angle of the sector is given in degrees, that's just the measure of that angle divided by 360 degrees, yielding: Asector = (θ/360°)(πr2) when θ is the central angle of the sector measured in degrees Similar to the arc length, we have to find the fraction of the circle in the sector and multiply this fraction by the area of the entire circle to find the area

f a sector

θ r

SLIDE 25

Slide 93 / 152 The Area of a Sector

Using our formula, we know that θ = 110°, and r = 20 cm. A B C D 110° 20 cm If we substitute these numbers into our formula, it will give us the area of our sector. Asector = (110°/360°)π(20)2 Asector = 383.97 cm2 In this case, we know that the measure

f our central angle, or m∠C = 110°.

But, if we are also told that the radius of the circle is 20 cm, we can determine the area of the sector enclosed by AB, AC & CB.

click click click click click

click

Slide 93 (Answer) / 152 The Area of a Sector

Using our formula, we know that θ = 110°, and r = 20 cm. A B C D 110° 20 cm If we substitute these numbers into our formula, it will give us the area of our sector. Asector = (110°/360°)π(20)2 Asector = 383.97 cm2 In this case, we know that the measure

f our central angle, or m∠C = 110°.

But, if we are also told that the radius of the circle is 20 cm, we can determine the area of the sector enclosed by AB, AC & CB.

click click click click click

click

[This object is a pull tab]

Math Practice

The blank spaces help the students "Create a representation of the problem" (MP2).

Slide 94 / 152 The Area of a Sector

Alternatively, we could just set this up as a proportion and solve it in one step. Area of Sector Area of Circle Central angle 360 = Asector πr2 110 360 = Asector = (πr2) 110 360 = π(20)2 110 360 Asector Asector = π cm2 = 383.97 cm2 1100 9 A B C D 110° 20 cm

click click click click

Slide 94 (Answer) / 152 The Area of a Sector

Alternatively, we could just set this up as a proportion and solve it in one step. Area of Sector Area of Circle Central angle 360 = Asector πr2 110 360 = Asector = (πr2) 110 360 = π(20)2 110 360 Asector Asector = π cm2 = 383.97 cm2 1100 9 A B C D 110° 20 cm

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

Math Practice

Questions to address the MP standards: How can you represent the problem with numbers and symbols? (MP2) What concepts that we have learned before were useful in solving this? (MP8) How is the area of a sector related to the arc length? (MP7)

Slide 95 / 152

35 Find the area of the sector. Leave your answer in terms

f π.

45° A B C 3

Asector = (θ/360°)(πr2)

Slide 95 (Answer) / 152

35 Find the area of the sector. Leave your answer in terms

f π.

45° A B C 3

Asector = (θ/360°)(πr2)

[This object is a pull tab]

Answer units2

SLIDE 26

Slide 96 / 152

36 Find the area of the major sector. Leave your answer in terms of π.

Asector = (θ/360°)(πr2)

C A T 8 cm 60°

Slide 96 (Answer) / 152

36 Find the area of the major sector. Leave your answer in terms of π.

Asector = (θ/360°)(πr2)

C A T 8 cm 60°

[This object is a pull tab]

Answer

cm2

Slide 97 / 152

37 Find the area of the minor sector of the circle. Round your answer to the nearest hundredth. C A T 5.5 cm 30°

Slide 97 (Answer) / 152

37 Find the area of the minor sector of the circle. Round your answer to the nearest hundredth. C A T 5.5 cm 30°

[This object is a pull tab]

Answer

cm2

Slide 98 / 152

38 Find the Area of the major sector for the circle. Round your answer to the nearest thousandth. C A T 12 cm 85°

Slide 98 (Answer) / 152

38 Find the Area of the major sector for the circle. Round your answer to the nearest thousandth. C A T 12 cm 85°

[This object is a pull tab]

Answer

cm2

SLIDE 27

Slide 99 / 152

39 What is the central angle for the major sector of the circle? C A G 15 cm 120°

Slide 99 (Answer) / 152

39 What is the central angle for the major sector of the circle? C A G 15 cm 120°

[This object is a pull tab]

Answer

Slide 100 / 152

40 Find the area of the major sector. Round to the nearest thousandth. C A G 15 cm 120°

Slide 100 (Answer) / 152

40 Find the area of the major sector. Round to the nearest thousandth. C A G 15 cm 120°

[This object is a pull tab]

Answer

cm2

Slide 101 / 152

41 If a circle is divided into 2 sectors, one major and one minor, then the sum of the areas of the 2 sectors is equal to the total area of the circle. True False

Slide 101 (Answer) / 152

41 If a circle is divided into 2 sectors, one major and one minor, then the sum of the areas of the 2 sectors is equal to the total area of the circle. True False

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Answer

True

SLIDE 28

Slide 102 / 152

42 A group of 10 students order pizza. They order 5 12" pizzas, that contain 8 slices each. If they split the pizzas equally, how many square inches of pizza does each student get (to the nearest hundredth)?

