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# NCERT Solutions for Class 9 Maths Chapter 13: Surface Areas and Volumes - Exercise 13.6

Last updated date: 02nd Aug 2024
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## NCERT Solutions for Class 9 Maths Chapter 13 (Ex 13.6)

Free PDF download of NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.6 and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.6 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all CBSE Solutions in your emails. You can also download Maths NCERT Solutions Class 9 to help you to revise complete syllabus and score more marks in your examinations. Students can also avail of NCERT Solutions for Class 9 Science from our website.

 Class: NCERT Solutions for Class 9 Subject: Class 9 Maths Chapter Name: Chapter 13 - Surface Areas and Volumes Exercise: Exercise - 13.6 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes
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## Access NCERT Solutions for Class 9 Maths Chapter 13 - Surface Areas and Volumes

Exercise 13.6

1. The circumference of the base of cylindrical vessel is $\text{132 cm}$ and its height is $\text{25 cm}$. How many litres of water can it hold? $\left( \text{1000 c}{{\text{m}}^{\text{3}}}\text{ = 1 litre} \right)$. $\left[ \text{Assume }\!\!\pi\!\!\text{ =}\frac{\text{22}}{\text{7}} \right]$

Ans:

Given the height of the vessel $\text{h = 25 cm}$

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

The circumference of the vessel $\text{= 132 cm}$

$\therefore \text{2 }\!\!\pi\!\!\text{ r = 132 cm}$

$\Rightarrow \text{r = }\left( \frac{\text{132 }\!\!\times\!\!\text{ 7}}{\text{2 }\!\!\times\!\!\text{ 22}} \right)\text{ cm}$

$\text{r = 21 cm}$

The volume of the cylindrical vessel $\text{V = }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\Rightarrow \text{V = }\frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ }{{\left( \text{21} \right)}^{\text{2}}}\text{ }\!\!\times\!\!\text{ 25}$

$\Rightarrow \text{V = 34650 c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\frac{\text{34650}}{\text{1000}}\text{ litres}$

$\Rightarrow \text{V = 34}\text{.65 litres}$

Therefore, the cylindrical vessel can hold $\text{34}\text{.65 litres}$ of water.

2. The inner diameter of a cylindrical wooden pipe is $\text{24 cm}$ and its outer diameter is $\text{28 cm}$. The length of the pipe is $\text{35 cm}$. Find the mass of the pipe, if $\text{1 c}{{\text{m}}^{\text{3}}}$ of wood has a mass of $\text{0}\text{.6 g}$. $\left[ \text{Assume }\!\!\pi\!\!\text{ =}\frac{\text{22}}{\text{7}} \right]$

Ans:

Given the inner diameter of cylindrical pipe $\text{= 24 cm}$

So, the inner radius $\text{r = }\frac{\text{24}}{\text{2}}\text{ = 12 cm}$

Given the outer diameter of cylindrical pipe $\text{= 28 cm}$

So, the outer radius $\text{R = }\frac{\text{28}}{\text{2}}\text{ = 14 cm}$

The height of the wooden pipe $\text{h = 35 cm}$.

The volume of the cylindrical pipe $\text{V = }\!\!\pi\!\!\text{ }\left( {{\text{R}}^{\text{2}}}\text{ - }{{\text{r}}^{\text{2}}} \right)\text{h}$

$\Rightarrow \text{V = }\left[ \frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ }\left( \text{1}{{\text{4}}^{\text{2}}}\text{ - 1}{{\text{2}}^{\text{2}}} \right)\text{ }\!\!\times\!\!\text{ 35} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\left( \text{110 }\!\!\times\!\!\text{ 52} \right)\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 5720 c}{{\text{m}}^{\text{3}}}$

It is given that the mass of $\text{1 c}{{\text{m}}^{\text{3}}}$ wood $\text{= 0}\text{.6 g}$

So, the mass of $\text{5720 c}{{\text{m}}^{\text{3}}}$ wood $\text{= }\left( \frac{\text{5720 }\!\!\times\!\!\text{ 0}\text{.6}}{\text{1000}} \right)\text{ kg = 3}\text{.432 kg}$

Therefore, the mass of the pipe is $\text{3}\text{.432 kg}$.

