The surface area of a pyramid represents the sum of the areas of all the faces of the pyramid. On the other hand, volume is a measure of the three-dimensional space occupied by the pyramid. We can calculate the surface area of a pyramid by adding the area of its base and the areas of its lateral faces, and we can calculate its volume by multiplying the area of its base by the height of the pyramid and dividing by 3.
In this article, we will learn all about the surface area and volume of a pyramid. We will explore their formulas and use them to solve some practice problems.
How to find the surface area of a pyramid?
The surface area of a pyramid can be calculated by adding the areas of all the faces of the pyramid. Depending on the type of pyramid we have, we will have a different number of lareal faces.
The most common pyramid is a pyramid with a square base, so let’s look at the surface area formula for this type of pyramid.
We can find the area of the base of the square pyramid by squaring the length of one of the sides of the base. Also, the area of the triangular faces is equal to one-half the length of the base multiplied by the height of the triangle.
Since all four lateral faces have the same dimensions, their areas are equal. Therefore, the following is the formula for the surface area of a square pyramid:
| $latex A_{s}={{l}^2}+2lh$ |
where, l is the length of the sides of the square base and h is the slant height of the triangular faces.
How to find the volume of a pyramid?
To find the volume of any pyramid, we have to multiply the area of the base of the pyramid by its height and divide the result by 3. That is, we have the following formula.
$latex \text{Volume}=\frac{1}{3}\text{Area base}\times \text{Height}$
Again, we can consider the most common pyramid to be a square pyramid. Therefore, considering that the area of a square is found by squaring one of its side lengths, we have the following formula:
| $latex V=\frac{1}{3}{{l}^2}\times h$ |
where, l is the length of one of the sides of the square base and h is the height of the pyramid.
Surface area and volume of pyramids – Examples with answers
In these examples, we will focus primarily on the surface area and volume of square pyramids, but the principles apply to any type of pyramid.
EXAMPLE 1
Find the surface area of a pyramid that has a square base with sides of 3 feet and triangular faces with a height of 4 feet.
Solution
We have the following:
- Sides of base, $latex l=3$
- Height of triangles, $latex h=4$
Using the formula for the surface area with the given lengths, we have:
$latex A_{s}={{l}^2}+2lh$
$latex A_{s}={{3}^2}+2(3)(4)$
$latex A_{s}=9+24$
$latex A_{s}=33$
The surface area is equal to 33 ft².
EXAMPLE 2
Find the volume of a square pyramid that has a height of 5 inches and sides with a length of 4 inches.
Solution
We have the following:
- Side of square, $latex l=4$
- Height, $latex h=5$
Using the volume formula, we have:
$latex V=\frac{1}{3}{{l}^2}\times h$
$latex V=\frac{1}{3}{{(4)}^2}\times (5)$
$latex V=\frac{1}{3}(16)\times (5)$
$latex V=26.67$
The volume is equal to 26.67 in³.
EXAMPLE 3
What is the surface area of a square pyramid that has sides with a length of 5 inches and triangular faces with a height of 6 inches?
Solution
We have the following:
- Sides of base, $latex l=5$
- Height of triangles, $latex h=6$
Using the formula for the surface area with these lengths, we have:
$latex A_{s}={{l}^2}+2lh$
$latex A_{s}={{5}^2}+2(5)(6)$
$latex A_{s}=25+60$
$latex A_{s}=85$
The surface area is equal to 85 in².
EXAMPLE 4
Find the volume of a pyramid that has a height of 6 yards and a square base with sides of 5 yards.
Solution
We have the following lengths:
- Sides of square, $latex l=5$
- Height, $latex h=6$
Using the volume formula with these values, we have:
$latex V=\frac{1}{3}{{l}^2}\times h$
$latex V=\frac{1}{3}{{(5)}^2}\times (6)$
$latex V=\frac{1}{3}(25)\times (6)$
$latex V=50$
The volume is equal to 50 yd³.
EXAMPLE 5
Find the surface area of a pyramid that has a square base with sides of 10 inches and triangular faces with a height of 7 inches.
Solution
We have the following dimensions:
- Sides of base, $latex l=10$
- Height of triangles, $latex h=7$
Using the formula for the surface area, we have:
$latex A_{s}={{l}^2}+2lh$
$latex A_{s}={{10}^2}+2(10)(7)$
$latex A_{s}=100+140$
$latex A_{s}=240$
The surface area is equal to 240 in².
EXAMPLE 6
Find the volume of a pyramid that has a height of 9 yards and a square base with sides of 8 yards.
Solution
We have the following:
- Sides of square, $latex l=8$
- Height, $latex h=9$
Applying the volume formula, we have:
$latex V=\frac{1}{3}{{l}^2}\times h$
$latex V=\frac{1}{3}{{(8)}^2}\times (9)$
$latex V=\frac{1}{3}(64)\times (9)$
$latex V=192$
The volume is equal to 192 yd³.
EXAMPLE 7
Find the surface area of a square pyramid that has sides with a length of 11 inches and triangular faces with a height of 12 inches.
Solution
We have the following:
- Sides of base, $latex l=11$
- Height of triangles, $latex h=12$
Using the formula for the surface area with the given lengths, we have:
$latex A_{s}={{l}^2}+2lh$
$latex A_{s}={{11}^2}+2(11)(12)$
$latex A_{s}=121+264$
$latex A_{s}=385$
The surface area is equal to 385 in².
EXAMPLE 8
Find the length of the height of a square pyramid that has a volume of 96 ft³ and a base with sides of 6 feet.
Solution
We have the following:
- Sides of square, $latex l=6$
- Volume, $latex V=96$
In this case, we know the volume of the pyramid, and we need to find the length of its height. Therefore, we use the volume formula and solve for h:
$latex V=\frac{1}{3}{{l}^2}\times h$
$latex 96=\frac{1}{3}{{(6)}^2}\times h$
$latex 96=\frac{1}{3}(36)\times h$
$latex 96=12 h$
$latex h=8$
The height is equal to 8 ft.
Surface area and volume of pyramids – Practice problems
Solve the following problems by using the formulas for the surface area and volume of a square pyramid.
See also
Interested in learning more about the area and volume of geometric figures? Take a look at these pages:
Jefferson Huera Guzman
Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.
