NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes (Ex 13.4) Exercise 13.4

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes (Ex 13.4) Exercise 13.4

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NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes (Ex 13.4) Exercise 13.4 part-1

Access NCERT Solutions for Class 10 Maths Chapter 13 – Surface Areas and Volumes

Exercise 13.4

1. A drinking glass is in the shape of a frustum of a cone of height\[\mathbf{14}\text{ }\mathbf{cm}\]. The diameters of its two circular ends are \[\mathbf{4}\text{ }\mathbf{cm}\] and \[\mathbf{2}\text{ }\mathbf{cm}\]. Find the capacity of the glass. $\left[ Use\,\,\pi =\dfrac{22}{7} \right]$

Ans:

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Radius (${{r}_{1}}$) of upper base of glass $=\dfrac{4}{2}=2cm$

Radius (${{r}_{2}}$) of lower base of glass$=\dfrac{2}{2}=1cm$ 

Height = $14cm$

Capacity of glass = Volume of frustum of cone

\[=\dfrac{1}{3}\pi\text{h}\left[ \text{r}_{1}^{2}+\text{r}_{2}^{2}+{{\text{r}}_{1}}{{\text{r}}_{2}} \right]\]

\[=\dfrac{1}{3}\pi \text{h}\left[ {{(2)}^{2}}+{{(1)}^{2}}+(2)(1) \right]\]

\[=\dfrac{1}{3}\times \dfrac{22}{7}\times 14(4+1+2)\]

\[=\dfrac{308}{3}\]

\[=102\dfrac{2}{3}~\text{c}{{\text{m}}^{3}}\]

Therefore, the capacity of the glass \[=102\dfrac{2}{3}~\text{c}{{\text{m}}^{3}}\]


2. The slant height of a frustum of a cone is \[\mathbf{4}\text{ }\mathbf{cm}\] and the perimeters (circumference) of its circular ends are \[\mathbf{18}\text{ }\mathbf{cm}\] and \[\mathbf{6}\text{ }\mathbf{cm}\]. Find the curved surface area of the frustum.

Ans:

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Perimeter of upper circular end of frustum $=18 cm$

\[\Rightarrow 2\pi {{r}_{1}}\text{ }=18\]

\[\Rightarrow {{r}_{1}}=\dfrac{9}{\pi }\]

Perimeter of lower circular end of frustum $=6 cm$

\[\Rightarrow 2\pi {{r}_{2}}\text{ }=6\]

\[\Rightarrow {{r}_{2}}=\dfrac{3}{\pi }\]

Slant height (l) of frustum $=4 cm$

CSA of frustum:

$\Rightarrow \pi \left( {{r}_{1}}+{{r}_{2}} \right)l$

$\Rightarrow \pi \left( \dfrac{9}{\pi }+\dfrac{3}{\pi } \right)4$

$\Rightarrow 12\left( 4 \right)=48c{{m}^{2}}$

Therefore, the curved surface area of the frustum$=48c{{m}^{2}}$


3. A fez, the cap used by the Turks, is shaped like the frustum of a cone (see the figure given below). If its radius on the open side is\[\mathbf{10}\text{ }\mathbf{cm}\], radius at the upper base is \[\mathbf{4}\text{ }\mathbf{cm}\] and its slant height is\[\mathbf{15}\text{ }\mathbf{cm}\], find the area of material used for making it. $\left[ Use\,\,\pi =\dfrac{22}{7} \right]$

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Ans:

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Radius (${{r}_{2}}$) at upper circular end \[=\text{ }4\text{ }cm\]

Radius (${{r}_{1}}$) at lower circular end \[=10\text{ }cm\]

Slant height (l) of frustum \[=15\text{ }cm\]

Area of material used for making the fez:

$\Rightarrow$ CSA of frustum+Area of upper circular end

$\Rightarrow \pi \left( {{r}_{1}}+{{r}_{2}} \right)l+\pi r_{2}^{2}$ 

$\Rightarrow \pi (10+4)15+\pi (4)^2$

$\Rightarrow \pi (14)15+16\pi $

$\Rightarrow 210\pi +16\pi =\dfrac{226\times 22}{7}$

$\Rightarrow 710\dfrac{2}{7}~\text{c}{{\text{m}}^{2}}$

Therefore, the area of material used for making it is $710\dfrac{2}{7}~\text{c}{{\text{m}}^{2}}$.


4. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height \[16\text{ }cm\] with radii of its lower and upper ends as \[\mathbf{8}\text{ }\mathbf{cm}\] and \[\mathbf{20}\text{ }\mathbf{cm}\] respectively. Find the cost of the milk which can completely fill the container, at the rate of \[\mathbf{Rs}.\mathbf{20}\]per liter. Also find the cost of metal sheet used to make the container, if it costs\[\mathbf{Rs}.\mathbf{8}\text{ }\mathbf{per}\text{ }\mathbf{100}\text{ }\mathbf{c}{{\mathbf{m}}^{2}}\]. Take \[\mathbf{\pi }\text{ }=\text{ }\mathbf{3}.\mathbf{14}\]

Ans:

Radius (\[{{r}_{1}}\]) of upper end of container \[=20\text{ }cm\]

Radius (\[{{r}_{2}}\]) of lower end of container \[=8\text{ }cm\]

Height (h) of container \[=16\text{ }cm\]

Slant height (l) of frustum $=\sqrt{{{\left( {{r}_{1}}-{{r}_{2}} \right)}^{2}}+{{h}^{2}}}$

$=\sqrt{{{(20-8)}^{2}}+{{(16)}^{2}}}$

$=\sqrt{{{(12)}^{2}}+{{(16)}^{2}}}=\sqrt{144+256}$

$=20~\text{cm}$

Capacity of container $=$ Volume of frustum 

$=\dfrac{1}{3}\pi \text{h}\left[ \text{r}_{1}^{2}+\text{r}_{2}^{2}+{{\text{r}}_{1}}{{\text{r}}_{2}} \right]$

$=\dfrac{1}{3}\times 3.14\times 16\times \left[ {{(20)}^{2}}+{{(8)}^{2}}+(20)(8) \right]$

$=\dfrac{1}{3}\times 3.14\times 16(400+64+160)$

$=\dfrac{1}{3}\times 3.14\times 16\times 624$

$=10449.92~\text{c}{{\text{m}}^{3}}$

$=10.45$ litres

Cost of 1 litre milk$=\operatorname{Rs}20$

Cost of \[10.45\] litre milk $=\operatorname{Rs} 209$ 

Area of metal sheet used to make the container $=\pi \left( {{\text{r}}_{1}}+{{\text{r}}_{2}} \right)\text{l}+\pi {{\text{r}}_{2}}^{2}$

$ =\pi (20+8)20+\pi {{(8)}^{2}} $ 

$=560\pi +64\pi =624\pi \text{c}{{\text{m}}^{2}}$

Cost of $100~\text{c}{{\text{m}}^{2}}$ metal sheet$=Rs\,8$

Cost of $624\pi \text{c}{{\text{m}}^{2}}$ metal sheet $= \dfrac{624\times 3.14\times 8}{100}=\operatorname{Rs}156.75$

Therefore, the cost of the milk which can completely fill the container is \[Rs\text{ }209\]  and the cost of metal sheet used to make the container is \[Rs156.75\]


5. A metallic right circular cone \[\mathbf{20}\text{ }\mathbf{cm}\] high and whose vertical angle is \[\mathbf{60}{}^\circ \] is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained is drawn into a wire of diameter$\dfrac{1}{16}cm$, find the length of the wire. $\left[ Use\,\,\pi =\dfrac{22}{7} \right]$

Ans:

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In $\vartriangle \text{AEG},$

$\dfrac{\text{EG}}{\text{AG}}=\tan {{30}^{{}^\circ }}$

$\text{EG}=\dfrac{10}{\sqrt{3}}~\text{cm}=\dfrac{10\sqrt{3}}{3}$

In \[\vartriangle \text{ABD},\]

\[\dfrac{\text{BD}}{\text{AD}}=\tan {{30}^{{}^\circ }}\]

\[\text{BD}=\dfrac{20}{\sqrt{3}}=\dfrac{20\sqrt{3}}{3}~\text{cm}\]

Radius $\left( {{r}_{1}} \right)$ of upper end of frustum 

$=\dfrac{10\sqrt{3}}{3}~\text{cm}$

Radius $\left( {{r}_{2}} \right)$ of lower end = $\dfrac{20\sqrt{3}}{3}$cm

 Height $(h)$  $=10~\text{cm}$

Volume of frustum:

$\Rightarrow\dfrac{1}{3}\pi\text{h}\left[\text{r}_{1}^{2}+\text{r}_{2}^{2}+{{\text{r}}_{1}}{{\text{r}}_{2}} \right]$

$\Rightarrow\left.\dfrac{1}{3}\times\pi\times10[{{\left(\dfrac{10\sqrt{3}}{3}\right)}^{2}}+{{\left(\dfrac{20\sqrt{3}}{3}\right)}^{2}}+\dfrac{(10\sqrt{3})(20\sqrt{3})}{3\times 3} \right]$

$\Rightarrow \dfrac{10}{3}\pi \left[ \dfrac{100}{3}+\dfrac{400}{3}+\dfrac{200}{3} \right]$

$\Rightarrow \dfrac{10}{3}\times \dfrac{22}{7}\times \dfrac{700}{3}$

$\Rightarrow \dfrac{22000}{9}~\text{c}{{\text{m}}^{3}}$

Radius \[(r)\] of wire$=\dfrac{1}{16}\times \dfrac{1}{2}=\dfrac{1}{32}~\text{cm}$

