Geometry (B): READ: Surface Area of a Pyramid Reading

Pyramids

Learning Objectives

Identify pyramids.
Find the surface area of a pyramid using a net or a formula.
Find the volume of a pyramid.

Introduction

A pyramid is a three-dimensional figure with a single base and a three or more non-parallel sides that meet at a single point above the base. The sides of a pyramid are triangles.

A regular pyramid is a pyramid that has a regular polygon for its base and whose sides are all congruent triangles.

Surface Area of a Pyramid Using Nets

You can deconstruct a pyramid into a net.

To find the surface area of the figure using the net, first find the area of the base:

$A &= s^2\\ &= (12)(12)\\ &= 144 \ \text{square\ units}$

Now find the area of each isosceles triangle. Use the Pythagorean Theorem to find the height of the triangles. This height of each triangle is called the slant height of the pyramid. The slant height of the pyramid is the altitude of one of the triangles. Notice that the slant height is larger than the altitude of the triangle.

We’ll call the slant height $n$ for this problem. Using the Pythagorean Theorem:

$(11.66)^2 &= 6^2 + n^2\\ 136 &= 36 + n^2\\ 100 &= n^2\\ 10 &= n$

Now find the area of each triangle:

$A &= \frac{1}{2} hb\\ &= \frac{1}{2} (10)(12)\\ &= 60 \ \text{square\ units}$

As there are $4$ triangles:

$A \text{(triangles)} &= 4(60)\\ &= 240 \ \mathrm{square\ units}$

Finally, add the total area of the triangles to the area of the base.

$A \text{(total)} &= A\text{(triangles)} + A\text{(base)}\\ &= 240 + 144\\ &= 384 \ \mathrm{square\ units}$

Example 1

Use the net to find the total area of the regular hexagonal pyramid with an apothem of $5.19$ . The dimensions are given in centimeters.

The area of the hexagonal base is given by the formula for the area of a regular polygon. Since each side of the hexagon measures $6 \;\mathrm{cm}$ , the perimeter is $6 \cdot 6$ or $36\;\mathrm{cm}$ . The apothem, or perpendicular distance to the center of the hexagon is $5.19\;\mathrm{cm}$ .

$A &= \frac{1}{2} \text{(apothem)}\text{(perimeter)}\\ &= \frac{1}{2} (5.19)(36)\\ &= 93.42 \ \text{square cm}$

Using the Pythagorean Theorem to find the slant height of each lateral triangle.

$(14)^2 &= 3^2 + n^2\\ 196 &= 9 + n^2\\ 187 &= n^2\\ 13.67 &= n$

Now find the area of each triangle:

$A &= \frac{1}{2} hb\\ &= \frac{1}{2} (13.67)(6)\\ &= 41 \ \text{square cm}$

Together, the area of all six triangles that make up the lateral sides of the pyramid are

$A &= 6 \text{(area of each triangle)}\\ &= 6 \cdot 41\\ &= 246 \ \text{square cm}$

Add the area of the lateral sides to the area of the hexagonal base.

$A \text{(total)} &= A\text{(triangles)} + A\text{(base)}\\ &= 246 + 93.42\\ &= 339.42 \ \text{square cm}$

Surface Area of a Regular Pyramid

To get a general formula for the area of a regular pyramid, look at the net for this square pyramid.

The slant height of each lateral triangle is labeled $l$ (the lowercase letter $L$ ), and the side of the regular polygon is labeled $s$ . For each lateral triangle, the area is:

$A = \frac{1}{2} l s$

There are $n$ triangles in a regular polygon—e.g., $n = 3$ for a triangular pyramid, $n = 4$ for a square pyramid, $n = 5$ for a pentagonal pyramid. So the total area, $L$ , of the lateral triangles is:

$L &= n \cdot \text{(area of each lateral triangle)}\\ &= n\left(\frac{1}{2}ls \right)$

If we rearrange the above equation we get:

$L = \left(\frac{1}{2}ln \cdot s \right)$

Notice that $n \cdot s$ is just the perimeter, $P$ , of the regular polygon. So we can substitute $P$ into the equation to get the following postulate.

$L = \left(\frac{1}{2}lP \right)$

To get the total area of the pyramid, add the area of the base, $B$ , to the equation above.

$A = \frac{1}{2}lP + B$

Area of a Regular Pyramid

The surface area of a regular pyramid is

$A =\frac{1}{2} l P + B$

where $l$ is the slant height of the pyramid and $P$ is the perimeter of the regular polygon that forms the pyramid’s base, and $B$ is the area of the base.

Example 2

A tent without a bottom has the shape of a hexagonal pyramid with a slant height $l$ of $30\;\mathrm{feet}$ . The sides of the hexagonal perimeter of the figure each measure $8\;\mathrm{feet}$ . Find the surface area of the tent that exists above ground.

For this problem, $B$ is zero because the tent has no bottom. So simply calculate the lateral area of the figure.

$A &= \frac{1}{2} l P + B\\ &= \frac{1}{2} l P + 0\\ &= \frac{1}{2} l P\\ &= \frac{1}{2} (30) (6 \cdot 8)\\ &= 720 \ \text{square feet}$

Example 3

A pentagonal pyramid has a slant height $l$ of $12\;\mathrm{cm}$ . The sides of the pentagonal perimeter of the figure each measure $9\;\mathrm{cm}$ . The apothem of the figure is $6.19\;\mathrm{cm}$ . Find the surface area of the figure.

First find the lateral area of the figure.

$L &= \frac{1}{2} l P\\ &= \frac{1}{2} (12) (5 \cdot 9)\\ &= 270 \ \text{square cm}$

Now use the formula for the area of a regular polygon to find the area of the base.

$A &= \frac{1}{2} \text{(apothem)} \text{(perimeter)}\\\ &= \frac{1}{2} (6.19) (5 \cdot 9)\\ &= 139.3605 \ \text{square cm}$

Finally, add these together to find the total surface area.

$139.3605 + 270 \approx 409.36\; \text{square centimeters}$

Last modified: Monday, July 5, 2010, 3:50 PM