Slide 102 (Answer) / 152

42 A group of 10 students order pizza. They order 5 12" pizzas, that contain 8 slices each. If they split the pizzas equally, how many square inches of pizza does each student get (to the nearest hundredth)?

[This object is a pull tab]

Answer Each student gets 4 pieces, which is 1/2 a pizza.

Slide 103 / 152

43 You have a circular sprinkler in your yard. The sprinkler has a radius of 25 ft. How many square feet does the sprinkler water if it only rotates 120°? Round your answer to the nearest hundredth.

Slide 103 (Answer) / 152

43 You have a circular sprinkler in your yard. The sprinkler has a radius of 25 ft. How many square feet does the sprinkler water if it only rotates 120°? Round your answer to the nearest hundredth.

[This object is a pull tab]

Answer

Slide 104 / 152

Area of Other Quadrilaterals

Return to Table of Contents Lab: Area of Other Quadrilaterals

Slide 105 / 152

So far, we have discussed calculating the area of rectangles, triangles, parallelograms, circles, sectors and regular polygons, and the relationships between each formula. In this section, we are going to determine how 3 additional area formulas are related to those that we already know and to each

ther.

Area of Other Quadrilaterals

SLIDE 29

Slide 106 / 152 Area of a Trapezoid

Let's start off by deriving the area formula of a trapezoid. If you will recall from our unit on Quadrilaterals, a trapezoid is a quadrilateral with 1 pair of opposite sides that are parallel, called bases and 1 pair of opposite sides that are not parallel, called legs. If we label our bases as b1 and b2 and draw an altitude, or the height (h), we have the figure given to the right. b1 b2 h

Slide 107 / 152 Area of a Trapezoid

b1 b2 h We can split the trapezoid into 2 triangles by drawing the diagonal from the upper left corner to the lower right

corner. Click on the bases to reveal

each new triangle (given in red). Now, we can separate these two triangles. Using the triangles, how could you calculate the area of a trapezoid? Add up the triangle areas

click

Slide 107 (Answer) / 152 Area of a Trapezoid

b1 b2 h We can split the trapezoid into 2 triangles by drawing the diagonal from the upper left corner to the lower right

corner. Click on the bases to reveal

each new triangle (given in red). Now, we can separate these two triangles. Using the triangles, how could you calculate the area of a trapezoid? Add up the triangle areas

click [This object is a pull tab]

Math Practice

The final question addresses MP1. Additional Questions that could be used: What connections do you see between the two triangles & the trapezoid? (MP4)

Slide 108 / 152 Area of a Trapezoid

b1 b2 h h b2 b1 h

+ =

We know that the area of a triangle is 1/2 bh and that the sum of these two triangles will be the area of the trapezoid. Create an equation using this fact and the variables provided. ATrapezoid = 1/2 b1h + 1/2 b2h ATrapezoid = 1/2 h(b1 + b2) What algebra facts can be used to simplify the equation? How does it simplify?

click click

Distributive Property or Factoring

click

Slide 108 (Answer) / 152 Area of a Trapezoid

b1 b2 h h b2 b1 h

+ =

We know that the area of a triangle is 1/2 bh and that the sum of these two triangles will be the area of the trapezoid. Create an equation using this fact and the variables provided. ATrapezoid = 1/2 b1h + 1/2 b2h ATrapezoid = 1/2 h(b1 + b2) What algebra facts can be used to simplify the equation? How does it simplify?

click click

Distributive Property or Factoring

click [This object is a pull tab]

Math Practice

The sentence "Create an equation..." addresses MP2. The final two questions address MP7.

Slide 109 / 152 Area of a Rhombus

Next, we are going to derive the area formula of a rhombus. If you will recall from our unit on Quadrilaterals, a rhombus is a parallelogram with congruent sides and diagonals that are perpendicular and bisect each other. Since it is a parallelogram, the formula A = bh will work for this shape. But what if we are given the diagonal lengths instead? What connections do you see? Can we figure

ut a formula for this case?

