Question

The perimeters of the ends of a frustum of a right circular cone are 44 cm and 33 cm. If the height of the frustum be 16 cm, find its volume, the slant surface and the total surface.

Answer 1

For the frustum, 

Perimeter of the larger end = 44 cm 

⇒ 2πr’ = 44 

⇒ r' = \(\frac{44}{{2}\times\frac{22}{7}}\) = 7 cm

Perimeter of the smaller end = 33 cm 

⇒ 2πr’’ = 33 

⇒ r" = \(\frac{33}{{2}\times\frac{22}{7}}\) = \(\frac{21}4\) cm or 5.25 cm

Height = 16 cm 

Its volume = \(\frac{1}3π(r'^2 + r"^2 + r'r")h\)

= \(\frac{1}3π(7^2 + (\frac{21}4)^2 + 7\times\frac{21}4)\times16\)

= \(\frac{1}3\times\frac{22}7(784 + 441 + 588)\)

≈ 1900 cub. cm 

Let l be the slant height of the cone, 

then 

⇒ l = \(\sqrt{(r' - r")^2 + h^2}\)

⇒ l = \(\sqrt{(7 - 5.25)^2 + 16^2}\)

⇒ l = 16.1 cm 

Curved surface area of frustum = π (r’ + r’’) × l 

= 3.14(7 + 5.25) × 16.1 

= 619.65 cm² 

Total surface area of the frustum = π (r’ + r’’) × l + πr’² + πr’’² 

= 619.65 + 3.14 × (7² + 5.25²) 

= 860.275 cm²