3. A soft drink is available in two packs − (i) a tin can with a rectangular base of length $\text{5 cm}$ and width $\text{4 cm}$, having a height of $\text{15 cm}$ and (ii) a plastic cylinder with circular base of diameter $\text{7 cm}$ and height $\text{10 cm}$. Which container has greater capacity and by how much? $\left[ \text{Assume }\!\!\pi\!\!\text{ =}\frac{\text{22}}{\text{7}} \right]$

Ans:

We will calculate for the cuboidal tin can.

Length of tin can $\text{l = 5 cm}$

Breadth of tin can $\text{b = 4 cm}$

Height of tin can $\text{h = 15 cm}$

The capacity of a tin can ${{\text{V}}_{1}}\text{ = l }\!\!\times\!\!\text{ b }\!\!\times\!\!\text{ h}$

$\Rightarrow {{\text{V}}_{1}}\text{ = }\left( \text{5 }\!\!\times\!\!\text{ 4 }\!\!\times\!\!\text{ 15} \right)\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow {{\text{V}}_{1}}\text{ = 300 c}{{\text{m}}^{\text{3}}}$

We will calculate for the plastic cylinder with circular base.

Height of plastic cylinder $\text{H = 10 cm}$

The diameter of circular base $\text{= 7 cm}$

So, the radius $\text{r = }\frac{\text{7}}{\text{2}}\text{ = 3}\text{.5 cm}$

The capacity of plastic cylinder ${{\text{V}}_{\text{2}}}\text{ = }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\Rightarrow {{\text{V}}_{\text{2}}}\text{ = }\left( \frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ }{{\left( \text{3}\text{.5} \right)}^{\text{2}}}\text{ }\!\!\times\!\!\text{ 10} \right)\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow {{\text{V}}_{\text{2}}}\text{ = }\left( \text{11 }\!\!\times\!\!\text{ 35} \right)\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow {{\text{V}}_{\text{2}}}\text{ = 385 c}{{\text{m}}^{\text{3}}}$

From both the values of capacities we found that ${{\text{V}}_{\text{2}}}\text{ }{{\text{V}}_{\text{1}}}\text{ }$.

The difference in capacity $\text{= }\left( \text{385 - 300} \right)\text{ c}{{\text{m}}^{\text{3}}}\text{ = 85 c}{{\text{m}}^{\text{3}}}$

Therefore, we can say that the capacity of plastic cylinder is greater than the tin can by $\text{85 c}{{\text{m}}^{\text{3}}}$.

4. If the lateral surface of a cylinder is $\text{94}\text{.2 c}{{\text{m}}^{\text{2}}}$ and its height is $\text{5 cm}$, then find

Ans:

We are given the height of cylinder $\text{h = 5 cm}$.

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

The curved surface area of cylinder $\text{A = 94}\text{.2 c}{{\text{m}}^{\text{2}}}$.

But we know the curved surface area of cylinder $\text{= 2 }\!\!\pi\!\!\text{ rh}$

$\therefore \text{2 }\!\!\pi\!\!\text{ rh = 94}\text{.2 c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \left( \text{2 }\!\!\times\!\!\text{ 3}\text{.14 }\!\!\times\!\!\text{ r }\!\!\times\!\!\text{ 5} \right)\text{ cm = 94}\text{.2 c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{r = }\frac{\text{94}\text{.2}}{\text{31}\text{.4}}\text{ cm}$

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

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

ii. its volume. $\left[ \text{Use }\!\!\pi\!\!\text{ = 3}\text{.14} \right]$

Ans:

The volume of the cylinder $\text{V = }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\Rightarrow \text{V = }\left( \text{3}\text{.14 }\!\!\times\!\!\text{ }{{\left( \text{3} \right)}^{\text{2}}}\text{ }\!\!\times\!\!\text{ 5} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{V = 141}\text{.3 c}{{\text{m}}^{\text{3}}}$

Therefore, the volume of the cylinder is $\text{141}\text{.3 c}{{\text{m}}^{\text{3}}}$.

5. It costs $\text{Rs}\text{. 2200}$to paint the inner curved surface of a cylindrical vessel $\text{10 m}$ deep. If the cost of painting is at the rate of $\text{Rs}\text{. 20}$ per ${{\text{m}}^{\text{2}}}$, find

i.Inner curved surface area of the vessel

Ans:

It is given that it requires $\text{Rs}\text{. 20}$ to paint an area of $\text{1 }{{\text{m}}^{\text{2}}}$.

