NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.4

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

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Access NCERT solutions Class 9 Maths Chapter 13 - Surface Areas and Volumes part-1

Access NCERT solutions Class 9 Maths Chapter 13 - Surface Areas and Volumes

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 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.

FAQs (Frequently Asked Questions)

Q1. What is meant by the term ‘surface area’?

Ans: The surface area of any solid or 3D shape can be defined as the sum of the areas of all its sides. It gives a measure of the area that a 3D figure occupies. There is a list of formulas that are commonly used to calculate the surface areas of 3D shapes, in maths. For example, there is a particular formula for calculating the surface area of a cube, whereas there is another formula for calculating the surface area of a sphere, and so on. Also, there are two types of surface areas for some 3D shapes, namely lateral surface area and total surface area.

Q2. What is meant by the ‘volume’ of a 3D shape?

Ans: The volume of a 3D shape or solid is the amount of space occupied by it. The volume of a 3D shape also gives the measure of the matter enclosed within the boundaries of the solid. If we measure the volume of a hollow or empty object, we can get the total volume of air or water that it can hold. Volume can be found only for 3-dimensional objects and nor for 2D objects, such as square or rectangle. In Maths, there are separate formulas to calculate the volume of various 3D objects. For example, the formula to find the volume of a cube is different from the formula to find the volume of a cuboid.

Q3. How many sums are there in the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes, exercise 13.4?

Ans: There are a total of 9 sums in the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes, exercise 13.4. Most of the sums in this exercise are based on the concept of surface areas and volumes of spheres and cylinders. When you refer to the solutions of these sums, you will be able to understand that you will have to write statements for solving these sums. All the 9 sums are solved and explained step by step in these NCERT solutions for Surface Areas and Volumes, exercise 13.4.

Q4. Are the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.4) Exercise 13.4 helpful?

Ans: Yes, the NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.4) Exercise 13.4 are very helpful for the exam preparation. These solutions are prepared by the highly experienced faculty at Vedantu, in a step by step manner. Thereby, these stepwise NCERT solutions will guide you to comprehend the concepts of Surface Areas and Volumes for various solid shapes. Therefore, these solutions make a self-explanatory study resource for practice and revision purposes for your exam preparation. Also, you can download these NCERT Solutions for free from our website as well as from our mobile application.

Q5. Where can I find the NCERT Solutions to Class 9 Maths Chapter 13 Exercise 13.4?

Ans: NCERT Solutions to Class 9 Maths Chapter 13 Exercise 13.4 are available on Vedantu. You can download NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.4 PDF free of cost. These NCERT Solutions are available in PDF format on the website as well as on the mobile app of Vedantu. It has helped thousands of students in their learning path and is continuing to do so. These solutions are created by highly experienced faculty and are explained in an easy-to-understand stepwise manner. 

Q6. Is Class 9 Maths Chapter 13 considered the toughest chapter?

Ans: No, not really, Chapter 13 of Class 9 Maths may be tricky so it requires good practice to understand the concepts covered in this chapter. If you practice regularly and with proper concentration then it will not be a tough chapter. Even if you don’t get the answer on your first try, solve it again, and eventually, you will be able to get the correct answer. You can verify your solutions from the link NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.4.

Q7. Can I seek help in solving the sums of Chapter 13 Exercise 13.4 from the experts at Vedantu?

Ans: Yes, our experts at Vedantu are always ready to make learning easy for you. They try to help you out as soon as possible when you reach out with any doubt. Our subject experts are highly professional and have years of teaching experience. Each expert has a wide knowledge of the specific subject. They will address your doubts and explain the sums to you. So you may post your doubt or get in contact with our experts for guidance in Chapter 13 Exercise 13.4.

Q8. What makes Vedantu’s NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.4 more reliable than other platforms?

Ans: Vedantu is one such platform that has a wholesome amount of brand value. It has been working for the best of students' interests and tries to update its platform as per the current education requirement. Queries are easily resolved without any barrier. The content is curated as per the latest format of CBSE. By applying the easy tricks and tips of problem-solving, students can solve the sums in Exercise 13.4 smoothly. Hence, you can download the PDF of NCERT Solutions of Class 9 Maths Chapter 13 Ex-13.4 from the Vedantu app or website.

Q9. Can I prepare for Class 9 in one month?

Ans: The time required for exam preparation entirely depends on the speed and accuracy of students. For some, one month can be a shorter period while for some it can be long enough. But the sooner the better, so it is advisable, to begin with the chapter as early as you can. 

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