# Exercise E13.4 Surface Areas and Volumes NCERT Solutions Class 9

Go back to  'Surface Areas and Volumes'

## Question 1

Assume $$\pi = \frac{{22}}{7}$$ unless stated otherwise

Find the surface area of a sphere of radius:

(i) $$10.5 \rm \, cm$$

(ii) $$5.6 \rm \, cm$$

(iii) $$14 \rm \, cm$$

### Solution

Reasoning:

Surface area of sphere of radius $$r$$ $$= 4\pi\,\rm{r^2}$$

What is known?

What is unknown?

Surface area of the sphere

Steps:

(i) Radius $$= 10. 5\rm\, cm$$

Surface area of the sphere

\begin{align}&= 4\pi {r^2}\\&= 4 \times \frac{{22}}{7} \times {(10.5)^2} \\&= 1386\,\rm{cm^2} \end{align}

(ii) Radius $$= 5.6\rm\, cm$$

Surface area of the sphere

\begin{align} &= 4\pi {r^2}\\&= 4 \times \frac{{22}}{7} \times {(5.6)^2}\\ &= 394.24\,\rm{cm^2} \end{align}

(iii) Radius $$= 14\rm\, cm$$

Surface area of the sphere

\begin{align} &= 4\pi {r^2} \\&= 4 \times \frac{{22}}{7} \times {(14)^2}\\ &= 2464\,\rm{cm^2} \end{align}

Surface area

\begin{align}&\text{(i) 1386}\,\rm{cm^2}\\&\text{(ii) 394.24}\,\rm{cm^2}\\&\text{(iii) 2464}\,\rm{cm^2} \end{align}

## Question 2

Find the surface area of a sphere of diameter:

i. $$14 \, \rm cm$$

ii. $$21 \, \rm cm$$

iii. $$3.5 \, \rm m$$

### Solution

Reasoning:

Surface area of the sphere of radius $$r$$ is equal $$4$$ times the area of the circle of radius $$r.$$

\begin{align}S = 4\pi {r^2} \end{align}

What is known?

Diameter of the sphere.

What is unknown?

Surface area of the sphere.

Steps:

(i)  Diameter ( $$2r$$ ) $$= 14{\rm\,{ cm}}$$

Radius ($$r$$) $$= 7\,\rm\,{cm}$$

Surface area

\begin{align}&= 4\pi {r^2}\\ & = 4 \times \frac{{22}}{7} \times 7 \times 7\\&= 616\,\rm{cm^2} \end{align}

(ii) Diameter($$2 r$$) = $$21 \rm\,{cm}$$

Radius ($$r$$) = $${\frac{21}{2} \, \mathrm {cm}}$$

Surface area

\begin{align}&={4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}} \\&={1386 \, \mathrm{cm}^{2}}\end{align}

(iii) Diameter ($$2 r$$$$= 3.5\rm\, m$$

Radius ($$r$$) $$= \frac{3.5}{2}\rm\, m$$

Surface area

\begin{align} &={4 \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2}} \\ &={38.5 \mathrm{m}^{2}}\end{align}

Surface areas

\begin{align}&\text{(i) 616} \rm\,{cm^2}\\&\text{(ii) 1386}\rm\,{cm^2}\\&\text{(iii) 38.5}\,\rm\,{m^2} \end{align}

## Question 3

Find the total surface area of a hemisphere of radius $$10\rm\, cm.$$ $$(\rm{Use\;\pi = 3.14})$$

### Solution

Reasoning:

A hemisphere is half of a sphere having one circular surface at the top.

$$∴$$  Total surface area of a hemisphere is the half area of sphere and the top circular $${\rm{area}} = 3\pi {r^2}$$

What is known?

What is unknown?

Total surface area of the hemisphere.

Steps:

Radius of the hemisphere$$(r) = 10 \rm\,cm$$

Total Surface area

\begin{align} &= 3\pi {r^2}\\&= 3 \times 3.14 \times 10 \times 10\\&= 942\,\,\rm{cm^2} \end{align}

Total surface area of the hemisphere $$= 942\,\rm{cm^2}$$

## Question 4

The radius of a spherical balloon increases from $$7 \rm\, cm$$ to $$14\rm\, cm$$ as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

### Solution

Reasoning:

The surface area of the sphere is\begin{align}\, = 4\pi {r^2} \end{align}

What is known?

What is unknown?

Ratio of the surface area.

Steps:

\begin{align}\frac{\text{CSA}_1 }{{\text{CSA}_2}} &= \frac{{4\pi r_1^2}}{{4\pi r_2^2}} \\&= \frac{{r_1^2}}{{r_2^2}} \\&= {\bigg(\frac{7}{{14}}\bigg)^2} \\&= {\bigg(\frac{1}{2}\bigg)^2} \\&= \frac{1}{4} \end{align}

The ratio of the surface areas of the balloon $$= 1:4$$

## Question 5

A hemispherical bowl made of brass has inner diameter $$10.5\rm\, cm.$$ Find the cost of Tin-plating it on the inside at the rate of $$16 \text{per} \,100 \, \rm{m^2}$$.

### Solution

Reasoning:

Hemisphere is half of a sphere so surface area is \begin{align}\frac{{4\pi {r^2}}}{2} = 2\pi {r^2} \end{align}.

What is known?

Ratio of tin plating per $$100\,\rm\,{m^2}$$ and inner diameter.

What is unknown?

