Pearson Geometry Common Core, 2011
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Pearson Geometry Common Core, 2011 View details
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Exercise 19 Page 753

391.6 in.^2

Practice makes perfect

The given solid is a pyramid.

To calculate the surface area of a pyramid, we can use the known formula where P is the perimeter of the base, l is the slant height, and B is the area of the base. S=1/2Pl+BThe base of the pyramid is a square with the side length of 11in. Let's calculate its perimeter! P=4s ⇒ P=4( 11)=44in. The area of the square equals the squared length of the side. B=s^2 ⇒ B= 11^2=121in.^2 To find the slant height, we can use the Pythagorean Theorem. When doing this, the slant height l is the hypotenuse. The height and apothem, which is half of the side length, of the pyramid are the legs. Let's use these given values to solve for l.
a^2+b^2=l^2
Solve for l
( 5.5)^2+ 11^2 = l^2
30.25 + 121 = l^2
151.25 = l^2
sqrt(151.25) = l
l = sqrt(151.25)
Let's substitute all of these values into the formula for the surface area and calculate it.
S=1/2Pl+B
S=1/2(44)(sqrt(151.25))+121
Simplify right-hand side
S=44/2sqrt(151.25)+121
S=22sqrt(151.25)+121
S = 391.56422...
S≈ 391.6
The surface area of the pyramid, to the nearest tenth, is 391.6 in.^2.