Question

A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of the hemisphere. If the radius of the base of the cone is \[21\] cm and its volume is \[\dfrac{2}{3}\] of the volume of the hemisphere, calculate the height of the cone and the surface area of the toy. (use \[\pi = \dfrac{{22}}{7}\])

Answer 1

Hint: Using the given base radius and condition on volume of both shapes we will find the slant height of the cone with the help of it we will find the height of the cone and also the surface area of the toy.

Formula used: The volume of a hemisphere with the radius \[r\]cm is \[V = \dfrac{2}{3}\pi {r^3}\]
The volume of a cone with radius \[r\] and height as \[h\] is \[V = \dfrac{1}{3}\pi {r^2}h\]
The surface area of the cone is \[\pi rl\] where, Slant height of the cone is \[l = \sqrt {{r^2} + {h^2}} \]
The surface area of the hemisphere is \[2\pi {r^2}\]

Complete step-by-step answer:
The shape of the toy is represented in the diagram below,
               

where r, h, l represents radius height and slant height respectively
It is given that a toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of the hemisphere.
The radius of the base of the cone is \[21\] cm and its volume is \[\dfrac{2}{3}\] of the volume of the hemisphere.
Since the base of the toy is hemisphere, the radius of the hemisphere is the same as the cone.
So, the radius of the hemisphere is \[21\] cm.
We know that, the volume of a hemisphere with the radius \[r\] cm is \[V = \dfrac{2}{3}\pi {r^3}\]
Substitute \[r = 21\] in the formula of the volume of the hemisphere we get,
\[V = \dfrac{2}{3} \times \dfrac{{22}}{7} \times {21^3}\]
By solving the above equation we get, the volume as \[V = 19404\] $cm^3$
According to the problem, the volume of the cone is \[\dfrac{2}{3}\] of the volume of the hemisphere.
We know that, the volume of a cone with radius \[r\] and height as \[h\] is \[V = \dfrac{1}{3}\pi {r^2}h\]
Let us consider, the height of the toy is \[h\]
So, we have,
\[\dfrac{1}{3} \times \dfrac{{22}}{7} \times {21^2} \times h = \dfrac{2}{3} \times \dfrac{2}{3} \times \dfrac{{22}}{7} \times {21^3}\]
Cancelling out the common terms in the above equation and on simplifying we get,
\[h = 28\]
Now, the surface area of the toy is the sum of the surface area of the hemisphere and the surface area of the cone.
For the surface area of the cone we need the slant height of the cone.
Slant height of the cone is found using the formula \[l = \sqrt {{r^2} + {h^2}} \]
Substitute the values we get,
\[l = \sqrt {{{21}^2} + {{28}^2}} = 35\]
So, the surface area of the cone is \[\pi rl\]
Let us substitute the values we know in the surface area formula we get,
\[ = \dfrac{{22}}{7} \times 21 \times 35 = 2310\] $cm^2$
Now, we will find the surface area of the hemisphere.
The surface area of the hemisphere is \[2\pi {r^2}\]
Substitute the values of radius we get,
The surface area as \[2 \times \dfrac{{22}}{7} \times {21^2} = 2772\] $cm^2$
Therefore, the total surface area of the toy is \[2310 + 2772 = 5082\] $cm^2$
Hence, the height of the cone is \[28\] cm and the surface area is \[5082\] $cm^2$

Note: Since, the base of the toy is hemisphere, the surface area of the toy is the sum of the surface area of the hemisphere and the surface area of the cone.