If we label our diagonals as d1 and d2, we have the figure given to the right. d1 d2

SLIDE 30

Slide 109 (Answer) / 152 Area of a Rhombus

Next, we are going to derive the area formula of a rhombus. If you will recall from our unit on Quadrilaterals, a rhombus is a parallelogram with congruent sides and diagonals that are perpendicular and bisect each other. Since it is a parallelogram, the formula A = bh will work for this shape. But what if we are given the diagonal lengths instead? What connections do you see? Can we figure

ut a formula for this case?

If we label our diagonals as d1 and d2, we have the figure given to the right. d1 d2

[This object is a pull tab]

Math Practice

The questions in the 3rd paragraph (Since it is a ...) address MP4.

Slide 110 / 152 Area of a Rhombus

The diagonals split the rhombus into 4 congruent

triangles. If we move two the

bottom triangles to the top, but

n opposite sides (i.e. move

the bottom right triangle to the upper left corner), we will create another shape and determine the area formula. d1 d2 Use the following steps to show the animation in the diagram: · Click on the hash marks in the 2 bottom triangles to reveal each new triangle (given in red). · Move the new red triangles to the opposite corners of the shape. · Click the bottom of the original rhombus (shown in black) to hide it.

Slide 110 (Answer) / 152 Area of a Rhombus

The diagonals split the rhombus into 4 congruent

triangles. If we move two the

bottom triangles to the top, but

n opposite sides (i.e. move

the bottom right triangle to the upper left corner), we will create another shape and determine the area formula. d1 d2 Use the following steps to show the animation in the diagram: · Click on the hash marks in the 2 bottom triangles to reveal each new triangle (given in red). · Move the new red triangles to the opposite corners of the shape. · Click the bottom of the original rhombus (shown in black) to hide it.

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Teacher Notes

Diagrams are given below to show the end diagram for each step in the bulleted list.

d1 d2 d1 d2 d1 d2

Slide 111 / 152

d1 d2 What shape has been made? Rectangle What is the area formula for this shape? A = bh or A = lw Using these connections and the variables given above, create an equation to represent the area of a rhombus. ARhombus = 1/2 d1d2

Area of a Rhombus

click click click

Slide 111 (Answer) / 152

d1 d2 What shape has been made? Rectangle What is the area formula for this shape? A = bh or A = lw Using these connections and the variables given above, create an equation to represent the area of a rhombus. ARhombus = 1/2 d1d2

Area of a Rhombus

click click click [This object is a pull tab]

Math Practice

The questions on this slide address MP2 & MP4. Additional Questions that could be used: Can anyone think of another method that might have worked? (MP3)

Sample Answer: Calculate the area
f 1 of the right triangles & multiply

by 4

Slide 112 / 152 Area of a Kite

Since the shape of a kite is very similar to a rhombus, you are going to explain how the Area of a Kite formula is the same as the Area of a Rhombus formula for homework. d1 d2 AKite = 1/2 d1d2

SLIDE 31

Slide 113 / 152

You are constructing a desk to fit into your bedroom using wood for the flat top surface and metal bars for the legs. In order to save space, you determine that an isosceles trapezoid would be the best shape. The bases of the trapezoid will be 3 feet and 7 feet and the angle formed by the short base and each leg of the trapezoid is 135°. How much wood is required to make the flat top surface of your desk?

Example

7 ft 3 ft 135° 135°

Slide 113 (Answer) / 152

You are constructing a desk to fit into your bedroom using wood for the flat top surface and metal bars for the legs. In order to save space, you determine that an isosceles trapezoid would be the best shape. The bases of the trapezoid will be 3 feet and 7 feet and the angle formed by the short base and each leg of the trapezoid is 135°. How much wood is required to make the flat top surface of your desk?

Example

7 ft 3 ft 135° 135°

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Answer A = 1/2(2)(3 + 7) A = 10 ft

7 ft 3 ft 135° 135° 45° 45° 2 ft 3 ft 2 ft 2 ft

Questions to address Math Practices: What is this problem asking? (MP1) How could you start this problem? (MP1) How could you use the drawing to show your thinking? (MP5) What do you know about right triangles & trapezoids that you can apply to this situation? (MP7) Can you find a shortcut to solve this problem? How could your shortcut make this problem easier? (MP8)

Slide 114 / 152

44 In order to make a kite, you need to cut enough wrapping paper, based on the lengths of the spars, or the sticks used as the frame of the kite (see diagram to the right). If the longest spar is 36 inches & the shortest spar is 24 inches, how much wrapping paper do you need to make your kite?

