A right circular cone has a base radius of 3 and a height of 4. A sphere is inscribed in the cone such that it is tangent to the base and the lateral surface of the cone. Find the radius of the sphere.
Correct: B
Let the cone have base radius r_c = 3 and height h_c = 4.
The slant height L of the cone can be found using the Pythagorean theorem:
L = sqrt(r_c^2 + h_c^2) = sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5.
Consider a 2D cross-section of the cone and the inscribed sphere, taken through the axis of the cone. This cross-section forms an isosceles triangle with base 2*r_c = 6 and height h_c = 4. The sides of this triangle are the two slant heights (each of length L=5) and the diameter of the base (length 6).
The inscribed sphere appears as an inscribed circle within this isosceles triangle. The radius of this inscribed circle is the radius of the sphere, let's call it R.
The formula for the inradius R of a triangle is R = Area / s, where 's' is the semi-perimeter of the triangle.
First, calculate the area of the triangular cross-section:
Area = 1/2 * base * height = 1/2 * (2*r_c) * h_c = 1/2 * 6 * 4 = 12.
Next, calculate the semi-perimeter 's' of the triangular cross-section:
The sides of the triangle are 5, 5, and 6.
s = (5 + 5 + 6) / 2 = 16 / 2 = 8.
Now, calculate the radius R of the inscribed sphere:
R = Area / s = 12 / 8 = 3/2.
Thus, the radius of the inscribed sphere is 3/2.