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# Surface Area and Volume of Pyramids

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# Surface Area and Volume of Pyramids

After about a week of working at the wrapping station, both Candice and Trevor were beginning to feel at ease. Neither one of them didn’t think that there was a package that was too difficult to wrap. They had wrapped lots of regular shaped boxes and also some odd shaped things too.

But after a week of wrapping, they felt like experts. Then one Saturday morning, an interesting dilemma came their way. A girl came up to the booth with a box in one hand.

“I wonder if you can help me,” she said. “I bought my brother a model pyramid and it needs paint. But the paint isn’t included. I need to know how much paint I’ll need.”

“Well, to figure that out, you need the dimensions of the pyramid to the amount of surface area that the paint will need to cover. Let me see the box and I’ll see if I can help you,” Candice said taking the box from the girl.

The box said that once the pyramid is built, that it will have a base of 7.7 cm, a height of 6.2 cm and a slant height of 7.3 cm.

“Okay, I know what to do,” Candice said smiling as she took out a piece of a paper and a calculator.

Do you know what to do? How can you use these measurements to find the surface area of the pyramid?

In this Concept you will learn how to find the surface area of a pyramid. Learn as you go through this Concept and at the end you can check out how Candice finds the surface area of the pyramid model.

### Guidance

A pyramid has sides that are triangular faces and a base. The base can be any shape. Let’s look at some pyramids.

Like pyramids, cones have a base and a point at the top. However, cones always have a circular base.

Surface area is the sum of all of the areas of the faces in a solid figure. Imagine you could wrap a pyramid or cone in wrapping paper, like a present. The amount of wrapping paper needed to cover the figure represents its surface area. To find the surface area, we must be able to calculate the area of each face and then add these areas together.

One way to find the surface area of a three-dimensional figure is using a net. A net is a two-dimensional diagram of a three-dimensional figure. Imagine you could unfold a pyramid so that it is completely flat. It might look something like this net.

With a net, we can see each face of the pyramid more clearly. To find the surface area, we need to calculate the area for each face in the net, the sides and the base.

The side faces of a pyramid are always triangles, so we use the area formula for triangles to calculate their area: $A = \frac{1}{2} bh$ .

The area of the base depends on what shape it is. Remember, pyramids can have bases in the shape of a triangle, square, rectangle, or any other polygon. We use whichever area formula is appropriate for the shape. Here are some common formulas to find area.

$\text{rectangle}: \quad \qquad A & = lw\\\text{square}: \qquad \qquad A & = s^2\\\text{triangle}: \quad \qquad \ A & = \frac{1}{2} bh$

Take a look at the pyramid below. Which formula should we use to find the area of the base?

The base of the pyramid is a square with sides of 6 centimeters, so we should use the area formula for a square: $A = s^2$ . Now we can find the area of the base.

$A & = 6^2 \\A&=36 \ sq.cm$

Next, we need to find the area of one of the triangles. There are four triangles, but we can work by finding the area of one of the triangles. We use the formula $A= \frac{1}{2} bh$ . The base width is 6 cm and the slant height, the height of the side is 4 cm.

$A & = \frac{1}{2} (6)(4) \\A & = \frac{1}{2}(24) \\A & = 12 \ sq.cm$

There are four triangles, so we can take this number and multiply it by four and then add it to the area of the square.

$SA & =4(12)+ 36 \\SA & =48+36 \\SA & =84 \ sq.cm$ Nets let us see each face of a pyramid so that we can calculate its area. However, we can also use a formula to represent the faces as we find their area. The formula is like a short cut, because we can put the measurements in for the appropriate variable in the formula and solve for $SA$ , surface area.

Here is the formula for finding the surface area of a pyramid.

$SA = \frac{1}{2} \ \text{perimeter} \times \text{slant height} + B$

The first part of the formula, $\frac{1}{2} \ \text{perimeter} \times \text{slant height}$ , is a quick way of finding the area of all of the triangular sides of the pyramid at once. Remember, the area formula for a triangle is $A = \frac{1}{2} bh$ . In the formula, $b$ stands for base. The perimeter of the pyramid’s bottom face represents all of the bases of the triangular faces at once, because it’s their sum. The height of each triangle is always the same, so we can just call this the slant height of the pyramid.

Therefore “ $\frac{1}{2} \ \text{perimeter} \times \text{slant height}$ ” is really the same as $\frac{1}{2} bh$ .

The $B$ in the formula represents the base’s area. Remember, pyramids can have bases of different shapes, so the area formula we use to find $B$ varies. We find the base’s area first and then put it into the formula in place of $B$ . Let’s give it a try.

What is the surface area of the pyramid below?

This is a square pyramid. The four sides of the base are all 8 inches, so the perimeter of the base is $8 \times 4 = 32 \ inches$ . We also know we will need to use the area formula for squares to find $B$ , the base’s area.

$B & = s^2\\B & = (8)^2\\B & = 64 \ in.^2$

Now we have all the information that we need. We can put it into the formula and solve for $SA$ , surface area.

$SA & = \frac{1}{2} \ \text{perimeter} \times \text{slant height} + B\\SA & = \left [ \frac{1}{2} (32) \times 3 \right ] + 64\\SA & = (16 \times 3) + 64\\SA & = 48 + 64\\SA & = 112 \ in.^2$

The surface area of the pyramid is 112 square inches.

