Question

The diagram shows a toy.
The shape of the toy is a cone, with radius $$4\ cm $$ and height $$9\ cm $$, on top of a hemisphere with radius $$4\ cm$$.
Calculate the volume of the toy.
Give your answer correct to the nearest cubic centimeter.
[The volume, V of a cone with radius $$r$$ and height $$h$$ is $$ V = \dfrac{1}{3} \pi r^{2} h $$]
[The volume, V of a sphere with radius $$r$$ is $$V = \dfrac{4}{3} \pi r^{3}$$].

Answer 1

Given that, a toy is in the shape of a cone on top of a hemisphere.

To find out: The volume of the toy.

$$1)$$ Volume of the cone -

Radius of cone, $$r=4\ cm$$

Height of the cone, $$h=9\ cm$$

We know that, the volume of a cone is given by,

$${ V }_{ 1 }=\dfrac { 1 }{ 3 } \pi { r }^{ 2 }h$$

$$\therefore \ { V }_{ 1 }=\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times { \left( 4 \right) }^{ 2 }\times 9$$

$$\therefore \ { V }_{ 1 }=\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times 16\times 9$$

$$\therefore \ { V }_{ 1 }= \dfrac { 22 }{ 7 } \times 16\times 3$$

$$\therefore \ { V }_{ 1 }=\dfrac { 1056 }{ 7 } $$

$$\therefore \ { V }_{ 1 }=150.857\ { cm }^{ 3 }$$

$$2)$$ Volume of the hemisphere -

Radius of hemisphere, $$r=4\ cm$$

We know that, the volume of a hemisphere is given by,

$${ V }_{ 2 }=\dfrac { 2 }{ 3 } \pi { r }^{ 3 }$$

$$\therefore \ { V }_{ 2 }=\dfrac { 2 }{ 3 } \times \dfrac { 22 }{ 7 } \times { \left( 4 \right) }^{ 3 }$$

$$\therefore\ { V }_{ 2 }=\dfrac { 2 }{ 3 } \times \dfrac { 22 }{ 7 } \times 64$$

$$\therefore \ { V }_{ 2 }=\dfrac { 2816 }{ 21 } $$

$$\therefore { V }_{ 2 }=143.095\ { cm }^{ 3 }$$

Now, the total volume of the toy $$=$$ volume of the cone $$+$$ volume of the hemisphere

$$\therefore \ V={ V }_{ 1 }+{ V }_{ 2 }$$

$$\therefore \ V=150.857+134.095$$

$$\therefore\ V=284.952\ { cm }^{ 3 }\approx 285\ cm^3$$

Hence, the required volume of the toy is $$285\ { cm }^{ 3 }$$.