+1839 Find the total surface area and the volume of the conical frustum below.
+1944 Let's make some obersvations. We have that
\(L = \sqrt {92^2 + (33-25)^2} = \sqrt{ 92^2 + 8^2} = 4\sqrt {533} \\ R = 33 \\ r =25\)
Surface area = \(\pi L (R + r) + \pi (R^2 + r^2) = \pi [ 4\sqrt{533} ](58) + \pi ( 33^2 + 25^2) ≈ 22211.49 units^2 \)
\(S1 = \pi * 33^2 \\ S2 = \pi * 25^2 \\ H = 92 \)
Volume =
\( H/3 * ( S1 + S2 + \sqrt { S1 * S2 }) = \\ 92/3 * ( \pi (33^2 + 25^2) + \sqrt{\pi*33^2 * \pi* 25^2} ) ≈ 244612.78 units^3\)
Thus, The answer is \( 244612.78\)
Thanks! :)
