A solid cone has a height of 24 cm and a base radius of 7 cm. It is cut by a plane parallel to the base at a point 16 cm from the vertex. After removing the smaller cone, the remaining frustum is obtained. Find the total surface area of the frustum.
168π cm2
Slant height of the full cone: \(L = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25\) cm.
The small cone (height 16 cm) is similar to the full cone with ratio \(\frac{16}{24} = \frac{2}{3}\), so its radius is \(r = 7 \times \frac{2}{3} = \frac{14}{3}\) cm and its slant height is \(25 \times \frac{2}{3} = \frac{50}{3}\) cm.
Slant height of the frustum: \(l = 25 - \frac{50}{3} = \frac{25}{3}\) cm; its radii are R = 7 cm and \(r = \frac{14}{3}\) cm.
Curved surface area: \(\pi (R + r) l = \pi \left(7 + \frac{14}{3}\right) \times \frac{25}{3} = \pi \times \frac{35}{3} \times \frac{25}{3} = \frac{875\pi}{9}\).
Areas of the two circular ends: \(\pi R^2 + \pi r^2 = 49\pi + \frac{196\pi}{9} = \frac{441\pi + 196\pi}{9} = \frac{637\pi}{9}\).
Total surface area: \(\frac{875\pi + 637\pi}{9} = \frac{1512\pi}{9} = 168\pi\) cm2.
Hence, the answer is 168π cm2.
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