McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
3. Surface Areas of Pyramids and Cones
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Exercise 39 Page 829

Consider a pyramid such that you can easily find the area of the base. Then, use the formula of the surface area to find the corresponding slant height.

Example Solution: A pyramid whose base is a square of side 6 units and slant height equal to 163 units.

Practice makes perfect
Let P and B be the perimeter and area of the base of a pyramid, and let l be its slant height. The surface area of this pyramid is given by the expression below. S = 1/2 P l + B We want a pyramid such that S=100 square units. To make computations easier, let's consider a pyramid such that its base is a square of side length 6 units. Then, its area is equal to 36 square units, and its perimeter is 24 units.
Let's substitute S=100 and B=36 into the initial equation and solve it for l.
S = 1/2 P l + B
100 = 1/2* 24* l + 36
â–Ľ
Solve for l
100 = 12 l + 36
64 = 12 l
12 l = 64
l = 64/12
l = 16/3
In consequence, a pyramid whose base is a square of side length 6 units and slant height equal to 163 units has a total surface area of 100 square units.

Keep in mind that there are other pyramids with surface area of 100 square units, and this is just a sample answer; so, your answer may vary.