So, it requires $\text{Rs}\text{. 2200}$ to paint an area of $\left( \frac{\text{1}}{\text{20}}\text{ }\!\!\times\!\!\text{ 2200} \right)\text{ }{{\text{m}}^{\text{2}}}\text{ = 110 }{{\text{m}}^{\text{2}}}$

Therefore, the inner surface area is $\text{110 }{{\text{m}}^{\text{2}}}$.

Ans:

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

The height of the vessel is $\text{h = 10 m}$.

The surface area of the vessel is $\text{110 }{{\text{m}}^{\text{2}}}$.

$\therefore \text{2 }\!\!\pi\!\!\text{ rh = 110 }{{\text{m}}^{\text{2}}}$

$\Rightarrow \left[ \text{2 }\!\!\times\!\!\text{ }\frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ r }\!\!\times\!\!\text{ 10} \right]\text{ m = 110 }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{r = }\frac{\text{7}}{\text{4}}\text{ m}$

$\Rightarrow \text{r = 1}\text{.75 m}$

Therefore, the radius of the base is $\text{1}\text{.75 m}$.

iii. Capacity of the vessel

Ans:

The volume of the vessel $\text{V = }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\Rightarrow \text{V = }\left[ \frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ }{{\left( \text{1}\text{.75} \right)}^{\text{2}}}\text{ }\!\!\times\!\!\text{ 10} \right]\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 96}\text{.25 }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\left( \text{96}\text{.25 }\!\!\times\!\!\text{ 1000} \right)\text{ litres}$

$\Rightarrow \text{V = 96250 litres}$

Therefore, the capacity of the vessel is $\text{96250 litres}$.

6. The capacity of a closed cylindrical vessel of height $\text{1 m}$ is $\text{15}\text{.4 litres}$. How many square metres of metal sheet would be needed to make it? $\left[ \text{Assume }\!\!\pi\!\!\text{ = }\frac{\text{22}}{\text{7}} \right]$

Ans:

Given height of cylindrical vessel $\text{h = 1 m}$

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

The volume of cylindrical vessel $\text{= 15}\text{.4 litres = 0}\text{.0154 }{{\text{m}}^{\text{3}}}$

$\therefore \text{ }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{h = 0}\text{.0154}$

$\Rightarrow \left( \frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ }{{\text{r}}^{\text{2}}}\text{ }\!\!\times\!\!\text{ 1} \right)\text{ m = 0}\text{.0154 }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{r = 0}\text{.07 m}$

The total surface area of the vessel $\text{A = 2 }\!\!\pi\!\!\text{ r }\left( \text{h + r} \right)$

$\Rightarrow \text{A = }\left[ \text{2 }\!\!\times\!\!\text{ }\frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ 0}\text{.07 }\left( \text{1 + 0}\text{.07} \right) \right]\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{0}\text{.44 }\!\!\times\!\!\text{ 1}\text{.07} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 0}\text{.4708 }{{\text{m}}^{\text{2}}}$

Therefore, the amount of metal sheet required to make the cylindrical vessel is $\text{0}\text{.4708 }{{\text{m}}^{\text{2}}}$.

7. A lead pencil consists of a cylinder of wood with solid cylinder of graphite filled in the interior. The diameter of the pencil is $\text{7 mm}$ and the diameter of the graphite is $\text{1 mm}$. If the length of the pencil is $\text{14 cm}$, find the volume of the wood and that of the graphite. $\left[ \text{Assume }\!\!\pi\!\!\text{ =}\frac{\text{22}}{\text{7}} \right]$

Ans:

Given the diameter of the pencil $\text{= 7 mm}$

So, the radius of pencil ${{\text{r}}_{\text{1}}}\text{ = }\frac{\text{7}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{10}}\text{ cm = 0}\text{.35 cm}$

The diameter of graphite $\text{= 1 mm}$

So, the radius of graphite ${{\text{r}}_{\text{2}}}\text{ = }\frac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{10}}\text{ cm = 0}\text{.05 cm}$

