NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.4

Exercise (13.4)

1. Find the surface area of a sphere of radius:

i. $\text{10}\text{.5 cm}$

Ans:

Given radius of the sphere $\text{r = 10}\text{.5 cm}$

The surface area of the sphere $\text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4  }\!\!\times\!\!\text{  }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{10}\text{.5} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{88  }\!\!\times\!\!\text{  1}\text{.5  }\!\!\times\!\!\text{  10}\text{.5} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 1386 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{1386 c}{{\text{m}}^{\text{2}}}$.

ii. $\text{5}\text{.6 cm}$

Ans:

Given radius of the sphere $\text{r = 5}\text{.6 cm}$

The surface area of the sphere $\text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4  }\!\!\times\!\!\text{  }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{5}\text{.6} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{88  }\!\!\times\!\!\text{  0}\text{.8  }\!\!\times\!\!\text{  5}\text{.6} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 394}\text{.24 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{394}\text{.24 c}{{\text{m}}^{\text{2}}}$.

iii. $\text{14 cm}$ $\left[ \text{Assume  }\!\!\pi\!\!\text{  = }\frac{\text{22}}{\text{7}} \right]$

Ans:

Given radius of the sphere $\text{r = 14 cm}$

The surface area of the sphere $\text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4  }\!\!\times\!\!\text{  }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{14} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{4  }\!\!\times\!\!\text{  44  }\!\!\times\!\!\text{  14} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 2464 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{2464 c}{{\text{m}}^{\text{2}}}$.

2. Find the surface area of a sphere of diameter:

i. $\text{14 cm}$

Ans:

Given diameter of the sphere $\text{= 14 cm}$

So, the radius of the sphere $\text{r = }\frac{\text{14}}{\text{2}}\text{ = 7 cm}$

The surface area of the sphere $\text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4  }\!\!\times\!\!\text{  }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{7} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{88  }\!\!\times\!\!\text{  7} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 616 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{616 c}{{\text{m}}^{\text{2}}}$.

ii. $\text{21 cm}$

Ans:

Given diameter of the sphere $\text{= 21 cm}$

So, the radius of the sphere $\text{r = }\frac{\text{21}}{\text{2}}\text{ = 10}\text{.5 cm}$

The surface area of the sphere $\text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4  }\!\!\times\!\!\text{  }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{10}\text{.5} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 1386 c}{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{1386 c}{{\text{m}}^{\text{2}}}$.

iii. $\text{3}\text{.5 m}$ $\left[ \text{Assume  }\!\!\pi\!\!\text{  = }\frac{\text{22}}{\text{7}} \right]$

Ans:

Given diameter of the sphere $\text{= 3}\text{.5 m}$

So, the radius of the sphere $\text{r = }\frac{\text{3}\text{.5}}{\text{2}}\text{ = 1}\text{.75 m}$

The surface area of the sphere $\text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{4  }\!\!\times\!\!\text{  }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{1}\text{.75} \right)}^{\text{2}}} \right]\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 38}\text{.5 }{{\text{m}}^{\text{2}}}$

Hence, the surface area of the sphere is $\text{38}\text{.5 }{{\text{m}}^{\text{2}}}$.

3. Find the total surface area of a hemisphere of radius $\text{10 cm}$. $\left[ \text{Use  }\!\!\pi\!\!\text{  = 3}\text{.14} \right]$

Ans:

Given the radius of hemisphere $\text{r = 10 cm}$

The total surface area of the hemisphere is the sum of its curved surface area and the circular base.

Total surface area of hemisphere $\text{A = 2 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{ +  }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = 3 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{3  }\!\!\times\!\!\text{  3}\text{.14  }\!\!\times\!\!\text{  }{{\left( \text{10} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 942 c}{{\text{m}}^{\text{2}}}$

Hence, the total surface area of the hemisphere is $\text{942 c}{{\text{m}}^{\text{2}}}$.

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

Ans:

Given the initial radius of the balloon ${{\text{r}}_{1}}\text{ = 7 cm}$

The final radius of the balloon ${{\text{r}}_{2}}\text{ = 14 cm}$

We have to find the ratio of surface areas of the balloon in the two cases.

The required ratio $\text{R = }\frac{\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}_{\text{1}}}^{\text{2}}}{\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}_{\text{2}}}^{\text{2}}}$

$\Rightarrow \text{R = }{{\left( \frac{{{\text{r}}_{\text{1}}}}{{{\text{r}}_{\text{2}}}} \right)}^{\text{2}}}$

$\Rightarrow \text{R = }{{\left( \frac{\text{7}}{\text{14}} \right)}^{\text{2}}}$

$\Rightarrow \text{R = }\frac{\text{1}}{\text{4}}$

Hence, the ratio of the surface areas of the balloon in both case is $\text{1 : 4}$.

5. A hemispherical bowl made of brass has inner diameter $\text{10}\text{.5 cm}$. Find the cost of tin plating it on the inside at the rate of $\text{Rs}\text{. 16}$ per $\text{100 c}{{\text{m}}^{\text{2}}}$. $\left[ \text{Assume  }\!\!\pi\!\!\text{  = }\frac{\text{22}}{\text{7}} \right]$

Ans:

Given the radius of inner hemispherical bowl $\text{r = }\frac{\text{10}\text{.5}}{\text{2}}\text{ = 5}\text{.25 cm}$

The surface area of the hemispherical bowl $\text{A = 2 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{2  }\!\!\times\!\!\text{  }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{5}\text{.25} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 173}\text{.25 c}{{\text{m}}^{\text{2}}}$

It is given that the cost of tin-plating $\text{100 c}{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. 16}$

So, the cost of tin-plating $173.25\text{ c}{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. }\left( \frac{\text{16}}{\text{100}}\text{  }\!\!\times\!\!\text{  173}\text{.25} \right)\text{ = Rs}\text{. 27}\text{.72}$

Hence, the cost of tin-plating the hemispherical bowl is $\text{Rs}\text{. 27}\text{.72}$.

