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A pyramid with base area and height has a 1 volume, = 3 . The - - PowerPoint PPT Presentation

Sep 08, 2022 320 likes β€’444 views

D AY 169 P YRAMID STRUCTURES I NTRODUCTION We have encountered structures made in the form of a pyramid such as tents and some roofs in our daily life. When designing these structures, one may want it to have the maximum space with the

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DAY 169 – PYRAMID

STRUCTURES

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INTRODUCTION

We have encountered structures made in the form of a pyramid such as tents and some roofs in

  • ur daily life. When designing these structures, one

may want it to have the maximum space with the available size of the material. One may also want a structure with a given volume to have the minimum size of the material to minimize the cost

  • f the material.

In this lesson, we will apply geometric methods to solve design problems of the pyramid structures or objects to satisfy physical constraints

  • r minimize cost.
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VOCABULARY

Pyramid This is a three dimensional solid with a rectangular, triangular or any other polygonal base and triangular faces which meet at a single apex.

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A pyramid with base area 𝐡 and height β„Ž has a volume,π‘Š =

1 3 𝐡 β„Ž.

The surface area of the pyramid is given by Base area + area of the slanting sides.

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We consider a square based pyramid with a surface area of 80 π‘—π‘œ2. The volume of the pyramid is π‘Š =

1 3 π‘š2β„Ž

The surface area, S.A = 80 = π‘š2 + 4(

1 2 π‘šπ‘‘)

From Pythagorean theorem, 𝑑 =

π‘š 2 2

+ β„Ž2 𝑑

π‘š π‘š

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Substituting for 𝑑 we get, S.A = 80 = π‘š2 + 2π‘š

π‘š 2 2

+ β„Ž2 Solving for h, we get h =

1600 π‘š2 βˆ’ 40

Substituting we have, π‘Š =

1 3 π‘š2 1600 π‘š2 βˆ’ 40

The volume as a function of π‘š is, π‘Š(π‘š) =

1 3 π‘š2 1600 π‘š2 βˆ’ 40

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Let’s consider a square pyramid with a volume of 60 π‘—π‘œ3. We want to find a function of the surface area that we can use to find the values of its sides and height that will give us the minimum surface area. The volume,π‘Š = 60 =

1 3 π‘š2β„Ž ⟹ β„Ž = 180 π‘š2

Let the surface area be K, K = π‘š2 + 4(

1 2 π‘š β„Ž2 + π‘š2 4 ) but β„Ž = 180 π‘š2

K = π‘š2 + 4(

1 2 π‘š ( 180 π‘š2 )2+ π‘š2 4,

𝐿 π‘š = π‘š2 + 2π‘š

32400 π‘š4

+

π‘š2 4

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Example 1 A rectangular based pyramid has a length of 3 π‘—π‘œ and a volume of 12 π‘—π‘œ3. Write the surface area as a function of its width. Solution Let the surface area be 𝑙 and vertical height be β„Ž and the slant heights be 𝑑1 and 𝑑2. 𝑙 = π‘šπ‘₯ + 2

1 2 π‘šπ‘‘ + 2 1 2 π‘₯𝑑

= 3π‘₯ + 3𝑑1 + π‘₯𝑑2 From Pythagorean theorem, 𝑑2 = 2.25 + β„Ž2 and 𝑑1 = π‘₯2 + β„Ž2

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Substituting these values we get, 𝑙 = 3π‘₯ + 3 π‘₯2 + β„Ž2 + π‘₯ 2.25 + β„Ž2 Volume, π‘Š = 12 =

1 3 Γ— 3 Γ— π‘₯ Γ— β„Ž ⟹ β„Ž = 12 π‘₯

Substituting the value of β„Ž in the surface area we have, 𝑙 π‘₯ = 3π‘₯ + 3 Γ— π‘₯2 + (

12 π‘₯)2+ π‘₯ Γ—

2.25 + (

12 π‘₯)2

𝑙 π‘₯ = 3π‘₯ + 3 Γ— π‘₯2 +

144 π‘₯2 + π‘₯ Γ—

2.25 +

144 π‘₯2

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HOMEWORK A man has 200 𝑔𝑒2 of canvas which he wants to use to make a square based pyramid tent. Write the volume of the tent as a function of its length.

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ANSWERS TO HOMEWORK

π‘Š π‘š =

1 3 π‘š2 4000 π‘š2 + π‘š2 4

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THE END