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

Mathematics can be perceived as a technical subject equipped with formulas and concepts. A majority of other subjects are either derived from Mathematics or have the use of mathematical formulas and equations. Mathematics acts as a foundation for a lot of fields. Having a substantial knowledge of mathematical concepts can open up avenues into many career fields like Banking, Business Analyst, Operations Research Analyst, Financial Manager, Numerical Analyst, Economist, etc. Mathematics is one of the core subjects that is taught from the beginning of a student’s academic life. Having a good grasp of Mathematics is crucial for the overall development of a student’s academic skills. Mathematical concepts taught in Class 10 hold a lot of importance in the academic progression of a student’s educational career. The mathematical concepts taught in Class 10 come in handy in the preparation for a number of competitive examinations. It is imperative that students work on developing their mathematical skills to score well not just in the CBSE Board examination of Class 10 but also in any competitive examinations that they may appear for in the future. Most competitive examinations assess students’ logical reasoning and quantitative aptitude, the basis of which lies in Mathematics. To gain a better understanding of mathematical concepts, students can use the NCERT Solutions for Class 10 Maths, Chapter 13, Exercise 13.4, which are available on the Extramarks website.

The CBSE Board Examination of Class 10 is one of the most significant examinations that a student appears for in their academic career. How students perform in the CBSE Board Examination of Class 10 helps determine the trajectory of their academic progression. It is crucial that students secure high marks in the CBSE Board Examination of Class 10. Students may use the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4 to perform well in the CBSE board examination. Class 10 Maths Ex 13.4 is one of the most important exercises enclosed in Chapter 13 of the NCERT Mathematics syllabus. Students are required to practise Class 10 Maths Exercise 13.4 Solution thoroughly to have a better understanding of the concepts introduced in the designated exercise. It is essential that students practice the NCERT exercise problem solutions regularly to secure high grades in the CBSE board examination as the Class 10 board examination, pattern is based on the NCERT curriculum. Students are suggested to refer to the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4 for their preparation for the Class 10 board examination of Mathematics. The NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4 can be obtainedby students from the Extramarks website or learning application. The NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4 available on the Extramarks website are elaborated in a manner that is easy to grasp and understand for students.

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

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

The NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4 have been thoroughly elaborated for students to be able to comprehend themes related to the designated chapter easily and efficiently. The NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.4 have been meticulously organised into smaller steps to allow for a thorough understanding of the concepts covered in the exercise in questionThe NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4 have been structured in a manner that is easy and convenient for students to study without any other assistance. Students can better understand the topics and concepts introduced in the NCERT syllabus with the help of learning resources provided by Extramarks. It is important that students of Class 10 devote more time to practising and understanding the topics that have more weightage in the Class 10 board examination of Mathematics. Students should be familiar with the latest examination pattern in order to obtain high marks in the CBSE board examination. Extramarks provides information about the marks distribution, question paper pattern, chapter weightage, examination structure, and marking scheme for the CBSE board examination. Students must take all of these things into consideration while preparing a timetable for their preparation process. Students may access the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4 from the Extramarks website for their preparation for the Class 10 board examination of Mathematics. The NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4 have been prepared in a systematic step-by-step manner for better understanding of students of Class 10. Students are advised to go through the NCERT Mathematics syllabus thoroughly and practise the exercise problems and solutions from the NCERT textbook regularly. Students can refer to the Extramarks website for the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4. Students may practise sample papers along with the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4 to better understand the examination question paper pattern. Students can also access past years’ papers from the Extramarks website. It is advantageous for students to go through past years’ papers for a better understanding of the question paper pattern, this can be beneficial for students in terms of getting familiar with recurring types of questions.

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

The range of study materials offered by Extramarks is extensive. Students can benefit from enhanced learning experiences with the help of Extramarks’ online live doubt-clearing sessions. On the Extramarks website, students may access past years’ papers for all CBSE examinations. Students frequently find it challenging to retain the knowledge they have acquired for an extended period of time. Extramarks offers revision notes that students may use in order to overcome this challenge and  ensure better retention of the knowledge they have acquired. Students may find the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4 on the Extramarks website for solutions that are more thoroughly explained and elaborated. Students can boost their speed of solving problems by practising regularly. By practising sample papers and solved examples, students may gain insight into the topics they are confident in and the topics they need to study more thoroughly. Students can ensure that they score high marks in the Class 10 board examination of Mathematics with the help of the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4. Students can acquire the necessary material to prepare for the Class 10 board examination of Mathematics from the Extramarks website and mobile application.