Spars

Slide 114 (Answer) / 152

44 In order to make a kite, you need to cut enough wrapping paper, based on the lengths of the spars, or the sticks used as the frame of the kite (see diagram to the right). If the longest spar is 36 inches & the shortest spar is 24 inches, how much wrapping paper do you need to make your kite?

Spars

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Answer

432 in2 or 3 ft2

Slide 115 / 152

45 One diagonal of a rhombus is 1/3 times as long as the

ther. The area of the rhombus is 0.24m2. What are the

lengths of the diagonals? A 1.2 m and 1.2 m B 1.2 m and 0.4 m C 1.8 m and 0.6 m D 0.6 m and 0.4 m

Slide 115 (Answer) / 152

45 One diagonal of a rhombus is 1/3 times as long as the

ther. The area of the rhombus is 0.24m2. What are the

lengths of the diagonals? A 1.2 m and 1.2 m B 1.2 m and 0.4 m C 1.8 m and 0.6 m D 0.6 m and 0.4 m

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Answer

B

SLIDE 32

Slide 116 / 152

46 A glass window in the shape of an isosceles trapezoid has bases that measure 8 inches and 12 inches. If the angle between the longest base and the legs is 60°, what is the area of the window? Round your answer to the nearest hundredth. 12 in. 8 in. 60°

Slide 116 (Answer) / 152

46 A glass window in the shape of an isosceles trapezoid has bases that measure 8 inches and 12 inches. If the angle between the longest base and the legs is 60°, what is the area of the window? Round your answer to the nearest hundredth. 12 in. 8 in. 60°

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Answer

34.64 in2

Slide 117 / 152 Area of Complex Figures

In most real-world problems, you will need to calculate the area of complex figures, or shapes that are a combination of 2 or more shapes. In order to calculate the area of these types of shapes, we need to either add or subtract the areas of the primary shapes involved. Add the area of a rectangle & 1/2 the area of a circle. Subtract the area of the rectangles from the area of the trapezoid.

Slide 118 / 152 Slide 118 (Answer) / 152 Slide 119 / 152 Example

In order to redesign your bedroom, you decide to paint one of the walls with an accent color, leaving the remaining walls with the same. The wall to be painted measures 13' 6" by 8' 6" and contains 2 doorways, each measuring 7' high and 3' wide. The doorways are not going to be painted with the accent color. · Find the amount of wall space that will be painted. · If one quart-size can of paint covers 87.5 ft2, then how many cans are required to paint 2 coats of the accent color? · If one quart-size can of paint costs $15, how much money do you need to spend to paint your wall?

SLIDE 33

Slide 120 / 152 Example

In order to redesign your bedroom, you decide to paint one of the walls with an accent color, leaving the remaining walls with the same. The wall to be painted measures 13' 6" by 8' 6" and contains 2 doorways, each measuring 7' high and 3' wide. The doorways are not going to be painted with the accent color. · Find the amount of wall space that will be painted. Wall Area: A = 13.5(8.5) A = 114.75 ft

Doorway Area: A = 7(3) = 21 ft

2 doorways, so total area not

included is 42 ft2 Painted Area: A = 114.75 - 42 A = 72.75 ft

2 click click click

Slide 120 (Answer) / 152 Example

In order to redesign your bedroom, you decide to paint one of the walls with an accent color, leaving the remaining walls with the same. The wall to be painted measures 13' 6" by 8' 6" and contains 2 doorways, each measuring 7' high and 3' wide. The doorways are not going to be painted with the accent color. · Find the amount of wall space that will be painted. Wall Area: A = 13.5(8.5) A = 114.75 ft

Doorway Area: A = 7(3) = 21 ft

2 doorways, so total area not

included is 42 ft2 Painted Area: A = 114.75 - 42 A = 72.75 ft

2 click click click

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Math Practice

Questions to address Math Practices: What information do you have? (MP1) What do you need to find out? (MP1) What strategies are you going to use? (MP1) Would it help to draw a picture? (MP4) How is this related to the different area formulas that we have learned about? (MP7) How can you represent the problem w/ symbols & numbers? (MP2) What tools do you need to solve this problem? (MP5) What labels could you use? (MP6) What concepts that we have learned before were useful in solving this? (MP8)

