# NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.7) Exercise 13.7

With regard to the NCERT academic textbooks, Extramarks seeks to provide students with a comprehensive collection of questions and solutions from the exercises. The subject matter specialists at Extramarks have carefully crafted the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7. Students can achieve excellent marks in various examinations by using NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7. By working through the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7 given on the Extramarks website and mobile application, students can put their mathematical abilities into practice and improve them. These NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7, comprise solutions to the exercises included in the NCERT textbooks in accordance with the curriculum. The questions in the Exercise 13.7 Class 9 Maths were primarily developed to help students perform well on the Class 9 CBSE examinations.

With the utmost care and while adhering to all CBSE guidelines, subject matter experts at Extramarks have provided NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7 for the questions relevant to the Class 9 Mathematics syllabus. Any student in Class 9 who is thoroughly familiar with all the ideas from Class 9 Maths Chapter 13 Exercise 13.7 and sufficiently knowledgeable about all the exercises provided in it can easily gain the highest possible score on the final test. Students may quickly grasp the types of questions that could be asked in the test from Maths Class 9 Chapter 13 Exercise 13.7 and learn the chapter’s weight in terms of overall grade by using the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7 so that they can adequately study for the final exam.

There are numerous exercises in this chapter that contain numerous questions in addition to the exercise encapsulated in NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7. As previously stated, highly qualified mentors from Extramarks have collaborated to prepare credible solutions to these exercises. Due to this, they are all guaranteed to be of the highest quality, and students can efficiently use them to study for exams. It is crucial to comprehend all the concepts in the textbooks and work through the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7 that are provided in order to attain excellence.

In NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7, the theme of the Volume of a Right Circular Cone is covered. Cones are three-dimensional shapes with a seamless transition from a flat base to a point. In Mathematics, Cones come in two varieties: Right Circular Cones and Oblique Cones. A Right Circular Cone is a type of Cone in which the Axis is parallel to the plane of the base, as demonstrated in the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7. The volume of the Cone is its holding capacity.

The slanted level of a Right Circular Cone is calculated using the equation l2 = r2 + h2. The Right Circular Cone’s dimensions, such as its radius and height, can be used to estimate its volume. 9 questions responded to in the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7 are about the volume of the Right Circular Cone.

**NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.7) Exercise 13.7 **

With the use of appropriate examples, the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7 based on Surface Areas and Volumes explain that a Right Circular Cone’s volume is one-third that of a Cylinder. The NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.7 provide comprehensive answers to nine questions. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7, are compiled in accordance with the guidelines of the CBSE.

The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7 pertaining to the NCERT textbook are accessible for free download in PDF format. The majority of the questions in CBSE exams are taken from NCERT textbooks, which CBSE recommends. The Extramarks website and Learning App offer access to NCERT solutions for the whole Mathematics Book.

Students can use the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7 based on “Surface Area and Volumes” as a helpful tool to perform well in their final examinations and other competitive examinations. The complete, detailed analysis of Class 9 Chapter 13 Maths Exercise 13.7 has been given on the Extramarks website and mobile application.

The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7 offered on Extramarks were created with the inclusion of the most recent exam and curriculum requirements. These NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7, were created by knowledgeable mentors of Extramarks using dependable research. Remarkably, Extramarks gives students access to more than 1500 exam practice problems from 67 different books.

**Access NCERT Solutions Class 9 Maths Chapter 13 – Surface Areas and Volumes**

The link between a figure’s length and breadth is explained in the text Surface Areas and Volumes. A Cuboid, for instance, is made up of six distinct Rectangles. Students must therefore calculate the areas of each Rectangle if they want to get the Cuboid’s surface area. They must add the surface areas of all the surfaces after determining their individual areas in order to obtain the cuboid’s surface area. Students can learn in detail about the exercise in the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7.

Students can similarly determine the Cylinder’s surface area. There are several examples and activities in this chapter. The right formulas must be known by the students, as well as how to use them. Thus, practising the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7 is the best method to grasp this chapter. The language used to explain the concepts in the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7 is very straightforward and precise. Below are a few points to remember for students before they dive further into the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7:

- When a cuboid has dimensions of length (l), breadth (b), and height (h), its surface area is equal to two (lb + bh + lh).
- The cuboid’s lateral surface area is equal to 2 (l + b) h.
- Cuboid’s volume is equal to l * b * h

Students can avail of these NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7 on the Extramarks website and mobile application.

