Question

A hemispherical bowl fits exactly on the open surface of a cone with height 5 cm and radius 3.5 cm.What is the volume of the solid formed?

Answer 1

Given,

Radius of the cone = Radius of the hemisphere = $r$ = 3.5 cm

Height of the cone, $h$ = 5 cm

Total volume of the solid = Volume of the cone + Volume of the hemisphere

$=\frac{1}{3}\pi r^{2}h+\frac{2}{3}\pi r^3$

$=(\frac{1}{3}\times \frac{22}{7}\times {3.5}^2\times 5)+(\frac{2}{3}\times \frac{22}{7}\times {3.5}^3$)
$=\frac{1}{3}\times \frac{22}{7}\times {3.5}^2(5+7)$

$=\frac{1}{3}\times \frac{22}{7}\times {3.5}^2\times 12$

= 154 cm³