Slide 121 / 152 Example

In order to redesign your bedroom, you decide to paint one of the walls with an accent color, leaving the remaining walls with the same. The wall to be painted measures 13' 6" by 8' 6" and contains 2 doorways, each measuring 7' high and 3' wide. The doorways are not going to be painted with the accent color. · If one quart-size can of paint covers 87.5 ft2, then how many cans are required to paint 2 coats of the accent color? Total painted Area = 72.75(2) = 145.5 ft2 Quart-sized Cans of Paint = 145.5/87.5 = 1.66 ≈ 2 cans

click click

Slide 121 (Answer) / 152 Example

In order to redesign your bedroom, you decide to paint one of the walls with an accent color, leaving the remaining walls with the same. The wall to be painted measures 13' 6" by 8' 6" and contains 2 doorways, each measuring 7' high and 3' wide. The doorways are not going to be painted with the accent color. · If one quart-size can of paint covers 87.5 ft2, then how many cans are required to paint 2 coats of the accent color? Total painted Area = 72.75(2) = 145.5 ft2 Quart-sized Cans of Paint = 145.5/87.5 = 1.66 ≈ 2 cans

click click [This object is a pull tab]

Math Practice

Questions to address Math Practices: What information do you have? (MP1) What do you need to find out? (MP1) What strategies are you going to use? (MP1) How can you represent the problem w/ symbols & numbers? (MP2) What tools do you need to solve this problem? (MP5) What labels could you use? (MP6)

Slide 122 / 152 Example

In order to redesign your bedroom, you decide to paint one of the walls with an accent color, leaving the remaining walls with the same. The wall to be painted measures 13' 6" by 8' 6" and contains 2 doorways, each measuring 7' high and 3' wide. The doorways are not going to be painted with the accent color. · If one quart-size can of paint costs $15, how much money do you need to spend to paint your wall? $15(2) = $30

click

Slide 122 (Answer) / 152 Example

In order to redesign your bedroom, you decide to paint one of the walls with an accent color, leaving the remaining walls with the same. The wall to be painted measures 13' 6" by 8' 6" and contains 2 doorways, each measuring 7' high and 3' wide. The doorways are not going to be painted with the accent color. · If one quart-size can of paint costs $15, how much money do you need to spend to paint your wall? $15(2) = $30

click [This object is a pull tab]

Math Practice

Questions to address Math Practices: What information do you have? (MP1) What do you need to find out? (MP1) What strategies are you going to use? (MP1) How can you represent the problem w/ symbols & numbers? (MP2) What tools do you need to solve this problem? (MP5) What labels could you use? (MP6)

SLIDE 34

Slide 123 / 152

47 What is the area of the shaded region? A 20 cm2 B 30 cm2 C 100 cm2 D 200 cm2

Slide 123 (Answer) / 152

47 What is the area of the shaded region? A 20 cm2 B 30 cm2 C 100 cm2 D 200 cm2

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Answer

C Slide 124 / 152

48 What is the area of the entire figure? Round your answer to the nearest tenth. A 25.1 yd2 B 28.9 yd2 C 54 yd2 D 79.1 yd2

Slide 124 (Answer) / 152

48 What is the area of the entire figure? Round your answer to the nearest tenth. A 25.1 yd2 B 28.9 yd2 C 54 yd2 D 79.1 yd2

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Answer

D Slide 125 / 152

49 A plant vase is formed by combining a square base with 4 isosceles trapezoids (see figure below). The square base has an area of 25 in2, the long bases of the trapezoids are 7 in., and the heights of the trapezoids are 4 in. How much glass was needed to make the entire vase? 7 in. 4 in.

Slide 125 (Answer) / 152

49 A plant vase is formed by combining a square base with 4 isosceles trapezoids (see figure below). The square base has an area of 25 in2, the long bases of the trapezoids are 7 in., and the heights of the trapezoids are 4 in. How much glass was needed to make the entire vase? 7 in. 4 in.