**Exercise – 13.7**

All of the exercises in the Class 9 NCERT Mathematics textbook have solutions in NCERT Solutions for Class 9 Mathematics. A great example of the same are the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7. On their smartphone or desktop computer, students can easily learn using the self-explanatory video tutorial format provided for the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7. A thorough step-by-step explanation has been provided for each problem in the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7. The most recent CBSE board curriculum and accurate syllabus are taken into consideration when creating the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7. These NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7 will help students understand the most recent exam format and scheme of marks distribution. The solutions, like the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7, are accompanied by credible solutions for all the chapters from Chapter 1 Number System to Chapter 15 Probability, which are available for download by students.

With the help of these NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.7, students can practise all types of questions from each exercise in the chapters. To make studying simpler and more engaging, competent subject matter experts have created the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7 in a step-by-step format. For their academic tests and to lay a better foundation for studies at a higher level, students are advised to completely practise all of these arithmetic problems from Class 9. Students should thoroughly read these NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.7.

**NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.7**

To make the process of learning more convenient for students, Extramarks has published the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7 PDF in an easy-to-understand format. The most recent edition of the NCERT Maths textbook is used in conjunction with the most recent CBSE syllabus to create the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7. For improved comprehension and quick review, students who find it challenging to solve complicated Mathematics problems can use these NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.7 explains how to calculate the volumes and surface areas of Cuboids, Cylinders, Cones, and Spheres.

**Chapter wise NCERT Solutions for Class 9 Maths **

NCERT solutions for Class 9 Mathematics Chapter 1 – Number System:

Students will study about a variety of subjects, including Rational and Irrational Numbers, in Class 9 Mathematics Chapter 1. They will also encounter the extended form of the Number Line and learn how to express numbers on the Number Line, including Integers, Rational, and Irrational Numbers. Students will be able to display terminating and non-terminating recurring decimals as well as the square roots of 2, 3, and other Non-Rational Integers in this Chapter 1 on the Number System.

NCERT solutions for Class 9 Mathematics Chapter 2 – Polynomials:

The topic of a Polynomial in Algebra is covered in Chapter 2: Polynomials. An expression made up of two or more Algebraic terms is called a Polynomial. Frequently, they are the product of numerous words with various variable powers. Students will also learn about the Factor Theorem and Remainder Theorem when learning about Polynomial Factorisation. The themes Polynomial, Coefficient, Degree, Zero, and Terms of a Polynomial are among those used in this chapter.

NCERT Solutions for Class 9 Mathematics Chapter 3 – Coordinate Geometry:

Students will study the coordinates of a point in the xy plane, terminology, and notations related to it in Chapter 3 of Coordinate Geometry. This chapter will also cover the terms abscissa and ordinates of a point as well as how to plot and name a point in the xy plane.

In Class 9 Maths Chapter 3 on Coordinate Geometry, there are three exercises that include questions on the chapter’s subjects. Exercise 3.1 Exercise 3.2 Cartesian System covers a variety of related subjects, including the Cartesian Plane, the positive and negative x- and y-axes, quadrants, and the signs of coordinates in the first, second, third, and fourth quadrants, among others. Plotting points in four quadrants and plotting points on the positive and negative x-axis and y-axis are only a few of the subtopics covered in Exercise 3.3.

NCERT solutions for Class 9 Mathematics Chapter 4 – Linear Equations in Two Variables:

Students will learn about the Linear Equation in Two Variables as well as its auxiliary rules in Class 9 Chapter 4 Mathematics. This is a chapter that will lay the groundwork for pupils to more successfully handle problems involving complex levels of Linear Equations. Students will learn how to plot the graph of a Linear Equation with Two Variables. There are four exercises in Chapter 4 of the Class 9 Mathematics book on linear equations in two variables. Exercises 4.1, 4.2, 4.3, and 4.4 deal with the Introduction, the Solution of Linear Equations, the graphing of Two-Variable Linear Equations, and Equations of Lines parallel to the x- and y-axes, respectively.

NCERT solutions for Class 9 Mathematics Chapter 5 – Introduction to Euclid’s Geometry:

There may be a perception that Chapter 5 Introductions to Euclid’s Geometry merely discusses the characteristics of Circles, Triangles, and Quadrilaterals. There are 4 exercises in Chapter 5 of the Class 9 Mathematics book, Introductions to Euclid’s Geometry.

