Question

A right circular cone has radius 8 cm and height 20 cm. A sphere is inscribed in the cone such that it touches the base and the lateral surface. Find the radius of the inscribed sphere.

• A. 3.35 cm
• B. 5.42 cm
• C. 2.8 cm
• D. 4.21 cm

Answer 1

Answer:

B. 5.42 cm

For a sphere inscribed in a right circular cone, the sphere touches the base and the slant surface. We can use the relation between the cone dimensions and the sphere radius.

Given:

• Cone radius (R = 8) cm

• Cone height (h = 20) cm

First find the slant height of the cone:

$l=\sqrt{R^2+h^2}$
$l=\sqrt{8^2+20^2}$

$l=\sqrt{64+400}=\sqrt{464}=4\sqrt{29}$

For an inscribed sphere in a cone:

$r=\frac{Rh}{R+l}$

Substitute the values:

$r=\frac{8\times20}{8+4\sqrt{29}}$

$r=\frac{160}{8+4\sqrt{29}}$

Factor out 4 from the denominator:

$r=\frac{40}{2+\sqrt{29}}$

Rationalizing:

$r=\frac{40(\sqrt{29}-2)}{29-4}$

$r=\frac{40(\sqrt{29}-2)}{25}$

$r=\frac{8}{5}(\sqrt{29}-2)$

Using $(\sqrt{29}\approx5.385):$

$r=\frac{8}{5}(5.385-2)$

$r=\frac{8}{5}(3.385)$

$r\approx5.42$

Answer:

$\boxed{r\approx5.42\text{ cm}}$