Question

A toy is in the form of cone mounted on a hemisphere as shown in figure if the radius of each solids is $\dfrac{7}{2}cm$ and height of the cone is $5cm$, find the volume of the toy.

Answer 1

Hint: For finding the volume of the toy which is in the form of a cone mounted on a hemisphere we will find the volume of each shape separately. We will find the volume of cone using its radius and height with the help of formula ${{V}_{c}}=\dfrac{1}{3}\pi {{r}^{2}}h$ where $r$ is the radius of the cone and $h$ is the height of the cone.
We will find the volume of the hemisphere ${{V}_{s}}=\dfrac{2}{3}\pi {{r}^{3}}$ where $r$ is the radius of the hemisphere.

Complete step by step answer:
Given that, the toy is in the form of a cone mounted on a hemisphere with radius $\dfrac{7}{2}cm$ and height $5cm$.
We know that the volume of cone is
${{V}_{c}}=\dfrac{1}{3}\pi {{r}^{2}}h$
Where $r$ is the radius of the cone and $h$ is the height of the cone.
Now the volume of cone with radius $\dfrac{7}{2}cm$ and height $5cm$ is
$\begin{align}
  & {{V}_{c}}=\dfrac{1}{3}\pi {{r}^{2}}h \\
 & =\dfrac{1}{3}\pi {{\left( 3.5 \right)}^{2}}\left( 5 \right) \\
 & =64.11c{{m}^{3}}
\end{align}$
We know the volume of the hemisphere is
${{V}_{s}}=\dfrac{2}{3}\pi {{r}^{3}}$
Where $r$ is the radius of the hemisphere.
Now the volume of hemisphere with radius $5cm$ is
$\begin{align}
  & {{V}_{s}}=\dfrac{2}{3}\pi {{r}^{3}} \\
 & =\dfrac{2}{3}\pi {{\left( 5 \right)}^{3}} \\
 & =89.75c{{m}^{3}}
\end{align}$
Area of the toy is calculated by adding the volume of cone and volume of hemisphere, so
$\begin{align}
  & {{V}_{t}}={{V}_{c}}+{{V}_{s}} \\
 & =64.11+89.75 \\
 & =153.86c{{m}^{3}}
\end{align}$

So, the correct answer is “153.86 $cm^3$”.

Note: We can also calculate the volume of cone directly by the following
$\begin{align}
  & {{V}_{t}}={{V}_{c}}+{{V}_{s}} \\
 & =\dfrac{1}{3}\pi {{r}^{2}}h+\dfrac{2}{3}\pi {{r}^{3}} \\
 & =\dfrac{1}{3}\pi {{r}^{2}}\left[ h+2r \right]
\end{align}$
Substituting the value of $r=\dfrac{7}{2}cm$ and $h=5cm$, then
$\begin{align}
  & {{V}_{t}}=\dfrac{1}{3}\pi {{\left( 3.5 \right)}^{2}}\left[ 5+2\left( 3.5 \right) \right] \\
 & =153.86c{{m}^{3}}
\end{align}$
From both these methods we got the same values.