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7.4: Base, Lateral and Surface Areas of Pyramids

Created by: CK-12

Learning Objectives

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

Pyramids

A pyramid is a 3-dimensional figure with a single base and 3 or more non-parallel sides that meet at a single point above the base called the apex. The sides of a pyramid are triangles.

A pyramid has only one ____________________________, while a prism has 2.

The sides of a pyramid are ______________________________.

A pyramid has 3 or more _________________________.

What is the name of the point on the very top of the pyramid? _______________

Below are examples of pyramids. Notice that each pyramid is named after the shape of its base (so the pyramid with a square base is called a square pyramid, and so on):

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

A regular pyramid has sides that are ___________________________ triangles.

The base of a regular pyramid is a regular ________________________.

Surface Area of a Pyramid Using Nets

You can deconstruct a pyramid into a net:

As you can see in the diagram, the easiest way to construct a net of a pyramid is to draw the base in the center, with each triangular side extending outwards from the base edges.

To find the surface area of the figure using the net, first find the area of the base. In this case, the base is a square:

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

Now find the area of each side of the pyramid. Each side is an isosceles triangle. Use the Pythagorean Theorem to find the height of the triangles.

The 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.

The height of the isosceles triangle side of the pyramid is labeled $n$ above, so we will call the slant height $n$ for this problem.

The slant height is the same as the _________________________ of the triangular side.

The triangular side of the pyramid is drawn again below:

We use the Pythagorean Theorem (with only half of the triangle, as shown on the right above) to find the height of the triangle:

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

Now find the area of each triangular side (using the formula for area of a triangle):

$A &= \frac{1}{2} bh \qquad \text{where}\ b = 12 \ \text{and}\ h = n = 10\\&= \frac{1}{2} (12)(10)\\&= 60 \ \text{square units}$

Since there are 4 triangular sides of the pyramid:

$\text{Area of all 4 triangles} &= 4(60)\\&= 240 \ \text{square units}$

Finally, add the total area of the 4 triangles to the area of the base:

$\text{Total area} &= \text{Area of the triangular sides} + \text{Area of the base}\\&= 240 + 144\\&= 384\ \text{square units}$

Review the steps for finding the surface area of a regular pyramid:

1. Find the area of the base of the pyramid.
2. Find the slant height of the triangular side of the pyramid.
3. Find the area of each triangular side and multiply it by the number of sides.
4. Add the area of the sides to the area of the base for total area!

Example 1

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

The base of the pyramid is a hexagon, which is a 6-sided figure. 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 cm, the perimeter is $6 \cdot 6$ or 36 cm.

The apothem, or perpendicular distance to the center of the hexagon, is 5.19 cm.

(Step #1)

$\text{Area of base} &= \frac{1}{2} (\text{perimeter})(\text{apothem})\\&= \frac{1}{2} ( \underline{\;\;\;\;\;\;\;\;\;\;\;} )( \underline{\;\;\;\;\;\;\;\;\;\;\;} ) \\&= 93.42 \ \text{square cm}$

Use the Pythagorean Theorem to find the slant height of each lateral triangle

(Step #2)

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

Now find the area of each triangle:

(Step #3)

$A &= \frac{1}{2} bh \qquad \text{where}\ b = 6 \ \text{and}\ h = n = 13.67\\&= \frac{1}{2} ( \underline{\;\;\;\;\;\;\;\;\;\;\;} )( \underline{\;\;\;\;\;\;\;\;\;\;\;} )\\&= 41 \ \text{ square cm}$

Together, the area of all 6 triangles that make up the lateral sides of the pyramid are:

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

Add the area of the 6 lateral triangular sides to the area of the hexagonal base

(Step #4)

$\text{Total area} &= \text{Area of the triangular sides} + \text{Area of the base}\\&= 246 + 93.42\\&= 339.42 \ \text{square cm}$

1. What is the slant height of a pyramid?

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2. What are the 4 steps for finding the surface area of a regular pyramid? Describe them in your own words.

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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 base is labeled $s$. For each lateral triangle, the area is:

$A = \frac{1}{2}\ sl$

There are $n$ triangles in a regular pyramid (for example, $n = 3$ for a triangular pyramid, $n = 4$ for a square pyramid, $n = 5$ for a pentagonal pyramid, and so on.)

The total area, $L$, of the lateral triangles is:

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

If we re-arrange the above equation we get:

$L = \frac{1}{2}\ nsl$

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

$L = \frac{1}{2}\ Pl$

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

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

Area of a Regular Pyramid

The surface area of a regular pyramid is:

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

Where $P$ is the perimeter of the regular polygon that forms the pyramid’s base, $l$ is the slant height of the pyramid, 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 feet. The sides of the hexagonal perimeter of the figure each measure 8 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:

$A &= \frac{1}{2}\ Pl + B\\&= \frac{1}{2}\ Pl + 0$

Now substitute the given information to find the perimeter and the given slant height:

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

In the formula for the surface area of a regular pyramid:

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

$A$ stands for the _____________________ ________________ of the pyramid,

$P$ stands for the ____________________________ of the base,

$l$ stands for the ________________ __________________ of the pyramid, and

$B$ stands for the ___________________ of the _________________.

Example 3

A pentagonal pyramid has a slant height $l$ of 12 cm. The sides of the pentagonal perimeter of the figure each measure 9 cm. The apothem of the figure is 6 cm.

Find the surface area of the figure.

If the base is a pentagon (which has 5 sides), and each side is 9 cm, then the perimeter of the base is ($5 \cdot 9$) or 45 cm.

First find the lateral area of the figure:

$L &= \frac{1}{2}\ Pl\\&= \frac{1}{2} (45)(12)\\&= 270 \ cm^2$

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

$A &= \frac{1}{2} \ \text{(perimeter)(apothem)}\\&= \frac{1}{2} (45)(6)\\&= 135 \ cm^2$

Finally, add these values together to find the total surface area:

$\text{Total area} &= \text{Lateral area} + \text{Base area}\\&= \underline{\;\;\;\;\;\;\;\;\;\;\;\;} cm^2 + \underline{\;\;\;\;\;\;\;\;\;\;\;\;} cm^2\\&= 405 \ cm^2$

Total surface area of a pyramid $= \ \text{Lateral} \underline{\;\;\;\;\;\;\;\;\;\;\;\;} + \underline{\;\;\;\;\;\;\;\;\;\;\;\;} \ \text{area}$

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Date Created:

Feb 23, 2012

May 12, 2014
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