For frustum:
Base radius, r’ = \(\frac{20}2\) = 10 cm
Top radius, r’’ = \(\frac{12}2\) = 6 cm
Height, h = 3 cm
Volume = \(\frac{1}3π(r'^2 + r"^2 + r'r")h\)
= \(\frac{1}3π(10^2 + 6^2 + 10\times6)\times3\)
= 616 cub. cm
Let l be the slant height of the cone,
then
⇒ l = \(\sqrt{(r' - r")^2 + h^2}\)
⇒ l = \(\sqrt{(10 - 6)^2 + 3^2}\)
⇒ l = 5 cm
Total surface area of the frustum = π (r’ + r’’) × l + πr’2 + πr’’2
= π (20 + 10) × 15.620 + π (10)2 + π (6)2
= 678.85 cm2