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Answer

121 in2

SLIDE 35

Slide 126 / 152

50 The Bisect Building Company has created a building plan for the new garage at the Heptagon home, shown in the

figure. The roof of the garage is made from 2 trapezoids

and 2 isosceles triangles. What is the area of the entire roof? A 700 ft2 B 600 ft2 C 250 ft2 D 100 ft2

20 ft 30 ft 20 ft 10 ft 10 ft

Slide 126 (Answer) / 152

50 The Bisect Building Company has created a building plan for the new garage at the Heptagon home, shown in the

figure. The roof of the garage is made from 2 trapezoids

and 2 isosceles triangles. What is the area of the entire roof? A 700 ft2 B 600 ft2 C 250 ft2 D 100 ft2

20 ft 30 ft 20 ft 10 ft 10 ft

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Answer

A Slide 127 / 152

51 The Bisect Building Company has created a building plan for the new garage at the Heptagon home, shown in the

figure. The roof of the garage is made from 2 trapezoids

and 2 isosceles triangles. Each bundle of shingles can cover approximately 40 ft2, and shingles must be purchased in full bundles. How many bundles of shingles are required to cover the roof? A 15 B 17 C 18 D 20

20 ft 30 ft 20 ft 10 ft 10 ft

Slide 127 (Answer) / 152

51 The Bisect Building Company has created a building plan for the new garage at the Heptagon home, shown in the

figure. The roof of the garage is made from 2 trapezoids

and 2 isosceles triangles. Each bundle of shingles can cover approximately 40 ft2, and shingles must be purchased in full bundles. How many bundles of shingles are required to cover the roof? A 15 B 17 C 18 D 20

20 ft 30 ft 20 ft 10 ft 10 ft

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Answer

C Slide 128 / 152

52 The Bisect Building Company has created a building plan for the new garage at the Heptagon home, shown in the figure. The roof of the garage is made from 2 trapezoids and 2 isosceles triangles. Each bundle of shingles costs $28.50. How much should the Bisect Building Company budget for the shingles?

20 ft 30 ft 20 ft 10 ft 10 ft

Slide 128 (Answer) / 152

52 The Bisect Building Company has created a building plan for the new garage at the Heptagon home, shown in the figure. The roof of the garage is made from 2 trapezoids and 2 isosceles triangles. Each bundle of shingles costs $28.50. How much should the Bisect Building Company budget for the shingles?

20 ft 30 ft 20 ft 10 ft 10 ft

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Answer

$513.00

SLIDE 36

Slide 129 / 152

Area & Perimeter of Figures in the Coordinate Plane

Return to Table of Contents

Slide 130 / 152

Some problems will show your shapes in a coordinate plane. In these cases, you will need to calculate the side lengths, or diagonal lengths,

f your shapes using the distance formula.

Remember that the distance formula is

Area & Perimeter of Figures in the Coordinate Plane

d =

After calculating the desired distances, you will either need to add them to calculate the perimeter or multiply them to calculate the area, using the appropriate formula.

Slide 131 / 152

Example

Calculate the perimeter and area of rhombus JKLM. 4 2 x y 6 8 4 2 6 8 J K L M 10

Since JKLM is a rhombus, all of the sides are congruent. How would we calculate the perimeter? Calculate one side length using the distance formula & then multiply that value by 4 to calculate our perimeter. JK = √(7 - 3)2 + (3 - 1)2 = √42 + 22 = √16 + 4 = √20 = 2√5 ≈ 4.47 Perimeter = 4(2√5) = 8√5 units ≈ 17.89 units

click click

click

Slide 131 (Answer) / 152

Example

Calculate the perimeter and area of rhombus JKLM. 4 2 x y 6 8 4 2 6 8 J K L M 10

click click

click

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Math Practice

Question about calculating the perimeter addresses MP1. Additional Q's to address Math Practices: What information do you have? (MP1) What do you need to find out? (MP1) How can you represent the problem w/ symbols & numbers? (MP2) What tools do you need to solve this problem? (MP5) What labels could you use? (MP6)

