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.

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Accepted Answer

1. Cross-section gives a right triangle with base radius $R=8$ and height $h=20$. The sphere center is at height $r$ above the base, so its distance from the apex is $y=h-r$.

2. The perpendicular distance from the center to the slant equals $r$. From the line equation this is $r=\frac{R(h-r)}{\sqrt{h^2+R^2}}$.

3. Solve for $r$: $r\big(\sqrt{h^2+R^2}+R\big)=Rh$, so $r=\frac{Rh}{R+\sqrt{h^2+R^2}}$.

4. Substitute $R=8$, $h=20$: $r=\frac{160}{8+\sqrt{464}}\approx5.42	ext{ cm}$. Thus option **(b)** is correct.

FINAL_ANSWER: b

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