10-Volume of Pyramids and Cones - Kuta Software - Free Printable

This is a worksheet on finding the volume of pyramids and cones. To solve it, I need to use the correct formulas for each shape and apply them to the given dimensions.

The general formula for the volume of a pyramid or cone is:
$V = \frac{1}{3} \times \text{Base Area} \times \text{Height}$

For a cone, the base is a circle, so the formula becomes:
$V = \frac{1}{3} \pi r^2 h$

For a pyramid, the base can be any polygon, so I'll need to calculate the area of that specific base (e.g., triangle, square, pentagon) and then multiply by the height and 1/3.

I will now solve each problem one by one.

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Problem 1: Cone

- Radius ($r$) = 2 mi
- Height ($h$) = 7 mi

Using the cone volume formula:
$V = \frac{1}{3} \pi r^2 h$
$V = \frac{1}{3} \pi (2)^2 (7)$
$V = \frac{1}{3} \pi (4)(7)$
$V = \frac{1}{3} \pi (28)$
$V = \frac{28\pi}{3} \approx \frac{28 \times 3.1416}{3} \approx \frac{87.9648}{3} \approx 29.32$

Rounded to the nearest tenth: 29.3 mi³

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Problem 2: Triangular Pyramid

- Base: Triangle with base = 5 mi and height = 4 mi
- Pyramid height = 3 mi

First, find the area of the triangular base:
$\text{Base Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 4 = 10 \text{ mi}^2$

Now, use the pyramid volume formula:
$V = \frac{1}{3} \times \text{Base Area} \times \text{Height} = \frac{1}{3} \times 10 \times 3 = 10$

Rounded to the nearest tenth: 10.0 mi³

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Problem 3: Square Pyramid

- Base: Square with side length = 11 cm
- Pyramid height = 12 cm

First, find the area of the square base:
$\text{Base Area} = \text{side}^2 = 11^2 = 121 \text{ cm}^2$

Now, use the pyramid volume formula:
$V = \frac{1}{3} \times \text{Base Area} \times \text{Height} = \frac{1}{3} \times 121 \times 12$

Calculate:
$V = \frac{1}{3} \times 1452 = 484$

Rounded to the nearest tenth: 484.0 cm³

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Problem 4: Rectangular Pyramid

- Base: Rectangle with sides = 2 in and 5 in
- Pyramid height = 5 in

First, find the area of the rectangular base:
$\text{Base Area} = \text{length} \times \text{width} = 2 \times 5 = 10 \text{ in}^2$

Now, use the pyramid volume formula:
$V = \frac{1}{3} \times \text{Base Area} \times \text{Height} = \frac{1}{3} \times 10 \times 5 = \frac{50}{3} \approx 16.666...$

Rounded to the nearest tenth: 16.7 in³

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Problem 5: Pentagonal Pyramid

- Base: Regular pentagon with side length = 8.3 yd
- Apothem (distance from center to midpoint of a side) = 11 yd (Note: The diagram shows "11 yd" as the apothem)
- Pyramid height = 12 yd

First, find the area of the pentagonal base. The area of a regular polygon is:
$\text{Base Area} = \frac{1}{2} \times \text{Perimeter} \times \text{Apothem}$

Perimeter of the pentagon = $5 \times 8.3 = 41.5 \text{ yd}$

So,
$\text{Base Area} = \frac{1}{2} \times 41.5 \times 11 = \frac{1}{2} \times 456.5 = 228.25 \text{ yd}^2$

Now, use the pyramid volume formula:
$V = \frac{1}{3} \times \text{Base Area} \times \text{Height} = \frac{1}{3} \times 228.25 \times 12$

Calculate:
$V = \frac{1}{3} \times 2739 = 913$

Rounded to the nearest tenth: 913.0 yd³

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Problem 6: Hexagonal Pyramid

- Base: Regular hexagon with side length = 5.2 m
- Apothem = 5 m (Note: The diagram shows "5 m" as the apothem)
- Pyramid height = 9 m

First, find the area of the hexagonal base. The area of a regular polygon is:
$\text{Base Area} = \frac{1}{2} \times \text{Perimeter} \times \text{Apothem}$

Perimeter of the hexagon = $6 \times 5.2 = 31.2 \text{ m}$

So,
$\text{Base Area} = \frac{1}{2} \times 31.2 \times 5 = \frac{1}{2} \times 156 = 78 \text{ m}^2$

Now, use the pyramid volume formula:
$V = \frac{1}{3} \times \text{Base Area} \times \text{Height} = \frac{1}{3} \times 78 \times 9$

Calculate:
$V = \frac{1}{3} \times 702 = 234$

Rounded to the nearest tenth: 234.0 m³

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Final Answers:

1. 29.3 mi³
2. 10.0 mi³
3. 484.0 cm³
4. 16.7 in³
5. 913.0 yd³
6. 234.0 m³