Math worksheet for calculating the volume of pyramids and cones with diagrams and word problems.

Here is the complete solution to the worksheet “Volume of Pyramids and Cones”, with clear explanations for each problem.

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## 🔷 SECTION A: Calculate the volume of the following pyramids and cones.

✔ General Formulas:

- Volume of a Pyramid:
\( V = \frac{1}{3} \times \text{Base Area} \times \text{Height} \)

- Volume of a Cone:
\( V = \frac{1}{3} \pi r^2 h \)

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1) Square Pyramid

- Base: square, side = 5 cm → Base Area = \( 5 \times 5 = 25 \, \text{cm}^2 \)
- Height = 8 cm

\[
V = \frac{1}{3} \times 25 \times 8 = \frac{200}{3} \approx 66.67 \, \text{cm}^3
\]

✔ Answer: 66.67 cm³

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2) Rectangular Pyramid

- Base: rectangle, 7 cm × 4 cm → Base Area = \( 7 \times 4 = 28 \, \text{cm}^2 \)
- Height = 12 cm

\[
V = \frac{1}{3} \times 28 \times 12 = \frac{336}{3} = 112.00 \, \text{cm}^3
\]

✔ Answer: 112.00 cm³

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3) Triangular Pyramid (Tetrahedron)

This is a pyramid with a triangular base. The base is a right triangle with legs 9 cm and 11 cm.

- Base Area = \( \frac{1}{2} \times 9 \times 11 = 49.5 \, \text{cm}^2 \)
- Height = 15.2 cm (given as vertical height from apex to base)

\[
V = \frac{1}{3} \times 49.5 \times 15.2 = \frac{752.4}{3} = 250.80 \, \text{cm}^3
\]

> Note: The 19.4 cm is the slant edge — not needed since we’re given the perpendicular height.

✔ Answer: 250.80 cm³

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4) Cone

- Radius \( r = 4.5 \, \text{cm} \)
- Height \( h = 13 \, \text{cm} \)

\[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (4.5)^2 (13)
\]

First, compute \( (4.5)^2 = 20.25 \)

Then: \( 20.25 \times 13 = 263.25 \)

Then: \( \frac{1}{3} \times 263.25 = 87.75 \)

Finally: \( V = 87.75 \pi \approx 87.75 \times 3.1416 \approx 275.73 \, \text{cm}^3 \)

✔ Answer: 275.73 cm³

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5) Cone

- Diameter = 21 cm → Radius \( r = 10.5 \, \text{cm} \)
- Height = 18 cm

\[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (10.5)^2 (18)
\]

\( (10.5)^2 = 110.25 \)

\( 110.25 \times 18 = 1984.5 \)

\( \frac{1}{3} \times 1984.5 = 661.5 \)

\( V = 661.5 \pi \approx 661.5 \times 3.1416 \approx 2078.17 \, \text{cm}^3 \)

✔ Answer: 2078.17 cm³

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6) Cone

- Diameter = 15 cm → Radius \( r = 7.5 \, \text{cm} \)
- Slant height = 25 cm → We need height, not slant height!

Use Pythagoras:

\[
h = \sqrt{25^2 - 7.5^2} = \sqrt{625 - 56.25} = \sqrt{568.75} \approx 23.85 \, \text{cm}
\]

Now compute volume:

\[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (7.5)^2 (23.85)
\]

\( (7.5)^2 = 56.25 \)

\( 56.25 \times 23.85 ≈ 1341.5625 \)

\( \frac{1}{3} \times 1341.5625 ≈ 447.1875 \)

\( V ≈ 447.1875 \pi ≈ 447.1875 \times 3.1416 ≈ 1404.85 \, \text{cm}^3 \)

✔ Answer: 1404.85 cm³

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## 🔷 SECTION B: Solve for missing dimensions.

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1) Find the height of a cone with volume 200 cm³ and diameter 10 cm.

- Diameter = 10 cm → Radius \( r = 5 \, \text{cm} \)
- Volume \( V = 200 \, \text{cm}^3 \)

Use formula:

\[
V = \frac{1}{3} \pi r^2 h
\]

Plug in values:

\[
200 = \frac{1}{3} \pi (5)^2 h = \frac{1}{3} \pi \cdot 25 \cdot h = \frac{25\pi}{3} h
\]

Solve for \( h \):

\[
h = \frac{200 \times 3}{25\pi} = \frac{600}{25\pi} = \frac{24}{\pi} \approx \frac{24}{3.1416} \approx 7.64 \, \text{cm}
\]

✔ Answer: 7.64 cm

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2) A cone is 7.3 cm high. It has a volume of 16 liters. Calculate the radius.

- First, convert liters to cm³:
\( 16 \, \text{liters} = 16 \times 1000 = 16000 \, \text{cm}^3 \)
- Height \( h = 7.3 \, \text{cm} \)

Use formula:

\[
V = \frac{1}{3} \pi r^2 h
\]

\[
16000 = \frac{1}{3} \pi r^2 (7.3)
\]

Multiply both sides by 3:

\[
48000 = \pi r^2 \cdot 7.3
\]

Divide both sides by \( 7.3\pi \):

\[
r^2 = \frac{48000}{7.3 \pi} \approx \frac{48000}{22.9336} \approx 2093.00
\]

Take square root:

\[
r \approx \sqrt{2093.00} \approx 45.75 \, \text{cm}
\]

✔ Answer: 45.75 cm

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## 📝 FINAL ANSWERS SUMMARY:

Section A:

1) 66.67 cm³
2) 112.00 cm³
3) 250.80 cm³
4) 275.73 cm³
5) 2078.17 cm³
6) 1404.85 cm³

Section B:

1) 7.64 cm
2) 45.75 cm

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