Cost of tin planting

Steps:

Inner Diameter $$(2r) = 10.5\rm\, cm$$

Inner Radius $$r = \frac{{10.5}}{2}$$

Surface area

\begin{align} &=2\pi {r^2} \\&= 2 \times \frac{{22}}{7} \times \frac{{10.5}}{2} \times \frac{{10.5}}{2}\\ &= 173.25\,\rm{cm^2} \end{align}

Cost of plating is $$16$$ per $$100$$$$\,\rm{cm^2}$$

$$∴$$Cost of plating for $$173.25 \,\rm{cm^2}$$

\begin{align} & = \frac{{173.25 \times 16}}{{100}}\\& = \rm{Rs}\,\,{{27}}.{\rm{ }}{{72}} \end{align}

Cost of tin plating $$= \rm Rs 27.72$$

## Question 6

Find the radius of a sphere whose surface area is $$154\,\rm\,{m^2}.$$

### Solution

Reasoning:

The surface area of the sphere is equal to $$4$$ times the surface area of the circle if $$r$$ is the sphere and circle. $$S = 4\pi {r^2}$$

What is known?

Surface area of the sphere.

What is unknown?

Steps:

Surface area of the sphere $$= 4\pi {r^2}$$

\begin{align} 154 &= 4 \times \frac{{22}}{7} \times {r^2}\\{r^2} &= \frac{{154}}{4} \times \frac{7}{{22}}\\{r^2} &= 7 \times 7 \times \frac{1}{2} \times \frac{1}{2}\\r\, &= 7 \times \frac{1}{2}\\&= 3.5\,\rm{cm} \end{align}

Radius of the sphere $$= 3.5 \, \rm cm$$

## Question 7

The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

### Solution

Reasoning:

Surface of the sphere having radius $$r$$ is equal to the $$4$$ times are of the circle having radius $$r.$$ \begin{align}S = 4\pi {r^2} \end{align}

What is known?

Ratio between the diameter of the moon and earth.

What is unknown?

Ratio of the surface areas.

Steps:

Let the diameter of the earth be $$2r.$$

Radius of the earth in $$= r$$

Radius of the moon is $$= \frac{r}{4}$$

Surface area of the earth $$= 4\pi \rm{r^2}$$

Surface area of the moon \begin{align}= 4 \pi\left(\frac{r}{4}\right)^{2} \end{align}

Therefore, Ratio of their surface area,

\begin{align}&=\frac{\text { Surface area of the moon }}{\text { Surface area of the earth }}\\&=\frac{4 \times \pi \times r^{2}}{4 \times \pi \times 4 \times 4 \times r^{2}}\\&=\frac{1}{16}\end{align}

\begin{align}\text{Ratio of their surface area = 1:16}\end{align}

## Question 8

A hemispherical bowl is made of steel, $$0.25 \rm\,cm$$ thick. The inner radius of the bowl is $$5\rm\, cm.$$ Find the outer curved surface area of the bowl.

### Solution

Reasoning:

The surface area of the hemisphere is half of the surface area of the sphere.

What is known?

Thickness of steel.

What is unknown?

Outer curved surface area.

Steps:

Inner radius bowl $$= 5\rm\, cm$$

Thickness of steel $$= 0.25\rm\, cm$$

$$= 5+0.25\\=5.25 \rm{cm}$$

Curve Surface area of the hemisphere,

\begin{align}&= 2\pi {r^2} \\&= 2 \times \frac{{22}}{7} \times {(5.25)^2}\\&= 173.25\,\,\rm{cm^2} \end{align}

The outer curved surface area of the hemisphere

\begin{align} = 173.25\,\rm{cm^2} \end{align}

## Question 9

A right circular cylinder just encloses a sphere of radius $$r$$ (see Fig. 13.22).

Find:

(i) Surface area of the sphere

(ii) Curved surface area of the cylinder

(iii) Ratio of the areas obtained in (i) and (ii).

### Solution

Reasoning:

The Curved surface area of the cylinder is given by \begin{align} = 2\pi rh \end{align} and Surface area of the sphere\begin{align}\, = 4\pi {r^2} \end{align}

What is known?

Radius of the sphere which touches the cylinder.

(i) Surface area of the sphere

Steps:

Radius of the sphere in $$= r$$

Surface area $$= 4\pi \rm{r^2}$$

(ii) Curved surface area of the cylinder.

Steps:

For cylinder: radius$$=r$$

Height $$=2r$$

Curved surface area of the cylinder

\begin{align}&= 2\pi rh \\&= 2 \times \pi \times r \times 2r \\&= 4\pi {r^2} \end{align}

(iii) Ratio of the areas obtained in (i) and (ii).

\begin{align} &{\text{Ratio of the area}} \\ &\qquad = \frac{{{\text{Surface area of sphere}}}}{{\left( \begin{array}{l} {\text{Curved surface}} \\ {\text{area of cylinder}} \\ \end{array} \right)}} \\ &\qquad = \frac{{4\pi r^2 }}{{4\pi r^2 }} \\ &\qquad = \frac{1}{1} \\ &\qquad = 1:1 \\ \end{align}

(i) Surface area of the sphere \begin{align} = 4\pi {r^2} \end{align}
(ii) Curved surface area of cylinder \begin{align} = 4\pi {r^2} \end{align}
(iii) Ratio between (i) and (ii) \begin{align} = 1:1 \end{align}