Slide 132 / 152

Example

Calculate the perimeter and area of rhombus JKLM. 4 2 x y 6 8 4 2 6 8 J K L M 10

Since JKLM is a rhombus, A = 1/2 d1d2 Therefore, we need to calculate the lengths of the diagonals using the distance formula. JL = √(5 - 3)2 + (-1 - 1)2 = √22 + (-2)2 = √4 + 4 = √8 = 2√2 KM = √(7 - 1)2 + (3 - (-3))2 = √62 + 62 = √36 + 36 = √72 = 6√2 A = 1/2 (2√2)(6√2) A = 12 units2

click click click click click

Slide 132 (Answer) / 152

Example

Calculate the perimeter and area of rhombus JKLM. 4 2 x y 6 8 4 2 6 8 J K L M 10

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

Math Practice

Questions about the area formula & the lengths of the diagonals address MP1. Additional Q's to address Math Practices: What information do you have? (MP1) What do you need to find out? (MP1) How can you represent the problem w/ symbols & numbers? (MP2) What tools do you need to solve this problem? (MP5) What labels could you use? (MP6) When calculating the distances, can anyone think of another method that might have worked or been easier? (MP7 & MP8)

Ans: 45-45-90 triangles could be used

SLIDE 37

Slide 133 / 152

2 1 x y 3 4 2 1 3 4 E F G H Miles (hundreds) Miles (hundreds) The figure shows trapezoid EFGH in the coordinate plane with point E at (0, 2.11), F at (3.30, 2.11), G at (3.40, 0) and H at the origin. Trapezoid EFGH can be used to approximate the size of the state of North Dakota with x and y scales representing hundreds of miles.

Example

· Based on the information given, how many miles is the perimeter of North Dakota? · At the end of 2010, the population of North Dakota was 672,591 people. Based on the information given, what was the population density at the end of 2010?

Slide 134 / 152

2 1 x y 3 4 2 1 3 4 E F G H Miles (hundreds) Miles (hundreds) The figure shows trapezoid EFGH in the coordinate plane with point E at (0, 2.11), F at (3.30, 2.11), G at (3.40, 0) and H at the origin. Trapezoid EFGH can be used to approximate the size of the state of North Dakota with x and y scales representing hundreds of miles.

Example

· Based on the information given, how many miles is the perimeter of North Dakota? How do we calculate the perimeter? add up all of the side lengths What strategies are you going to use? counting & distance formula

click click

Slide 134 (Answer) / 152

2 1 x y 3 4 2 1 3 4 E F G H Miles (hundreds) Miles (hundreds) The figure shows trapezoid EFGH in the coordinate plane with point E at (0, 2.11), F at (3.30, 2.11), G at (3.40, 0) and H at the origin. Trapezoid EFGH can be used to approximate the size of the state of North Dakota with x and y scales representing hundreds of miles.

Example

· Based on the information given, how many miles is the perimeter of North Dakota? How do we calculate the perimeter? add up all of the side lengths What strategies are you going to use? counting & distance formula

click click [This object is a pull tab]

Math Practice

Questions about the perimeter & the strategies used address MP1. Additional Q's to address Math Practices: How can you represent the problem w/ symbols & numbers? (MP2) What tools do you need to solve this problem? (MP5) What labels could you use? (MP6)

Slide 135 / 152

2 1 x y 3 4 2 1 3 4 E F G H Miles (hundreds) Miles (hundreds) The figure shows trapezoid EFGH in the coordinate plane with point E at (0, 2.11), F at (3.30, 2.11), G at (3.40, 0) and H at the origin. Trapezoid EFGH can be used to approximate the size of the state of North Dakota with x and y scales representing hundreds of miles.

Example

· Based on the information given, how many miles is the perimeter of North Dakota? EF = 3.3 = 330 miles GH = 3.4 = 340 miles EH = 2.11 = 211 miles FG = √(3.4 - 3.3)2 + (0 - 2.11)2 = √(0.1)2 + (-2.11)2 = √0.01 + 4.4521 = √4.4621 ≈ 2.11 = 211 miles Perimeter: 330 + 340 + 211 + 211 = 1092 miles

click click click click click

Slide 135 (Answer) / 152

2 1 x y 3 4 2 1 3 4 E F G H Miles (hundreds) Miles (hundreds) The figure shows trapezoid EFGH in the coordinate plane with point E at (0, 2.11), F at (3.30, 2.11), G at (3.40, 0) and H at the origin. Trapezoid EFGH can be used to approximate the size of the state of North Dakota with x and y scales representing hundreds of miles.