Q.1 A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

Ans

$\begin{array}{l}\text{Radius of upper base of glass}={\mathrm{r}}_{1}=\frac{4}{2}=2\text{\hspace{0.17em}\hspace{0.17em}cm}\\ \text{Radius of lower base of glass}={\mathrm{r}}_{2}=\frac{2}{2}=1\text{\hspace{0.17em}\hspace{0.17em}cm}\\ \mathrm{Capacity}\text{of glass}=\mathrm{Volume}\text{of frustum of cone}\\ \text{}=\frac{1}{3}\mathrm{\pi h}\left({{\mathrm{r}}_{1}}^{2}+{{\mathrm{r}}_{2}}^{2}+{\mathrm{r}}_{1}{\mathrm{r}}_{2}\right)\\ \text{}=\frac{14}{3}\mathrm{\pi }\left({2}^{2}+{1}^{2}+2×1\right)=102\frac{2}{3}{\text{cm}}^{3}\end{array}$

Q.2 The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

Ans

$\begin{array}{l}\text{S}=\mathrm{l}=\text{}\\ \text{P i.e.,}\\ \text{2}\mathrm{\pi }{\text{r}}_{1}=18\text{cm and 2}\mathrm{\pi }{\text{r}}_{2}=6\text{cm}\\ {\text{i.e., r}}_{1}=\frac{9}{\mathrm{\pi }}{\text{cm and r}}_{2}=\frac{3}{\mathrm{\pi }}\text{cm}\\ \text{C}=\mathrm{\pi }\text{ }\left({\mathrm{r}}_{1}+{\mathrm{r}}_{2}\right)\mathrm{l}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{\pi }\text{ }\left(\frac{9}{\mathrm{\pi }}+\frac{3}{\mathrm{\pi }}\right)×\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=48{\text{cm}}^{2}\end{array}$

Q.3 A fez, the cap used by the Turks, is shaped like the frustum of a cone (see the following figure). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Ans

$\begin{array}{l}\text{R}\mathbf{adius}\text{}\mathbf{on}\text{}\mathbf{the}\text{}\mathbf{open}\text{}\mathbf{side}={\mathrm{r}}_{1}=\mathbf{10}\text{}\mathbf{cm}\\ \text{R}\mathbf{adius}\text{}\mathbf{at}\text{}\mathbf{the}\text{}\mathbf{upper}\text{}\mathbf{base}={\mathrm{r}}_{2}=\mathbf{4}\text{}\mathbf{cm}\\ \text{S}\mathbf{lant}\text{}\mathbf{height}\text{}\mathbf{of}\text{}\mathbf{frustum}=\mathrm{l}=15\text{}\mathbf{cm}\\ \text{A}\mathbf{rea}\text{}\mathbf{of}\text{}\mathbf{material}\text{}\mathbf{used}\text{}\mathbf{for}\text{}\mathbf{making}\text{fez}\\ \text{}=\text{CSA}\mathbf{of}\text{}\mathbf{frustum}+\text{area of the}\mathbf{upper}\text{}\mathbf{base}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{\pi }\left({\mathrm{r}}_{1}+{\mathrm{r}}_{2}\right)\mathrm{l}+{{\mathrm{\pi r}}_{2}}^{2}=\mathrm{\pi }\left(10+4\right)×15+\mathrm{\pi }{4}^{2}=710\frac{2}{7}{\text{cm}}^{2}\end{array}$

Q.4

$\begin{array}{l}\text{A container, opened from the top and made up of a metal sheet,}\\ \text{is in the form of a frustum of a cone of height 16 cm with radii of its}\\ \text{lower and upper ends as 8 cm and 20 cm, respectively.}\\ \text{Find the cost of the milk which can completely fill the container, at}\\ \text{the rate of ₹ 20 per litre. Also find the cost of metal sheet used to}\\ {\text{make the container, if it costs ₹\hspace{0.17em}8 per 100 cm}}^{\text{2}}\text{. (Take}\mathrm{\pi }\text{= 3.14)}\end{array}$

Ans

$\begin{array}{l}\text{Height of frustum = h = 16 cm}\\ {\text{Radius of lower end \hspace{0.17em}= r}}_{\text{1}}\text{ = 8 cm}\\ {\text{Radius of upper end = r}}_{\text{2}}\text{ = 20 cm}\\ \text{Slant height of frustum = l =}\sqrt{{\left({\text{r}}_{\text{1}}{\text{-r}}_{\text{2}}\right)}^{\text{2}}{\text{+h}}^{\text{2}}}\\ \text{ = }\sqrt{{\left(\text{8-20}\right)}^{\text{2}}{\text{+16}}^{\text{2}}}\text{\hspace{0.17em} = 20 cm}\\ \text{Capacity of container = Volume of frustum}\\ \text{ = }\frac{\text{1}}{\text{3}}{{\text{πh (r}}^{\text{2}}}_{\text{1}}{{\text{+r}}_{\text{2}}}^{\text{2}}{\text{+r}}_{\text{1}}{\text{r}}_{\text{2}}\text{)}\\ \text{ = }\frac{\text{16}}{\text{3}}{\text{π (8}}^{\text{2}}{\text{+20}}^{\text{2}}\text{+8×20)}\\ \text{ = 10.45 litres}\\ \text{Cost of 10.45 litre milk = 20×10.45 = ₹ 209}\\ \text{Area of metal sheet used to make the container}\\ {\text{ = π(r}}_{\text{1}}{\text{+r}}_{\text{2}}{{\text{)l+πr}}_{\text{1}}}^{\text{2}}\\ {\text{ = π(8+20)×20+π×8}}^{\text{2}}\\ {\text{ = 624π cm}}^{\text{2}}\\ {\text{Cost of 624π cm}}^{\text{2}}\text{metal sheet = 624×3.14×}\frac{\text{8}}{\text{100}}\text{ = ₹ 156.75}\end{array}$

Q.5 A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.