The height of the pencil $\text{h = 14 cm}$

The volume of wood in the pencil $\text{V = }\!\!\pi\!\!\text{ }\left( \text{r}_{\text{2}}^{\text{2}}\text{ - r}_{\text{1}}^{\text{2}} \right)\text{h}$

$\Rightarrow \text{V =}\left[ \frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ }\left\{ {{\left( \text{0}\text{.35} \right)}^{\text{2}}}\text{ - }{{\left( \text{0}\text{.05} \right)}^{\text{2}}} \right\}\text{ }\!\!\times\!\!\text{ 14} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\left[ \frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ }\left\{ \text{0}\text{.1225 - 0}\text{.0025} \right\}\text{ }\!\!\times\!\!\text{ 14} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\left( \text{44 }\!\!\times\!\!\text{ 0}\text{.12} \right)\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 5}\text{.28 c}{{\text{m}}^{\text{3}}}$

The volume of graphite $\text{{V}' = }\!\!\pi\!\!\text{ r}_{\text{2}}^{\text{2}}\text{h}$

$\Rightarrow \text{{V}' =}\left[ \frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ }{{\left( \text{0}\text{.05} \right)}^{\text{2}}}\text{ }\!\!\times\!\!\text{ 14} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{{V}' =}\left( \text{44 }\!\!\times\!\!\text{ 0}\text{.0025} \right)\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{{V}' = 0}\text{.11 c}{{\text{m}}^{\text{3}}}$

Therefore, the volume of wood is $\text{5}\text{.28 c}{{\text{m}}^{\text{3}}}$ and the volume of graphite is $\text{0}\text{.11 c}{{\text{m}}^{\text{3}}}$.

8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter $\text{7 cm}$. If the bowl is filled with soup to a height of $\text{4 cm}$, how much soup the hospital has to prepare daily to serve $\text{250}$ patients? $\left[ \text{Assume }\!\!\pi\!\!\text{ = }\frac{\text{22}}{\text{7}} \right]$

Ans:

It is given that the diameter of cylindrical bowl $\text{= 7 cm}$

So, the radius $\text{r = }\frac{\text{7}}{\text{2}}\text{ = 3}\text{.5 cm}$

Let $\text{h}$ be the height of the soup in the bowl and $\text{h = 4 cm}$.

The volume of the soup in the bowl $\text{V = }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}\text{h}$

$\Rightarrow \text{V = }\left[ \text{ }\frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ }{{\left( \text{3}\text{.5} \right)}^{\text{2}}}\text{ }\!\!\times\!\!\text{ 4} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = }\left[ \text{ 11 }\!\!\times\!\!\text{ 3}\text{.5 }\!\!\times\!\!\text{ 4} \right]\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 154 c}{{\text{m}}^{\text{3}}}$

So, the total volume of the soup given to $250$ patients $\text{= }\left( \frac{\text{250 }\!\!\times\!\!\text{ 154}}{\text{1000}} \right)\text{ litres = 38}\text{.5 litres}$

### Important Formulas for NCERT Solutions Maths Class 9 Exercise 13.6

Exercise 13.6 NCERT Maths Class 9 is based on the volume of a cylinder. There are a total of 8 questions asked in the 6th exercise of the chapter Surface Areas and Volumes. There is one most important formula related to the volume of the cylinder as given below:

Volume of cylinder = area of circular base × height = π r2 h

where r is the radius of the circular base and h is the height of the cylinder.

You also need to know some other basic formulas to solve the problems asked in Exercise 13.6.

For Question 1: Circumference of circular base = 2π r

For Question 2: Radius = $\frac{Diameter}{2}$

Mass of an object = Density of an object × volume of an object

For Question 3: Rectangular base of some length forms a cuboid. Therefore, we need to find the volume of the cuboid and compare it with the volume of the cylinder.

Volume of cuboid = length × width × height

For Question 4: Lateral surface of cylinder = 2π r h

For Question 5: To calculate capacity, calculate volume in m3, and then used 1 m3 = 1000 l to find the capacity of the cylindrical vessel.

