A frustum of a right circular cone has a diameter of base 20 cm, of to

1.	A frustum of a right circular cone has a diameter of base 20 cm, of top 12 cm, and height 3 cm. Find the area of its whole surface and volume.
Answer» 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

A frustum of a right circular cone has a diameter of base 20 cm, of top 12 cm, and height 3 cm. Find the area of its whole surface and volume.

Answer»

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