Question

A right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere. What is the ratio of the height of the cone to the radius of the hemisphere?

• A. \( \sqrt{3}:1 \)
• B. \( 1:1 \)
• C. \( \sqrt{2}:1 \)
• D. \( 2:\sqrt{1} \)
• E. \( 2:1 \)

Hint:

For problems involving inscribed figures, use the Pythagorean theorem to relate the dimensions of the figure.

Answer 1

The Correct Option is A

Step 1: Understand the geometry of the problem.
Let the radius of the hemisphere be \( r \), and the height of the cone be \( h \). The cone is inscribed in the hemisphere such that its vertex touches the top of the hemisphere, and its base coincides with the base of the hemisphere.
Step 2: Apply Pythagoras’ theorem.
The radius of the base of the cone is \( r \), and the height of the cone is \( h \). The slant height \( l \) of the cone is the distance from the vertex to the point on the circumference of the base. This forms a right triangle with the radius and height of the cone. According to the Pythagorean theorem: \[ l^2 = r^2 + h^2 \] Since the cone is inscribed in the hemisphere, the slant height \( l \) is also the radius of the hemisphere, so: \[ l = r \] Step 3: Solve for the ratio \( \frac{h}{r} \).
Substitute \( l = r \) into the Pythagorean theorem: \[ r^2 = r^2 + h^2 \] This simplifies to: \[ h^2 = \frac{r^2}{3} \implies h = \frac{r}{\sqrt{3}} \] Thus, the ratio of the height of the cone to the radius of the hemisphere is: \[ \frac{h}{r} = \frac{1}{\sqrt{3}} \implies \boxed{\sqrt{3}:1} \]