A right circular cone is inscribed in a sphere.

Question

A right circular cone is inscribed in a sphere. The base of the cone lies in the bottom half of the sphere. A dashed line segment goes from the vertex of the cone to a point in the center of the sphere. It is labeled 12 inches.

The radius of the sphere has a length of 12 in.

If the slant height of the cone has a length equal to that of its diameter, find the length (in inches) of the following.

(a) the radius of the base of the cone (inches)

(b) the altitude of the cone (inches)

Answer 1

If we make a cross section of these solids with front plane perpendicular to the ground, we will get a triangle, inscribed into the circle. The triangle is associated with the cone and the circle is associated with the sphere. Vertices of the triangle A, B and C will lie on the circle.

The triangle is equilateral since AB = BC = AC, where AC is the diameter of the cone.

Angles A, B and C have the same measure of 60°.

We will find the length of AB by this formula (Sine Rule):

AB/sin(60°) = 2R_circle

AB = 2 • 12 • sin(60°) ≅ 20.7846

R_cone = (1/2)AB ≅ 10.7846

BE is the altitude of the cone (point "E" is located half-way between vertices "A" and "C").

ΔABE is the right triangle.

From ΔABE:

BE = AB • sin(60°) = 20.7846 • sin(60°) ≅ 18 (ft)

Answer 2

Hi Maddie U.,

You should be able to draw the relevant cross-section of the right circular cone: since the diameter (on the bottom) is the same length as the (slant) side, that sets up two back-to-back 30-60-90 triangles. Therefore, the cone crossectional profile is an equilateral triangle; you should be able to complete the required distances, etc. from what you know about an equilateral triangle inscribed in a circle --

-- Cheers, --Mr. d.