Geometry Worksheets | Volume Worksheets - Free Printable

Of course! Let's solve each problem step by step. The worksheet asks for the volume of pyramids and cones, rounded to the nearest hundredth if necessary.

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🔷 General Formulas:

- Volume of a Pyramid = $\frac{1}{3} \times \text{Base Area} \times \text{Height}$
- Volume of a Cone = $\frac{1}{3} \pi r^2 h$

We’ll apply these formulas to each figure.

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## ✔ Problem 1: Square Pyramid

- Base: square with side = 3 ft
- Height = 11 ft

Base Area = $3 \times 3 = 9$ ft²
Volume = $\frac{1}{3} \times 9 \times 11 = 3 \times 11 = 33$ ft³

> Answer: 33.00 ft³

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## ✔ Problem 2: Rectangular Pyramid

- Base: rectangle with sides 8 cm and 2 cm → area = $8 \times 2 = 16$ cm²
- Height = 10 cm (given as slant height? Wait — check diagram!)

⚠️ Important Note: In pyramid diagrams, if a dimension is drawn from apex to base center (perpendicular), that’s the height. If it’s along the face, it’s slant height.

In problem 2, the 10 cm is labeled on the triangular face — this is slant height, not vertical height. But we need vertical height to compute volume.

However, looking at the diagram, the 10 cm is drawn perpendicular to the base edge? Actually, in many worksheets, if no right triangle is shown, and only one height is given, they often intend it as the vertical height.

But here, the 10 cm is drawn along the lateral face — so it’s slant height. We need to find the actual height using Pythagoras.

Let’s assume the base is 8 cm by 2 cm. The distance from the center of the base to the middle of the 8 cm side is half of 2 cm = 1 cm.

So, we have a right triangle:
- Slant height = 10 cm (hypotenuse)
- Base leg = 1 cm (half the shorter side)
- Vertical height = ?

$h = \sqrt{10^2 - 1^2} = \sqrt{100 - 1} = \sqrt{99} ≈ 9.95$ cm

Base Area = $8 \times 2 = 16$ cm²
Volume = $\frac{1}{3} \times 16 \times 9.95 ≈ \frac{1}{3} \times 159.2 ≈ 53.07$ cm³

> Answer: 53.07 cm³

*(Note: Some teachers may expect you to treat 10 cm as vertical height. If so, answer would be $\frac{1}{3} \times 16 \times 10 = 53.33$. But strictly speaking, since it’s drawn on the face, it’s slant height. We’ll go with the more accurate 53.07.)*

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## ✔ Problem 3: Triangular Pyramid (Tetrahedron)

- Base: triangle with base 12 cm, height 6 cm → area = $\frac{1}{2} \times 12 \times 6 = 36$ cm²
- Height of pyramid = 10 cm (drawn perpendicular from apex to base)

Volume = $\frac{1}{3} \times 36 \times 10 = 12 \times 10 = 120$ cm³

> Answer: 120.00 cm³

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## ✔ Problem 4: Rectangular Pyramid

- Base: 10 mm × 4 mm → area = $10 \times 4 = 40$ mm²
- Height = 12 mm (assuming vertical height — drawn from apex to base center)

Volume = $\frac{1}{3} \times 40 \times 12 = \frac{480}{3} = 160$ mm³

> Answer: 160.00 mm³

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## ✔ Problem 5: Rectangular Pyramid

- Base: 10 in × 4 in → area = $40$ in²
- Height = 12 in (drawn perpendicular)

Volume = $\frac{1}{3} \times 40 \times 12 = 160$ in³

> Answer: 160.00 in³

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## ✔ Problem 6: Triangular Pyramid

- Base: triangle with base 5 in, height 11 in → area = $\frac{1}{2} \times 5 \times 11 = 27.5$ in²
- Height of pyramid = 3 in (given as perpendicular height)

Volume = $\frac{1}{3} \times 27.5 \times 3 = 27.5$ in³

> Answer: 27.50 in³

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## ✔ Problem 7: Square Pyramid

- Base: square with side 5 mm → area = $25$ mm²
- Height = 13 mm

Volume = $\frac{1}{3} \times 25 \times 13 = \frac{325}{3} ≈ 108.33$ mm³

> Answer: 108.33 mm³

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## ✔ Problem 8: Cone

- Radius = 4 yd
- Height = 11 yd

Volume = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (4)^2 (11) = \frac{1}{3} \pi (16)(11) = \frac{176\pi}{3}$

≈ $\frac{176 \times 3.1416}{3} ≈ \frac{552.92}{3} ≈ 184.31$ yd³

> Answer: 184.31 yd³

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## ✔ Problem 9: Cone

- Radius = 2 yd
- Height = 10 yd

Volume = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (2)^2 (10) = \frac{1}{3} \pi (4)(10) = \frac{40\pi}{3}$

≈ $\frac{40 \times 3.1416}{3} ≈ \frac{125.664}{3} ≈ 41.89$ yd³

> Answer: 41.89 yd³

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## 📝 Final Answers Summary:

| Problem | Volume |
|---------|---------------------|
| 1 | 33.00 ft³ |
| 2 | 53.07 cm³ |
| 3 | 120.00 cm³ |
| 4 | 160.00 mm³ |
| 5 | 160.00 in³ |
| 6 | 27.50 in³ |
| 7 | 108.33 mm³ |
| 8 | 184.31 yd³ |
| 9 | 41.89 yd³ |

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✔ All answers are rounded to the nearest hundredth where needed.

Let me know if you want to see diagrams or step-by-step calculations for any specific problem!