Geometry EOC worksheet on volume of pyramids, cones, and spheres with diagrams and measurements.

It looks like you've uploaded a worksheet titled "Volume of Pyramids, Cones and Spheres", containing six problems involving 3D geometric shapes. Let's go through each problem one by one, solve them step by step, and explain the reasoning.

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🔷 Problem 1: Pyramid (Triangular Base)

Given:
- Base triangle with base = 6 cm, height = 4 cm
- Height of pyramid = 9 cm

We are to find the volume of a pyramid.

Formula for volume of a pyramid:
\[
V = \frac{1}{3} \times \text{Base Area} \times \text{Height}
\]

First, calculate the area of the triangular base:
\[
\text{Base Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 4 = 12 \text{ cm}^2
\]

Now use the pyramid volume formula:
\[
V = \frac{1}{3} \times 12 \times 9 = 36 \text{ cm}^3
\]

✔ Answer: 36 cm³

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🔷 Problem 2: Cone

Given:
- Radius \( r = 5 \) m
- Slant height = 13 m
- Height is not directly given, but we can find it using the Pythagorean Theorem

In a cone:
\[
\text{Slant height}^2 = \text{radius}^2 + \text{height}^2
\]
\[
13^2 = 5^2 + h^2 \Rightarrow 169 = 25 + h^2 \Rightarrow h^2 = 144 \Rightarrow h = 12 \text{ m}
\]

Now use the volume of a cone formula:
\[
V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (5)^2 (12) = \frac{1}{3} \pi (25)(12) = 100\pi \text{ m}^3
\]

Using \( \pi \approx 3.14 \):
\[
V \approx 100 \times 3.14 = 314 \text{ m}^3
\]

✔ Answer: \( 100\pi \) m³ or approximately 314 m³

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🔷 Problem 3: Square Pyramid

Given:
- Base side = 11 cm
- Height of pyramid = 12 cm

Base area (square):
\[
A = 11 \times 11 = 121 \text{ cm}^2
\]

Volume:
\[
V = \frac{1}{3} \times 121 \times 12 = \frac{1}{3} \times 1452 = 484 \text{ cm}^3
\]

✔ Answer: 484 cm³

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🔷 Problem 4: Sphere

Given:
- Diameter = 3 ft → radius \( r = \frac{3}{2} = 1.5 \) ft

Volume of a sphere:
\[
V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1.5)^3 = \frac{4}{3} \pi (3.375) = 4.5\pi \text{ ft}^3
\]

Approximate:
\[
4.5 \times 3.14 \approx 14.13 \text{ ft}^3
\]

✔ Answer: \( 4.5\pi \) ft³ or approximately 14.13 ft³

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🔷 Problem 5: Square Pyramid (with slant height)

Given:
- Base side = 8.5 yd
- Slant height = 11 yd
- But height is not given — need to find it.

This is a square pyramid. The slant height is the distance from the midpoint of a base edge to the apex.

We can use the right triangle formed by:
- Half the base: \( \frac{8.5}{2} = 4.25 \) yd
- Slant height = 11 yd
- Height of pyramid = \( h \)

Use Pythagorean Theorem:
\[
\text{Slant height}^2 = h^2 + \left(\frac{\text{base}}{2}\right)^2
\]
\[
11^2 = h^2 + 4.25^2 \Rightarrow 121 = h^2 + 18.0625 \Rightarrow h^2 = 102.9375 \Rightarrow h \approx \sqrt{102.9375} \approx 10.145 \text{ yd}
\]

Now compute volume:
\[
\text{Base area} = 8.5 \times 8.5 = 72.25 \text{ yd}^2
\]
\[
V = \frac{1}{3} \times 72.25 \times 10.145 \approx \frac{1}{3} \times 732.1 \approx 244.03 \text{ yd}^3
\]

✔ Answer: Approximately 244 yd³

*(Note: This answer depends on rounding; exact value may vary slightly.)*

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🔷 Problem 6: Sphere

Given:
- Diameter = 12 cm → radius \( r = 6 \) cm

Volume:
\[
V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \pi (216) = 288\pi \text{ cm}^3
\]

Approximate:
\[
288 \times 3.14 \approx 904.32 \text{ cm}^3
\]

✔ Answer: \( 288\pi \) cm³ or approximately 904.32 cm³

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | 36 cm³ |
| 2 | \( 100\pi \) m³ ≈ 314 m³ |
| 3 | 484 cm³ |
| 4 | \( 4.5\pi \) ft³ ≈ 14.13 ft³ |
| 5 | ≈ 244 yd³ |
| 6 | \( 288\pi \) cm³ ≈ 904.32 cm³ |

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