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

## 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

**Video Solution**

**Reasoning:**

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

**What is known?**

Radius of the sphere.

**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

**Video 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

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

**Answer:**

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

**Video 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?**

Radius of the hemisphere.

**What is unknown?**

Total surface area of the hemisphere.

**Steps:**

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

Total Surface area

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

**Answer:**

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

**Video Solution**

**Reasoning:**

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

**What is known?**

The radius of balloon.

**What is unknown?**

Ratio of the surface area.

**Steps:**

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

**Answer:**

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

**Video 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 \(173.25 \,\rm{cm^2}\)

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

**Answer:**

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

**Video 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?**

Radius of the sphere.

**Steps:**

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

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

**Answer:**

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

**Video 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,

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

**Answer:**

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

## 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

**Video 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 \(= 0.25\rm\, cm\)

Outer radius of the bowl

\(= 5+0.25\\=5.25 \rm{cm}\)

Curve Surface area of the hemisphere,

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

**Answer:**

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

**Video 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)**.

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

**Answer:**

(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) \(\begin{align} = 1:1 \end{align}\)

- Live one on one classroom and doubt clearing
- Practice worksheets in and after class for conceptual clarity
- Personalized curriculum to keep up with school