arhumsid wrote:

What is the radius of the largest sphere that can fit inside a right circular cone of base radius \(6 m\) and slant height \(12 m\)?

A. \(2 m\)

B. \(2\sqrt{3} m\)

C. \(4\sqrt{3} m\)

D. \(6 m\)

E. \(6\sqrt{3} m\)

You can use the properties of equilateral triangle, if you remember them.

Imagine cutting a vertical cross section of the cone with a sphere inscribed in it. The front will look like the figure above (

https://gmatclub.com/forum/what-is-the- ... l#p1580927)

The radius of the cone is 6 so the base of the triangle is 12. The slant height of the cone is 12 so both other sides are also 12. Hence it is an equilateral triangle.

In an equilateral triangle, the inradius is \(s*\sqrt{3}/6 = 12*\sqrt{3}/6 = 2*\sqrt{3}\)

Answer (B)

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Karishma

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