Let the length of wire be $l$

Volume of wire $=Area\,\,of\,cross\,section\times Length$

$=\left( \pi {{r}^{2}} \right)\times (l)$

$=\pi \times {{\left( \dfrac{1}{32} \right)}^{2}}\times l$

Volume of frustum = Volume of wire

$\Rightarrow \dfrac{22000}{9}=\dfrac{22}{7}{{\left( \dfrac{1}{32} \right)}^{2}}l$

$\Rightarrow \dfrac{7000}{9}\left( 1024 \right)=l$

\[\Rightarrow l=796444.44~\text{cm}\]

\[\Rightarrow l=7964.44~\text{m}\]


NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes   

Exercise 13.4

Opting for the NCERT solutions for Ex 13.4 Class 10 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 10 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 10 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 10 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 10 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 10 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)

1. How many questions are there in the NCERT Class 10 Maths Chapter 13 Exercise 13.4?

There are five questions in total in the NCERT Class 10 Maths Chapter 13 Exercise 13.4. Answers to these questions are provided in the readymade NCERT Solutions by Vedantu, which can be found on Vedantu website and mobile application.

2. How does Vedantu’s NCERT Solutions Class 10 Maths Chapter 13 Exercise 13.4 help to score well in the exam?

Vedantu’s NCERT Solutions Class 10 Maths Chapter 13 Exercise 13.4 are prepared by the highly qualified subject matter experts. Hence these NCERT solutions are considered as the best aid for the students. By referring to these NCERT Solutoons, students of CLss 10 will be well versed with all the concepts thoroughly. Thus, every student can score the highest possible marks in the exam.

3. How can you make your Class 10 Maths preparation easier?

You can always use the NCERT Solutions created by Vedantu’s experts to make your Class 10th exam preparation easier.

4. Can I download these NCERT Solutions by Vedantu at free of cost?

Yes, you can download these NCERT Solutions by Vedantu at free of cost. You need to visit the official website of Vedantu and register yourself by enlisting your email address and phone number. Then you will be able to download not only the NCERT Solutions but all the study materials for absolutely free of cost.

5. How many topics does Chapter 13 of Class 10 Maths consist of?

Chapter 13 “Surface Areas And Volumes” of Class 10 Maths consists of the following listed topics:

  • Introduction

  • Surface Area of a Combination of Solids

  • The volume of a Combination of Solids

  • Conversion of Solid from One Shape to Another

  • Frustum of a Cone

  • Summary

The points mentioned above are very important from an examination point of view. To score good marks in Chapter 13 of Class 10 Maths, students must prepare these topics thoroughly through the NCERT book.

6. What points should I obey to achieve great marks in Chapter 13 Exercise 13.4 of Class 10 Maths?

Exercise 13.4 of Chapter 13 of Class 10 Maths is based on the topic “Frustum of a Cone” which is a little bit difficult. The given steps will help you in preparing this exercise:

  • The first step is to understand the theory related to the topic.

  • Learn all the formulas used in this topic.

  • Solve all the examples.

  • After solving examples, try to solve the questions given in the exercise.

  • Ask the question from your teacher whose answer you are not able to find.

  • You can get the solutions of this exercise on the page NCERT Solutions for Class 10 Maths Chapter 13.

7. Explain why the NCERT books help in studying Chapter 13 Exercise 13.4 of Class 10 Maths?

Underneath are justifications why students should use the NCERT books to study Chapter 13 Exercise 13.4 of Class 10 Maths:

  • The CBSE has prescribed the Maths NCERT book for students of Class 10.

  • Through this book, the students will get thorough knowledge about Chapter 13 Exercise 13.4 of Class 10 Maths.

  • There are many questions available in this book.

  • In this book, the content is created in a very simple dialect so that students can comprehend.

  • You will discover that most of the problems asked in the exam are directly taken from the NCERT book.

8. Can I build a successful study plan for Exercise 13.4 Chapter 13 of Class 10 Maths?

Yes, it is easy to invent a triumphant study plan for Chapter 13 Exercise 13.4 of Class 10 Maths. The steps discussed below will help students for the same:

  • Have a timetable with balanced activities.

  • Analyze Maths subject for at least two hours.

  • Sample papers and mock tests will help you to understand the exercise in a better way.

  • Work out on the questions given in the guidebooks.

  • Keep your mind fresh while solving questions and do not panic if your answer is not correct.

  • Learn all the formulas.

9. Is Chapter 13 Exercise 13.4 of Class 10 Maths easy or difficult?

The reply to this query depends upon the grasping power of the students. If a student is excellent in academics, then for him Chapter 13 Exercise 13.4 of Class 10 Maths. But for average students, it becomes difficult to solve questions. Therefore, to make Exercise 13.4 Chapter 13 of Class 10 Maths easy, toppers must practice extra questions to have a strong grip on the concepts. Others must learn the formulas first and should understand the theory. After that, they should centralize on the NCERT questions and examples. By solving these, they can gain confidence for solving different types of questions. These resources are available on the Vedantu website and on the Vedantu app free of cost.

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