11.3. Surface Area of Pyramids and Cones http://www.ck12.org
Example 2:Find the surface area of the pyramid from Example 1.
Solution:The surface area of the four triangular faces are 4
( 1
2 bl)
= 2 ( 16 )
(
8
√
10
)
= 256
√
- To find the total
surface area, we also need the area of the base, which is 16^2 =256. The total surface area is 256
√
10 + 256 ≈
1065 .54.
From this example, we see that the formula for a square pyramid is:
SA= (area o f the base)+ 4 (area o f triangular f aces)SA=B+n(
1
2
bl)
Bis the area of the base andnis the number of triangles.SA=B+
1
2
l(nb) Rearranging the variables,nb=P,the perimeter of the base.SA=B+1
2
PlSurface Area of a Regular Pyramid:IfBis the area of the base andPis the perimeter of the base andlis the slant
height, thenSA=B+^12 Pl.
If you ever forget this formula, use the net. Each triangular face is congruent, plus the area of the base. This way,
you do not have to remember a formula, just a process, which is the same as finding the area of a prism.
Example 3:Find the area of the regular triangular pyramid.
Solution:The area of the base isA=^14 s^2
√
3 because it is an equilateral triangle.B=
1
4
82
√
3 = 16
√
3
SA= 16
√
3 +
1
2
( 24 )( 18 ) = 16
√
3 + 216 ≈ 243. 71
Example 4:If the lateral surface area of a square pyramid is 72f t^2 and the base edge is equal to the slant height,
what is the length of the base edge?
Solution: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.