A cone of height h is inscribed in a ball with radius R. Find the area of the axial section of the cone.

The axial section of the cone is the isosceles triangle ABC.

The area of the triangle will be equal to: Sавс = АС * ВН / 2.

Since the triangle is isosceles, then the height of BH is also the median of the triangle, then AH = CH = AC / 2.

Then Saws = 2 * AH * BH = 2 * AH * h.

Let’s draw the radius of the AO. In a right-angled triangle AON, AO = R, OH = (h – R), then, by the Pythagorean theorem, AH ^ 2 = R ^ 2 – (h – R) ^ 2 = R ^ 2 – h ^ 2 + 2 * R * h – R ^ 2 = (2 * R * h – h ^ 2).

AH = √ (2 * R * h – h ^ 2).

Savs = 2 * h * √ (2 * R * h – h ^ 2) cm2.

Answer: The axial section area is 2 * h * √ (2 * R * h – h ^ 2) cm2.