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

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Chapter 13 Ex.13.4 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

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

Chapter 13 Ex.13.4 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

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

Chapter 13 Ex.13.4 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 \( = 942\,\rm{cm^2} \)

Chapter 13 Ex.13.4 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 \( = 1:4 \)

Chapter 13 Ex.13.4 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 \( = \rm Rs 27.72 \)

Chapter 13 Ex.13.4 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 \(= 3.5 \, \rm cm\)

Chapter 13 Ex.13.4 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:

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

Chapter 13 Ex.13.4 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

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

Chapter 13 Ex.13.4 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) \(\begin{align} = 1:1 \end{align}\)

  
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