Here is another one.

Find the surface area of the figure below.

First of all, what kind of pyramid is this? It is a triangular pyramid because its base is a triangle. That means we need to use the area formula for triangles to find $B$ . The base’s sides are all the same length, so we can calculate the perimeter by multiplying $16 \times 3 = 48$ . Now let’s find $B$ .

$B & = \frac{1}{2} bh\\B & = \frac{1}{2} (16) (13.86)\\B & = 8 (13.86)\\B & = 110.88 \ cm^2$

Now we’re ready to put all of the information into the formula. Let’s see what happens.

$SA & = \frac{1}{2} \ \text{perimeter} \times \text{slant height} + B \\SA & = \left [ \frac{1}{2} (48) \times 6 \right ] + 110.88\\SA & = (24 \times 6) + 110.88\\SA & = 144 + 110.88\\SA & = 254.88 \ cm^2$

The surface area of this triangular pyramid is 254.88 square centimeters.

Write this formula down in your notebook.

Try a few of these on your own.

Find the surface area of each pyramid.

#### Example A

A square pyramid with side of 6 cm, slant height of 5 cm

Solution: $96 \ cm^2$

#### Example B

A rectangular pyramid with a length of 6 in, a width of 4 in and a slant height of 3 in

Solution: $54 \ in^2$

#### Example C

A square pyramid with side of 8 in, slant height of 9 in

Solution: $208 \ in^2$

Here is the original problem once again.

After about a week of working at the wrapping station, both Candice and Trevor were beginning to feel at ease. Neither one of them didn’t think that there was a package that was too difficult to wrap. They had wrapped lots of regular shaped boxes and also some odd shaped things too.

But after a week of wrapping, they felt like experts. Then one Saturday morning, an interesting dilemma came their way. A girl came up to the booth with a box in one hand.

“I wonder if you can help me,” she said. “I bought my brother a model pyramid and it needs paint. But the paint isn’t included. I need to know how much paint I’ll need.”

“Well, to figure that out, you need the dimensions of the pyramid to the amount of surface area that the paint will need to cover. Let me see the box and I’ll see if I can help you,” Candice said taking the box from the girl.

The box said that once the pyramid is built, that it will have a base of 7.7 cm, a height of 6.2 cm and a slant height of 7.3 cm.

“Okay, I know what to do,” Candice said smiling as she took out a piece of a paper and a calculator.

First, Candice needs to find the surface area of the base. She can use this formula.

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

$A = \frac{1}{2} (7.7)(6.2)$

$A = 23.87 \ sq. cm$

Now she can use the formula for surface area.

$SA = \frac{1}{2}P \times sh + B$

$SA = \frac{1}{2} (23.1) \times 7.3 + 23.87$

$SA = 84.315 + 23.87$

$SA = 108.19 \ cm^2$

The surface area of the pyramid is 108.19 sq. cm.

“You need to find paint that stretches across 108.19 sq. cm. That will give him just enough. Try the hobby shop. They will help you,” Candice said writing the figure on a piece of paper as she handed it to the girl.

The girl smiled and thanked her. Candice felt good about her ability to use math to figure out the problem.

### Vocabulary

Pyramid
a three-dimensional object with a polygon for a base and triangles for sides that join at a single vertex.
Surface area
the measurement of the outer covering or surface of a three-dimensional figure.

### Guided Practice

Here is one for you to try on your own.

Find the surface area of the following triangular pyramid.

Base of 8 cm, height of 7 cm, and a slant height of 6 cm.

First, we need to find the surface area of each of the four lateral faces. She can use this formula.

$A & = \frac{1}{2} bh\\A & = 4 \left [ \frac{1}{2} (8)(7) \right ]\\A & = 112 \ sq. cm$

Now she can find the area of the base.

$A & = s^2\\A & = 8(8)\\A & = 64 \ sq. cm$

The surface area of the pyramid is 176 sq. cm.

### Practice

Directions: Find the surface area of each square pyramid. Remember that $b$ means base and $sh$ means slant height.

1. $b = 4 \ in, \ sh = 5 \ in$

2. $b = 4 \ in, \ sh = 6 \ in$

3. $b = 6 \ in, \ sh = 8 \ in$

4. $b = 5 \ in, \ sh = 7 \ in$

5. $b = 7 \ m, \ sh = 9 \ m$

6. $b = 8 \ m, \ sh = 10 \ m$

7. $b = 9 \ cm, \ sh = 11 \ cm$

8. $b = 11 \ m, \ sh = 13 \ m$

9. $b = 6 \ in, \ sh = 9 \ in$

10. $b = 10 \ cm, \ sh = 12 \ cm$

Directions : Find the surface area of each rectangular pyramid.

11. $l = 10 \ in, w = 8 \ in, sh = 6 \ in$

12. $l = 12 \ ft, w = 8 \ ft, sh = 10 \ ft$

13. $l = 4 \ in, w = 2 \ in, sh = 3 \ in$

14. $l = 16 \ m, w = 12 \ m, sh = 11 \ m$

15. $l = 15 \ in, w = 10 \ in, sh = 12 \ in$