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. \(\frac{1}{2}:1\)
• D. \(\sqrt{2}:1\)
• E. 2:1

Hint:

For geometry problems involving inscribed shapes, drawing a simple 2D cross-section can make the relationships between dimensions much clearer. In this case, a cross-section would show a semicircle with an isosceles triangle inscribed in it, making it obvious that the height of the triangle equals the radius of the semicircle.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem requires visualizing the geometric setup described. A hemisphere is half of a sphere. A cone is inscribed within it, sharing the same circular base.
Step 2: Detailed Explanation:
Let's define the dimensions based on the description.

Let the radius of the hemisphere be R. The base of the hemisphere is a circle with radius R.
The problem states that the base of the cone coincides with the base of the hemisphere. This means the cone's base is also a circle with radius R.
For the cone to be "inscribed in a hemisphere," its vertex must lie on the surface of the hemisphere. Since the bases coincide, the vertex of the cone must be at the "top" point of the hemisphere, which is directly above the center of the base.
The height of the cone (h) is the perpendicular distance from its vertex to the center of its base.
This distance, from the top point of the hemisphere down to the center of its circular base, is precisely the radius of the hemisphere, R.
Therefore, the height of the cone is equal to the radius of the hemisphere: \(h = R\).
Step 3: Final Answer:
The question asks for the ratio of the height of the cone (h) to the radius of the hemisphere (R). \[ \text{Ratio} = h:R = R:R = 1:1 \]