Answer to Question #149316 in Trigonometry for yui

Question

PROBLEM 2. A block of wood is in the form of a right circular cone. The altitude is 12 cm and the radius of the base is 5 cm. A cylindrical hole of 5 cm is bored completely through the solid, the axis of the hole coinciding with the axis of the cone. Find the amount of wood left after the hole is bored.
FINAL ANSWER: _______________________________________

Answer 1

$\displaystyle h_1 = \textsf{Height of cone}\\ r_1 = \textsf{Radius of cone}\\ h_2 = \textsf{Height of cylinder}\\ r_1 = \textsf{Radius of cylinder}\\ \textsf{Taking the ratios.}\\ \frac{h_1}{r_1} = \frac{h_2}{r_2}\\ \frac{12}{5} = \frac{5}{r_2}\\ r_2 = \frac{25}{12}\\ \begin{aligned} \textsf{Volume of cone}\, &= \frac{\pi {r_1}^2h_1}{3} \\&= \frac{1}{3}\times \pi \times 5^2 \times 12 = 100\pi \end{aligned} \\ \begin{aligned} \textsf{Volume of cylinder}\,&=\pi {r_2}^2 h_2 \\&= \pi \times 5 \times \frac{25^2}{12^2} = \frac{3125\pi}{144}\\ \end{aligned}\\ \textsf{Amount of wood left}\, = 100\pi - \frac{3125\pi}{144} = \frac{11275\pi}{144} = 245.98\, cm^3$