Ans

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{AEO,}\\ \text{}\frac{\mathrm{EO}}{\mathrm{AO}}=\mathrm{tan}30\mathrm{°}\\ \text{or}\mathrm{EO}=\frac{1}{\sqrt{3}}\mathrm{AO}=\frac{10}{\sqrt{3}}=\frac{10\sqrt{3}}{3}\text{cm}\\ \text{In}\mathrm{\Delta }\text{ABD,}\\ \text{}\frac{\mathrm{BD}}{\mathrm{AD}}=\mathrm{tan}30\mathrm{°}\\ \text{or}\mathrm{BD}=\frac{1}{\sqrt{3}}\mathrm{AD}=\frac{20}{\sqrt{3}}=\frac{20\sqrt{3}}{3}\text{cm}\\ \text{R}\mathrm{adius}\text{of upper end of frustum}={\mathrm{r}}_{1}=\frac{10\sqrt{3}}{3}\text{}\mathrm{cm}\\ \text{R}\mathrm{adius}\text{of lower end of frustum}={\mathrm{r}}_{2}=\frac{20\sqrt{3}}{3}\text{}\mathrm{cm}\\ \text{Height of frustum}=\mathrm{h}=10\text{cm}\end{array}$

$\begin{array}{l}\text{Volume of}\mathrm{frustum}=\frac{1}{3}\mathrm{\pi h}\left({{\mathrm{r}}^{2}}_{1}+{{\mathrm{r}}_{2}}^{2}+{\mathrm{r}}_{1}{\mathrm{r}}_{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{10}{3}\mathrm{\pi }\left({\left(\frac{10\sqrt{3}}{3}\right)}^{2}+{\left(\frac{20\sqrt{3}}{3}\right)}^{2}+\frac{10\sqrt{3}}{3}×\frac{20\sqrt{3}}{3}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{22000}{9}{\text{cm}}^{3}\\ \text{R}\mathrm{adius}\text{of wire}=\mathrm{r}=\frac{1}{16}×\frac{1}{2}=\frac{1}{32}\text{cm}\\ \text{Let length of wire be}\mathrm{l}.\\ \text{Volume of wire}=\text{Area of cross section}×\text{length}=\mathrm{\pi }{\left(\frac{1}{32}\right)}^{2}\mathrm{l}\\ \text{\hspace{0.17em} Volume of}\mathrm{frustum}=\text{Volume of wire}\\ \text{or}\frac{22000}{9}=\frac{22}{7}×{\left(\frac{1}{32}\right)}^{2}\mathrm{l}\\ \text{or}\mathrm{l}=7964.44\text{m}\end{array}$

## FAQs (Frequently Asked Questions)

### 1. From where can students obtain learning material for the Class 10 CBSE Board Examination in Mathematics?

Students may visit the Extramarks website and mobile application to access the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4 for their preparation. In order to be well-prepared for their board exams, students must make sure that they regularly study with the aid of NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4. Students would benefit from revising the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4 on a regular basis. For their preparation, students can utilise the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4 as the primary asset of reference.

### 2. How can students improve their preparation with the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4?

Students can benefit immensely from the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4 when preparing for the CBSE board examination. Students must use the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4 for their preparation as the NCERT curriculum is strictly adhered to in the CBSE board examinations. Students are advised to utilise the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4, as these can help students have a thorough knowledge of mathematical concepts. Students need constant practice in order to score well in the CBSE Board Examination of Mathematics.

### 3. What strategies can students use to perform better in the CBSE board examination?

By using the following study tips, students can attain higher scores in the Class 10 board exams:-

1. Students must review the NCERT textbooks carefully and thoroughly. The NCERT textbook serves as the primary text for the CBSE board examination. Students can use the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4 from the Extramarks website to prepare for Mathematics Class 10 board examination.
2. In order to do well in the CBSE board examination, students must fully comprehend the designated topics in accordance with the prescribed syllabus. Revision of the syllabus is necessary to ensure that the fundamental concepts are understood. Students may acquire the NCERT Solutions For Class 10 Maths Chapter 13 Exercise 13.4, on the Extramarks website. Students may also download revision notes in PDF format from the Extramarks website.