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

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

V OCABULARY 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.

A pyramid with base area 𝐵 and height ℎ has a 1 volume, 𝑊 = 3 𝐵 ℎ . The surface area of the pyramid is given by Base area + area of the slanting sides.

We consider a square based pyramid with a surface area of 80 𝑗𝑜 2 . 𝑡 𝑚 𝑚 1 The volume of the pyramid is 𝑊 = 3 𝑚 2 ℎ The surface area, S.A = 80 = 𝑚 2 + 4( 1 2 𝑚𝑡) 2 𝑚 From Pythagorean theorem, 𝑡 = + ℎ 2 2

2 Substituting for 𝑡 we get, S.A = 80 = 𝑚 2 + 2𝑚 𝑚 + ℎ 2 2 1600 Solving for h, we get h = 𝑚 2 − 40 1 1600 Substituting we have, 𝑊 = 3 𝑚 2 𝑚 2 − 40 1 1600 The volume as a function of 𝑚 is, 𝑊(𝑚) = 3 𝑚 2 𝑚 2 − 40

Let’s consider a square pyramid with a volume of 60 𝑗𝑜 3 . 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. 1 180 The volume, 𝑊 = 60 = 3 𝑚 2 ℎ ⟹ ℎ = 𝑚 2 Let the surface area be K, 𝑚 2 K = 𝑚 2 + 4( 1 180 2 𝑚 ℎ 2 + 4 ) but ℎ = 𝑚 2 𝑚 2 K = 𝑚 2 + 4( 1 180 4 , 𝑚 2 ) 2 + 2 𝑚 ( 𝑚 2 𝐿 𝑚 = 𝑚 2 + 2𝑚 32400 + 𝑚 4 4

Example 1 A rectangular based pyramid has a length of 3 𝑗𝑜 and a volume of 12 𝑗𝑜 3 . Write the surface area as a function of its width. Solution Let the surface area be 𝑙 and vertical height be ℎ and the slant heights be 𝑡 1 and 𝑡 2 . 1 1 𝑙 = 𝑚𝑥 + 2 2 𝑚𝑡 + 2 2 𝑥𝑡 = 3𝑥 + 3𝑡 1 + 𝑥𝑡 2 2.25 + ℎ 2 and From Pythagorean theorem, 𝑡 2 = 𝑥 2 + ℎ 2 𝑡 1 =

Substituting these values we get, 𝑙 = 3𝑥 + 3 𝑥 2 + ℎ 2 + 𝑥 2.25 + ℎ 2 1 12 Volume, 𝑊 = 12 = 3 × 3 × 𝑥 × ℎ ⟹ ℎ = 𝑥 Substituting the value of ℎ in the surface area we have, 12 12 𝑥 2 + ( 𝑥 ) 2 + 𝑥 × 𝑥 ) 2 𝑙 𝑥 = 3𝑥 + 3 × 2.25 + ( 144 144 𝑥 2 + 𝑙 𝑥 = 3𝑥 + 3 × 𝑥 2 + 𝑥 × 2.25 + 𝑥 2

HOMEWORK A man has 200 𝑔𝑢 2 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.

A NSWERS TO HOMEWORK 𝑚 2 1 4000 3 𝑚 2 𝑊 𝑚 = 𝑚 2 + 4

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