A toy is in the form of a hemisphere surmounted by a right circular cone of the same badse radius - Maths - Surface Areas and Volumes

Question

A toy is in the form of a hemisphere surmounted by a right
circular cone of the same badse radius as thst of hemisphere If the
radius of base of the cone is 21cm and the volume is 2/3 of the
volume of the hemisphere, calculate the height of the cone and the
surface area of the toy

Ajanta Trivedi · Accepted Answer

let the radius of the base of the cone be r=21 cm

let the height of the cone be h cm.

volume of the cone =2/3 volume of the hemisphere.

$\frac{1}{3} π r^{2} h = \frac{2}{3} * \frac{2}{3} π r^{3} \Rightarrow h = \frac{4}{3} r = \frac{4}{3} * 21 = 28 c m$

surface area of cone = lateral surface area of cone + curved surface area of sphere

= $π r \sqrt{r^{2} + h^{2}} + 2 π r^{2}$

$= \frac{22}{7} * 21 * \sqrt{21^{2} + 28^{2}} + 2 * \frac{22}{7} * 21 * 21 = 66 * \sqrt{441 + 784} + 2772 = 66 * 35 + 2772 = 2310 + 2772 = 5082 c m^{2}$

A toy is in the form of a hemisphere surmounted by a right circular cone of the same badse radius as thst of hemisphere. If the radius of base of the cone is 21cm and the volume is 2/3 of the volume of the hemisphere, calculate the height of the cone and the surface area of the toy.