Example

· Based on the information given, how many miles is the perimeter of North Dakota? EF = 3.3 = 330 miles GH = 3.4 = 340 miles EH = 2.11 = 211 miles FG = √(3.4 - 3.3)2 + (0 - 2.11)2 = √(0.1)2 + (-2.11)2 = √0.01 + 4.4521 = √4.4621 ≈ 2.11 = 211 miles Perimeter: 330 + 340 + 211 + 211 = 1092 miles

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

Math Practice

Questions about the perimeter & the strategies used address MP1. Additional Q's to address Math Practices: How can you represent the problem w/ symbols & numbers? (MP2) What tools do you need to solve this problem? (MP5) What labels could you use? (MP6)

Slide 136 / 152

2 1 x y 3 4 2 1 3 4 E F G H Miles (hundreds) Miles (hundreds) The figure shows trapezoid EFGH in the coordinate plane with point E at (0, 2.11), F at (3.30, 2.11), G at (3.40, 0) and H at the origin. Trapezoid EFGH can be used to approximate the size of the state of North Dakota with x and y scales representing hundreds of miles.

Example

· At the end of 2010, the population

f North Dakota was 672,591
people. Based on the information

given, what was the population density at the end of 2010? What does population density mean? ratio that represents the number

f people living per square mile.

How do we find it? dividing the total population by the total area

click click

SLIDE 38

Slide 136 (Answer) / 152

2 1 x y 3 4 2 1 3 4 E F G H Miles (hundreds) Miles (hundreds) The figure shows trapezoid EFGH in the coordinate plane with point E at (0, 2.11), F at (3.30, 2.11), G at (3.40, 0) and H at the origin. Trapezoid EFGH can be used to approximate the size of the state of North Dakota with x and y scales representing hundreds of miles.

Example

· At the end of 2010, the population

f North Dakota was 672,591
people. Based on the information

given, what was the population density at the end of 2010? What does population density mean? ratio that represents the number

f people living per square mile.

How do we find it? dividing the total population by the total area

click click [This object is a pull tab]

Math Practice

Questions about population density address MP6. Additional Q's to address Math Practices: How can you represent the problem w/ symbols & numbers? (MP2) What tools do you need to solve this problem? (MP5) What labels could you use? (MP6)

Slide 137 / 152

2 1 x y 3 4 2 1 3 4 E F G H Miles (hundreds) Miles (hundreds) The figure shows trapezoid EFGH in the coordinate plane with point E at (0, 2.11), F at (3.30, 2.11), G at (3.40, 0) and H at the origin. Trapezoid EFGH can be used to approximate the size of the state of North Dakota with x and y scales representing hundreds of miles.

Example

· At the end of 2010, the population

f North Dakota was 672,591
people. Based on the information

given, what was the population density at the end of 2010? Area = 1/2(211)(340 + 330) = 1/2(211)(670) = 70,685 square miles Population Density: 672,591 70,685 = 9.51 people per square mile

click click

Slide 137 (Answer) / 152

2 1 x y 3 4 2 1 3 4 E F G H Miles (hundreds) Miles (hundreds) The figure shows trapezoid EFGH in the coordinate plane with point E at (0, 2.11), F at (3.30, 2.11), G at (3.40, 0) and H at the origin. Trapezoid EFGH can be used to approximate the size of the state of North Dakota with x and y scales representing hundreds of miles.

Example

· At the end of 2010, the population

f North Dakota was 672,591
people. Based on the information

given, what was the population density at the end of 2010? Area = 1/2(211)(340 + 330) = 1/2(211)(670) = 70,685 square miles Population Density: 672,591 70,685 = 9.51 people per square mile

click click [This object is a pull tab]

Math Practice

Questions about population density address MP6. Additional Q's to address Math Practices: How can you represent the problem w/ symbols & numbers? (MP2) What tools do you need to solve this problem? (MP5) What labels could you use? (MP6)

Slide 138 / 152

53 Calculate the perimeter of square PQRS. A 10 units B 20 units C 25 units D 28.28 units

4 2 x y 6 8 4 2 6 8 P Q S R 10

Slide 138 (Answer) / 152

53 Calculate the perimeter of square PQRS. A 10 units B 20 units C 25 units D 28.28 units

4 2 x y 6 8 4 2 6 8 P Q S R 10

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Answer

B Slide 139 / 152

54 Calculate the area of square PQRS. A 28.28 units2 B 20 units2 C 25 units2 D 50 units2

4 2 x y 6 8 4 2 6 8 P Q S R 10

SLIDE 39

Slide 139 (Answer) / 152

54 Calculate the area of square PQRS. A 28.28 units2 B 20 units2 C 25 units2 D 50 units2

4 2 x y 6 8 4 2 6 8 P Q S R 10

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