6. Find the radius of a sphere whose surface area is $\text{154 c}{{\text{m}}^{\text{2}}}$. $\left[ \text{Assume  }\!\!\pi\!\!\text{  = }\frac{\text{22}}{\text{7}} \right]$

Ans:

Let us assume the radius of the sphere be $\text{r}$.

We are given the surface area of the sphere, $\text{A = 154 c}{{\text{m}}^{\text{2}}}$.

$\therefore \text{4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{ = 154 c}{{\text{m}}^{\text{2}}}$

$\Rightarrow r^{2}=\left(\frac{154 \times 7}{2 \times 22}\right) \mathrm{cm}^{2}$

$\Rightarrow \text{r = }\left( \frac{\text{7}}{\text{2}} \right)\text{ cm}$

$\Rightarrow \text{r = 3}\text{.5 cm}$

Therefore, the radius of the sphere is $\text{3}\text{.5 cm}$.

7. The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface area.

Ans:

Let us assume the diameter of earth is $\text{d}$.

So, the diameter of the moon will be $\frac{\text{d}}{\text{4}}$.

The radius of the earth ${{\text{r}}_{\text{1}}}\text{ = }\frac{\text{d}}{\text{2}}$

The radius of the moon ${{\text{r}}_{\text{2}}}\text{ = }\frac{\text{1}}{\text{2}}\text{  }\!\!\times\!\!\text{  }\frac{\text{d}}{\text{2}}\text{ = }\frac{\text{d}}{\text{8}}$

The ratio of surface area of moon and earth $\text{R = }\frac{\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}_{\text{2}}}^{\text{2}}}{\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}_{\text{1}}}^{\text{2}}}$

$\Rightarrow \text{R = }\frac{\text{4 }\!\!\pi\!\!\text{ }{{\left( \frac{\text{d}}{\text{8}} \right)}^{\text{2}}}}{\text{4 }\!\!\pi\!\!\text{ }{{\left( \frac{\text{d}}{\text{2}} \right)}^{\text{2}}}}$

$\Rightarrow \text{R = }\frac{\text{4}}{\text{64}}$

$\Rightarrow \text{R = }\frac{\text{1}}{\text{16}}$

Therefore, the ratio of surface area of the moon and earth is $\text{1 : 16}$.

8. A hemispherical bowl is made of steel, $\text{0}\text{.25 cm}$ thick. The inner radius of the bowl is $\text{5 cm}$. Find the outer curved surface area of the bowl. $\left[ \text{Assume  }\!\!\pi\!\!\text{  = }\frac{\text{22}}{\text{7}} \right]$

Ans:

Given the inner radius $\text{= 5 cm}$

The thickness of the bowl $\text{= 0}\text{.25 cm}$

So, the outer radius of the hemispherical bowl is $\text{r = }\left( \text{5 + 0}\text{.25} \right)\text{ cm = 5}\text{.25 cm}$

The outer curved surface area of the hemispherical bowl $\text{A = 2 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A =}\left[ \text{ 2  }\!\!\times\!\!\text{  }\frac{\text{2}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{5}\text{.25} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 173}\text{.25 c}{{\text{m}}^{\text{2}}}$

Therefore, the outer curved surface area of the hemispherical bowl is $\text{173}\text{.25 c}{{\text{m}}^{\text{2}}}$.

9. A right circular cylinder just encloses a sphere of radius $\text{r}$ (see figure). Find

(Image Will Be Updated Soon)

i. surface area of the sphere, 

Ans:

The surface area of the sphere is $\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$.

ii. curved surface area of the cylinder, 

Ans:

(Image Will Be Updated Soon)

Given the radius of cylinder $\text{= r}$

The height of cylinder $\text{= r + r = 2r}$

The curved surface area of cylinder $\text{A = 2 }\!\!\pi\!\!\text{ rh}$

$\Rightarrow \text{A = 2 }\!\!\pi\!\!\text{ r }\left( \text{2r} \right)$

$\Rightarrow \text{A = 4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

Therefore the curved surface area of the cylinder is  $\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$.

iii. ratio of the areas obtained in i. and ii.

The ratio of surface area of the sphere and curved surface area of cylinder  $\text{R = }\frac{\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}}{\text{4 }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}}$

$\text{R = }\frac{\text{1}}{\text{1}}$

Therefore, the required ratio is $\text{1 : 1}$.

NCERT Solutions for Class 9 Maths

• Chapter 1 - Number System

• Chapter 2 - Polynomials 

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introductions to Euclid's Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelogram and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Heron's formula

• Chapter 13 - Surface area and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

NCERT Solution Class 9 Maths of Chapter 13 All Exercises

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.4

Opting for the NCERT solutions for Ex 13.4 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 13.4 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 9 students who are thorough with all the concepts from the Subject Surface Areas and Volumes textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 9 Maths Chapter 13 Exercise 13.4 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 9 Maths Chapter 13 Exercise 13.4, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 9 Maths Chapter 13 Exercise 13.4 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.