The base of the pyramid is an isosceles triangle, the base of which is 16 cm, and the lateral side is 10 cm. A cone is inscribed in the pyramid. Find the area of the axial section of the cone if its height is 9 cm.
Determine the radius of the inscribed circle in the isosceles triangle ABC.
The semi-perimeter of the triangle is: p = (AB + AC + BC) / 2 = (16 + 10 + 10) / 2 = 18 cm.
Then R = (AB / 2) * √ ((2 * AC – AB) / (2 * AC + AB)) = (16/2) * √ ((2 * 10 – 16) / (2 * 10 + 16 )) = 8 * √ (4/36) = 8 * (2/6) = 16/6 = 8/3 = 2 (1/3) cm.
Then the diameter of the inscribed circle is: D = MK = 2 * R = 16/3 = 5 (1/3) cm.
The axial section of the inscribed cone is the DMC triangle, in which the cone height is the triangle height, and the cone diameter is the base of the triangle.
Then Ssec = MK * DO / 2 = (16/3) * 9/2 = 24 cm2.
Answer: The area of the axial section of the cone is 24 cm2.
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