Question

Figure 7.13 shows a toy. Its lower part is a hemisphere and the upper part is a cone. Find the volume and the surface area the toy from the measures shows in the figure. $ \left( {\pi = 3.14} \right) $

Answer 1

Hint: This question is based on Geometry. In this question a toy is given and the lower part of the toy is a hemisphere and the upper part is a cone. So, the toy is a combination of two three dimensional shapes – a cone and a hemisphere and the volume of the toy will be the sum of the volume of the cone and the hemisphere, similarly the surface area of the toy will be the sum of the surface area of the cone and the hemisphere.

Complete step-by-step answer:

Given:
Let us assume the height of the cone is $ h $ , the radius of the base of the cone is $ r $ and the slant height of the cone is $ l $ .
The height of the cone $ h = 4{\rm{ cm}} $
The radius of the base of the cone $ r = 3{\rm{ cm}} $
So, using the Pythagoras theorem, the slant height of the cone
 $
\Rightarrow = \sqrt {{r^2} + {h^2}} \\
\Rightarrow l = \sqrt {{3^2} + {4^2}} \\
\Rightarrow l = \sqrt {9 + 16} \\
\Rightarrow l = 5{\rm{ cm}}
 $
And,
The radius of the hemisphere is equal to the radius of the base of the cone. So,
The radius of the hemisphere $ r = 3{\rm{ cm}} $
Let us assume the volume of the cone is $ {V_1} $ , volume of the hemisphere is $ {V_2} $ and the total volume of the toy is $ V $ .
Then, $ V = {V_1} + {V_2} $
We know the volume of the cone is given by the formula –
 $ {V_1} = \dfrac{1}{3}\pi {r^2}h $
Substituting the values in the formula we get,
 $
\Rightarrow {V_1} = \dfrac{1}{3} \times 3.14 \times {3^2} \times 4\\
\Rightarrow {V_1} = 37.68{\rm{ c}}{{\rm{m}}^3}
 $
And the volume of the hemisphere is given by the formula –
 $ {V_2} = \dfrac{2}{3}\pi {r^3} $
Substituting the values in the formula we get,
 $
\Rightarrow {V_2} = \dfrac{2}{3} \times 3.14 \times {3^3}\\
\Rightarrow {V_2} = 56.52{\rm{ c}}{{\rm{m}}^3}
 $
So, the total volume of the toy = Volume of the cone + Volume of the hemisphere
 $
\Rightarrow V = {V_1} + {V_2}\\
\Rightarrow V = 37.68 + 56.52\\
\Rightarrow V = 94.2{\rm{ c}}{{\rm{m}}^3}
 $

Let us assume the surface area of the cone is $ {A_1} $ , the surface area of the hemisphere is $ {A_2} $ and the total surface area of the toy is $ A $ .
Then, $ A = {A_1} + {A_2} $
We know the surface area of the cone is given by the formula –
 $
\Rightarrow {A_1} = \pi rl\\
\Rightarrow {A_1} = 3.14 \times 3 \times 5\\
\Rightarrow {A_1} = 47.1{\rm{ c}}{{\rm{m}}^2}
 $
The surface area of the hemisphere is given by the formula –
 $
\Rightarrow {A_2} = 2\pi {r^2}\\
\Rightarrow {A_2} = 2 \times 3.14 \times {3^2}\\
\Rightarrow {A_2} = 56.52{\rm{ c}}{{\rm{m}}^2}
 $
So, the total surface area of the toy = surface area of the cone + surface area of the hemisphere
 $
\Rightarrow A = {A_1} + {A_2}\\
\Rightarrow A = 47.1 + 56.52\\
\Rightarrow A = 103.62{\rm{ c}}{{\rm{m}}^2}
 $

Therefore, the total volume of the toy is $ 94.2{\rm{ c}}{{\rm{m}}^3} $ and the total surface area of the toy is $ 103.62{\rm{ c}}{{\rm{m}}^2} $ .

Note: It should be noted that while calculating the total volume of the toy and the total surface area of the toy the units of all the dimensions of the cone and the hemisphere should be the same otherwise the calculation process gets complex.