11.5. Pyramids http://www.ck12.org
- In the formula for surface area, the lateral surface area is^12 Plor^12 nbl. We know thatn=4 andb=l. Let’s solve
forb.
1
2
nbl= 72 f t^2
1
2
( 4 )b^2 = 72
2 b^2 = 72
b^2 = 36
b= 6Therefore, the base edges are all 6 units and the slant height is also 6 units.
- To find the area of the base, we need to find the apothem. If the base edges are 10 units, then the apothem is 5
√
3
for a regular hexagon. The area of the base is^12 asn=^12
(
5
√
3
)
( 10 )( 6 ) = 150
√
- The total surface area is:
SA= 150
√
3 +
1
2
( 6 )( 10 )( 22 )
= 150
√
3 + 660 ≈ 919. 81 units^2- In this example, we are given the slant height. For volume, we need the height, so we need to use the Pythagorean
Theorem to find it.
72 +h^2 = 252
h^2 = 576
h= 24Using the height, the volume is^13 ( 142 )( 24 ) = 1568 units^3.
- The base of this pyramid is a right triangle. So, the area of the base is^12 ( 14 )( 8 ) = 56 units^2.
V=
1
3
( 56 )( 17 )≈ 317. 33 units^3- The formula for the volume of a pyramid works for any pyramid, as long as you can find the area of the base.
224 = 56 h
4 =hPractice
Fill in the blanks about the diagram below.
1.xis the ___________.- The slant height is ____.