Volume of Pyramids, Cones, and Spheres worksheet with problems to calculate volumes and find missing dimensions.

Let's solve each part of the problem step by step.

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Section A: Volume of Geometric Solids

We'll use the following formulas:

- Pyramid:
$$
V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
$$

- Cone:
$$
V = \frac{1}{3} \pi r^2 h
$$

- Sphere:
$$
V = \frac{4}{3} \pi r^3
$$

Use $ \pi = 3.14 $ where needed.

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#### a) Triangular Pyramid (Tetrahedron)

Given:
- Base is a triangle with base = 6 cm, height = 7 cm (from diagram)
- Height of pyramid = 5 cm

Step 1: Find base area
$$
\text{Base Area} = \frac{1}{2} \times 6 \times 7 = 21\ \text{cm}^2
$$

Step 2: Volume
$$
V = \frac{1}{3} \times 21 \times 5 = 35\ \text{cm}^3
$$

✔ Volume = 35 cm³

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#### b) Cone

Given:
- Diameter = 10 cm → Radius $ r = 5 $ cm
- Height $ h = 12 $ cm

$$
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times 3.14 \times 5^2 \times 12
$$
$$
= \frac{1}{3} \times 3.14 \times 25 \times 12 = 3.14 \times 25 \times 4 = 314\ \text{cm}^3
$$

✔ Volume = 314 cm³

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#### c) Square-based Pyramid

Given:
- Base side = 3 cm → Area = $ 3 \times 3 = 9 $ cm²
- Height = 4 cm

$$
V = \frac{1}{3} \times 9 \times 4 = 12\ \text{cm}^3
$$

✔ Volume = 12 cm³

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#### d) Rectangular-based Pyramid

Given:
- Base: 8 cm × 11 cm → Area = $ 8 \times 11 = 88 $ cm²
- Height = 3 cm

$$
V = \frac{1}{3} \times 88 \times 3 = 88\ \text{cm}^3
$$

✔ Volume = 88 cm³

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#### e) Cone

Given:
- Diameter = 6 cm → Radius $ r = 3 $ cm
- Height $ h = 11 $ cm

$$
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times 3.14 \times 3^2 \times 11
$$
$$
= \frac{1}{3} \times 3.14 \times 9 \times 11 = 3.14 \times 3 \times 11 = 3.14 \times 33 = 103.62\ \text{cm}^3
$$

✔ Volume = 103.62 cm³

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#### f) Square-based Pyramid

Given:
- Base side = 8 cm → Area = $ 8 \times 8 = 64 $ cm²
- Height = 6 cm

$$
V = \frac{1}{3} \times 64 \times 6 = 128\ \text{cm}^3
$$

✔ Volume = 128 cm³

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#### g) Sphere

Given:
- Diameter = 8 cm → Radius $ r = 4 $ cm

$$
V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.14 \times 4^3 = \frac{4}{3} \times 3.14 \times 64
$$
$$
= \frac{4}{3} \times 200.96 = 267.946... \approx 267.95\ \text{cm}^3
$$

✔ Volume = 267.95 cm³ (to 2 d.p.)

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#### h) Sphere

Given:
- Diameter = 6 cm → Radius $ r = 3 $ cm

$$
V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.14 \times 27 = \frac{4}{3} \times 84.78 = 113.04\ \text{cm}^3
$$

✔ Volume = 113.04 cm³

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✔ Section A Summary:

| Letter | Volume (cm³) |
|--------|---------------|
| a) | 35 |
| b) | 314 |
| c) | 12 |
| d) | 88 |
| e) | 103.62 |
| f) | 128 |
| g) | 267.95 |
| h) | 113.04 |

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Section B: Find Missing Length

#### a) Cone with volume 270 cm³, height = 15 cm, find radius $ x $

Use:
$$
V = \frac{1}{3} \pi r^2 h
$$
Given:
- $ V = 270 $
- $ h = 15 $
- $ \pi = 3.14 $

$$
270 = \frac{1}{3} \times 3.14 \times r^2 \times 15
$$
$$
270 = 3.14 \times 5 \times r^2 = 15.7 \times r^2
$$
$$
r^2 = \frac{270}{15.7} \approx 17.197
$$
$$
r = \sqrt{17.197} \approx 4.147\ \text{cm}
$$

Round to 3 significant figures:
✔ x ≈ 4.15 cm

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✔ Final Answers

#### Section A:
a) 35
b) 314
c) 12
d) 88
e) 103.62
f) 128
g) 267.95
h) 113.04

#### Section B:
a) $ x = 4.15 $ cm

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