# A hemisphere of lead of radius 9 cm is cast into a right circular cone of height 72 cm. Find the radius of the base of the cone.

## Question

A hemisphere of lead of radius 9 cm is cast into a right circular cone of height 72 cm. Find the radius of the base of the cone.

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## Answer

**:**

Let $ r \mathrm{~cm} $ be the radius of the cone.

Given:

- Radius of the hemisphere, $ r_1 = 9 \mathrm{~cm} $
- Height of the cone, $ h = 72 \mathrm{~cm} $

Since the **volume of the hemisphere** is being cast into the **volume of the cone**, we have:

$$ \text{Volume of the hemisphere} = \text{Volume of the cone} $$

The volume of a hemisphere is given by:

$$ \text{Volume of hemisphere} = \frac{2}{3} \pi r_1^3 $$

The volume of a cone is given by:

$$ \text{Volume of cone} = \frac{1}{3} \pi r^2 h $$

Equating the two volumes:

$$ \frac{2}{3} \pi (9)^3 = \frac{1}{3} \pi r^2 (72) $$

Simplifying the equation:

$$ \frac{2}{3} \pi \times 9^3 = \frac{1}{3} \pi \times r^2 \times 72 $$

$$ \Rightarrow \frac{2 \times 729}{3} = \frac{72 r^2}{3} $$

$$ \Rightarrow 2 \times 243 = 24 r^2 $$

$$ \Rightarrow 486 = 24 r^2 $$

$$ \Rightarrow r^2 = \frac{486}{24} $$

$$ \Rightarrow r^2 = \frac{81}{4} $$

Therefore:

$$ r = \sqrt{\frac{81}{4}} = \frac{9}{2} = 4.5 \mathrm{~cm} $$

Thus, the **radius of the base of the cone is** $ \mathbf{4.5 \mathrm{~cm}} $.

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