Problems involving the volume of cones with varying known dimensions and variables.

Let's solve each of these cone volume problems step by step.

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🔷 Formula for the Volume of a Cone:

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

Where:
- $ V $ is the volume,
- $ r $ is the radius of the base,
- $ h $ is the height (vertical height, not slant height).

We are given volumes and either radius or height, and we need to find the missing dimension.

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✔ Problem 1:

Given:
- Volume = $ 140\pi \, \text{cm}^3 $
- Radius $ r = 5 \, \text{cm} $
- Height = $ x \, \text{cm} $

Use the formula:

$$
140\pi = \frac{1}{3} \pi (5)^2 x
$$

Simplify:

$$
140\pi = \frac{1}{3} \pi (25) x
$$

Cancel $ \pi $ from both sides:

$$
140 = \frac{25}{3} x
$$

Multiply both sides by 3:

$$
420 = 25x \Rightarrow x = \frac{420}{25} = 16.8 \, \text{cm}
$$

✔ Answer: $ x = 16.8 \, \text{cm} $

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✔ Problem 2:

Given:
- Volume = $ 140\pi \, \text{cm}^3 $
- Height $ h = 16.8 \, \text{cm} $
- Radius = $ r \, \text{cm} $

$$
140\pi = \frac{1}{3} \pi r^2 (16.8)
$$

Cancel $ \pi $:

$$
140 = \frac{1}{3} r^2 (16.8)
$$

Multiply both sides by 3:

$$
420 = 16.8 r^2
$$

Divide:

$$
r^2 = \frac{420}{16.8} = 25 \Rightarrow r = \sqrt{25} = 5 \, \text{cm}
$$

✔ Answer: $ r = 5 \, \text{cm} $

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✔ Problem 3:

Given:
- Volume = $ 140\pi \, \text{cm}^3 $
- Radius $ r = 10 \, \text{cm} $
- Height = $ x \, \text{cm} $

$$
140\pi = \frac{1}{3} \pi (10)^2 x = \frac{1}{3} \pi (100) x
$$

Cancel $ \pi $:

$$
140 = \frac{100}{3} x \Rightarrow x = \frac{140 \times 3}{100} = \frac{420}{100} = 4.2 \, \text{cm}
$$

✔ Answer: $ x = 4.2 \, \text{cm} $

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✔ Problem 4:

Given:
- Volume = $ 280\pi \, \text{cm}^3 $
- Radius $ r = 5 \, \text{cm} $
- Height = $ x \, \text{cm} $

$$
280\pi = \frac{1}{3} \pi (5)^2 x = \frac{1}{3} \pi (25) x
$$

Cancel $ \pi $:

$$
280 = \frac{25}{3} x \Rightarrow x = \frac{280 \times 3}{25} = \frac{840}{25} = 33.6 \, \text{cm}
$$

✔ Answer: $ x = 33.6 \, \text{cm} $

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✔ Problem 5:

Given:
- Volume = $ 280\pi \, \text{cm}^3 $
- Radius $ r = 10 \, \text{cm} $
- Height = $ x \, \text{cm} $

$$
280\pi = \frac{1}{3} \pi (10)^2 x = \frac{1}{3} \pi (100) x
$$

Cancel $ \pi $:

$$
280 = \frac{100}{3} x \Rightarrow x = \frac{280 \times 3}{100} = \frac{840}{100} = 8.4 \, \text{cm}
$$

✔ Answer: $ x = 8.4 \, \text{cm} $

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✔ Problem 6:

Given:
- Volume = $ 280\pi \, \text{cm}^3 $
- Radius $ r = 0.1 \, \text{m} = 10 \, \text{cm} $ (since 1 m = 100 cm → 0.1 m = 10 cm)
- Height = $ x \, \text{cm} $

Wait! Let's be careful with units.

But the volume is in cm³, so convert $ r = 0.1 \, \text{m} = 10 \, \text{cm} $. So same as Problem 5.

So again:

$$
280\pi = \frac{1}{3} \pi (10)^2 x = \frac{100}{3} \pi x
$$

Cancel $ \pi $:

$$
280 = \frac{100}{3} x \Rightarrow x = \frac{840}{100} = 8.4 \, \text{cm}
$$

✔ Answer: $ x = 8.4 \, \text{cm} $

> Note: Even though it says "0.1 m", since the volume is in cm³, we must convert to cm.

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✔ Problem 7:

Given:
- Volume = $ 280\pi \, \text{cm}^3 $
- Radius $ r = \frac{1}{10} \, \text{m} = 0.1 \, \text{m} = 10 \, \text{cm} $
- Height = $ x \, \text{cm} $

Same as above — radius = 10 cm

So:

$$
280\pi = \frac{1}{3} \pi (10)^2 x = \frac{100}{3} \pi x
\Rightarrow x = \frac{280 \times 3}{100} = 8.4 \, \text{cm}
$$

✔ Answer: $ x = 8.4 \, \text{cm} $

> Note: $ \frac{1}{10} \, \text{m} = 0.1 \, \text{m} = 10 \, \text{cm} $ — correct.

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✔ Problem 8:

Given:
- Volume = $ 280\pi \, \text{cm}^3 $
- Height $ h = 8.4 \, \text{cm} $
- Radius = $ r \, \text{cm} $

$$
280\pi = \frac{1}{3} \pi r^2 (8.4)
$$

Cancel $ \pi $:

$$
280 = \frac{8.4}{3} r^2 = 2.8 r^2
$$

$$
r^2 = \frac{280}{2.8} = 100 \Rightarrow r = \sqrt{100} = 10 \, \text{cm}
$$

✔ Answer: $ r = 10 \, \text{cm} $

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✔ Problem 9:

Given:
- Volume = $ 280\pi \, \text{cm}^3 $
- Height $ h = 84 \, \text{cm} $
- Radius = $ r \, \text{cm} $

$$
280\pi = \frac{1}{3} \pi r^2 (84)
$$

Cancel $ \pi $:

$$
280 = \frac{84}{3} r^2 = 28 r^2
\Rightarrow r^2 = \frac{280}{28} = 10
\Rightarrow r = \sqrt{10} \approx 3.16 \, \text{cm}
$$

✔ Answer: $ r = \sqrt{10} \, \text{cm} \approx 3.16 \, \text{cm} $

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✔ Final Answers Summary:

| Problem | Missing Value | Answer |
|--------|----------------|--------|
| 1 | $ x $ (height) | $ 16.8 \, \text{cm} $ |
| 2 | $ r $ (radius) | $ 5 \, \text{cm} $ |
| 3 | $ x $ (height) | $ 4.2 \, \text{cm} $ |
| 4 | $ x $ (height) | $ 33.6 \, \text{cm} $ |
| 5 | $ x $ (height) | $ 8.4 \, \text{cm} $ |
| 6 | $ x $ (height) | $ 8.4 \, \text{cm} $ |
| 7 | $ x $ (height) | $ 8.4 \, \text{cm} $ |
| 8 | $ r $ (radius) | $ 10 \, \text{cm} $ |
| 9 | $ r $ (radius) | $ \sqrt{10} \, \text{cm} \approx 3.16 \, \text{cm} $ |

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Let me know if you'd like a diagram explanation or want to see how the formulas were derived!