For Question 6: Total metal sheet required = Total surface area of the vessel =  2π r (h + r)

For Question 8: Total amount of the soup the hospital has to prepare (in litres)

= $\frac{Number \, of \, patients \ast Volume\, of \, the \, bowl (by \, 4 cm \, height \, only)}{1000}$

## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.6

Opting for the NCERT solutions for Ex 13.6 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 13.6 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 Maths 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 13.6 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 13.6, 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 13.6 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 on NCERT Solutions for Class 9 Maths Chapter 13: Surface Areas and Volumes - Exercise 13.6

1. What is a combination of solids?

Combination of solids consists of 2 sections in it. They are:

• Surface Area of a Combination of Solids

In NCERT Class 9 Maths, you will learn how to break down a solid and analyse its shape in an easy way. You can find the areas of the unknown shapes with known shapes.

• The Volume of Combined Solids

In this segment of NCERT Solutions for Class 9 Maths, you will be learning how to find the volumes of complex objects with visualising in a simplified way of shapes.

The volume of combined solids

V(solid) = V(Cone) + V(hemisphere)

2. How many questions are there in this exercise?

This exercise of  NCERT Class 9 Maths book consists of 9 questions.  Question 1 is based on the cube, where you will have to find the surface area of it. Question 2, 3, 4, 5, 6 and 9 are hemisphere based questions. In which you will have to find the surface area of the given shape. Question 7 is based on the cylinder, in which you have to find the area of the canvas. Question 8 is asked from the solid cylinder, where you will be finding the surface area of the solid.

3. What will I learn in the cuboid and surface area?

In this part of NCERT Solutions for Class 9 Maths Chapter 13, you will be taught how to find out the surface area of a cuboid. You will also understand how it is equal to the sum of the areas of its six rectangular faces.

cuboid = 2(l×b)+2(b×h)+2(l×h)=2(lb+bh+lh)

You will also learn about the cube and its surface area, cylinder and its surface area, right circular cone and its surface area and sphere and it’s surface area.

Volume of Cube

Here, you will learn that all the dimensions of the cube are identical, Volume = l3

Here, l will be the length of the cube of the edge

The volume of a cube = base area×height

4. How do I increase my ranks with Vedantu’s study guide?

You can improve your grades with our help. We at Vedantu have a team of experts who understand the academic needs and requirements of students. The study material provided by Vedantu is crafted as per the latest syllabus and is made available to everybody through our online portal.

Our solutions and study guide are designed by our subject-matter experts in a step by step way for easy and better understanding of the concept. We made sure that CBSE and NCERT guidelines are strictly followed while drafting these solutions. All the topics are explained in detail to make it more clear for you.

5. Which is more difficult, Surface Area and Volumes or Geometry?

It is quite a tough question as not all will find difficulty in Surface Area and Volumes and not all will find difficulty in Geometry. It is basically up to you. However, for most students, Geometry is quite difficult as it has figures, measurements, derivations, and more, which are not easy in comparison to Surface Area and Volumes, which have the necessary formulas for solving. But, it is all in the mind. Practice will make you an expert in both topics.

6. What is the minimum number of hours required to study for Class 9 Chapter 13 Maths?

The minimum number of hours required to study for Class 9 Maths Chapter 13 varies amongst students. But on average, if you are preparing for Class 9 Maths Chapter 13 then you should study for at least three hours a day and then do the revision again. However, you shouldn’t try to do it at a stretch and take breaks in between.

7. Does Chapter 13 matter in Class 10 Maths?

Yes, it does matter. This chapter forms a base, and it is important to be thorough here as it continues even till Class 12. To know more about the chapter, you can visit Vedantu. The solutions are created by experts and are completely free.

8. How can I top in  Class 9 Maths Chapter 13 during the class test?

To top in Class 9 Chapter 13 Maths you have to follow some easy tips which include proper concentration. Place your mind towards one question and don’t waste your time looking around other questions. Practice your coursebook questions one by one, when you think you have completed them, pick up the sample papers of the past year’s and see if you can find questions from Chapter 13. You can check NCERT Solutions for Class 9 Maths free of cost for further knowledge. These solutions are available free of cost on Vedantu’s website(vedantu.com) and mobile app.

9. How many questions are there in Chapter 13 Exercise 13.6 of Class 9 Maths?

In Chapter 13 Exercise 13.6 of Class 9 Maths, there are only eight questions, however, these eight questions are long and are difficult too. These questions can be solved by applying the formulas of the geometrical objects. In some of the questions, there are diagrams also, and you have to solve the questions by seeing the dimensions from the diagrams given.