- Exercise 5.1: Introduction,
- Exercise 5.2: Euclid’s Definitions, Axioms, And Postulates,
- Exercise 5.3: Equivalent Forms Of Euclid’s Fifth Postulate,
- Exercise 5.4: Summary

NCERT solutions for Class 9 Mathematics Chapter 6 – Lines and Angles:

In NCERT Chapter 6 Lines and Angles, the definitions and fundamental characteristics of Lines and Angles in Geometry are introduced. This chapter will lay the groundwork for several higher-level comprehension ideas in Geometry. Students will study a variety of intriguing ideas, such as Intersecting and Non-Intersecting Lines, Pairs of Angles, Parallel and Transversal Lines, Lines Parallel to the Same Line, the Angle Sum Feature of a Triangle, and much more. There are three exercises in Class 9 Mathematics Chapter 6 Lines and Angles, which encompasses four axioms and eight theorems.

NCERT solutions for Class 9 Mathematics Chapter 7 – Triangles:

Students will discover in-depth information on Triangles in Chapter 7. The topics will include Triangle Congruence, Triangle Characteristics, And Triangle Inequality. Class 9 Mathematics Chapter 7 Triangles contains 5 exercises. Students will learn how to demonstrate the qualities they studied in past sessions in this course. This chapter deals with eight theorems. Exercise 7.1 is centered around Triangle Congruence, Exercise 7.2 focuses on Triangular Congruence Criteria, and exercises 7.3 and 7.4 provide explanations about additional requirements for a Triangle’s Congruence and some features of Triangles, respectively. The concept of Triangle Inequality is encapsulated in exercise 7.5.

NCERT solutions for Class 9 Mathematics Chapter 8 – Quadrilaterals:

Students will learn about various crucial geometrical concepts in this chapter, including the Quadrilateral’s Angle Sum Property, Different Forms of Quadrilaterals, the Mid-Point Theorem, and the Characteristics of a Parallelogram, among others. The exercise-by-exercise problems and topics associated with this chapter are fully covered in the CBSE Class 9 Mathematics NCERT solutions provided on the Extramarks learning platform. To assist students in better internalising the subject, all of the solutions are offered in a step-by-step fashion. There are two exercises in Class 9 Mathematics Chapter 8 Quadrilaterals, one of which requires proving a theorem. Extramarks also provides NCERT Solutions for Class 9 Mathematics chapters 9 to 15.

**Class 9 Maths Chapter 13 Includes:**

The chapter begins with a review of Solid and Planar Figures. The chapter illustrates how the link between a figure’s length and breadth may help in calculating the area and how adding the idea of heights can produce an object’s volume. The surface areas of a Cuboid and a Cube are discussed in the first section. This also applies to a Cuboid’s lateral surface area. The same applies to Exercise 13.1, which has numerous word problems. Surface area of a Right Circular Cylinder is the theme that comes next. The topic also takes into account the total surface area of the Right Circular Cylinder and the curved surface area. The problems based on the surface area of a Cylinder are covered in Exercise 13.2. Right Circular Cone is the next solid shape discussed, after which the methods for calculating its total surface area and curved surface area are given. Eight questions in Exercise 13.3 are based on the surface area of a cone.

- A Sphere is a three-dimensional solid object made up of all points in space that are situated at a given position known as the Sphere’s centre and at a predetermined distance from that point, termed the radius.
- A solid Sphere is divided into two equal portions, known as Hemispheres when it is precisely cut “in the middle” by a plane that runs through its centre.
- The chapter discusses the total surface area of Spheres and Hemispheres as well as their curved surface area.
- If an object is solid, its volume is determined by measuring the amount of space it takes up.

First, the volumes of the Cuboid and Cube are examined. Next, the volumes of the Cylinder and Right Circular Cone are described. Finally, students will know about the volumes of Spheres and Hemispheres. The easiest method to master this chapter is to practice the questions. The chapter’s summary is covered at the end.

**NCERT Solutions for Class 9**

With regards to the NCERT syllabus, Extramarks aims to provide students with a comprehensive collection of questions and solutions from the exercises. The subject matter specialists and skilled mentors have carefully crafted the NCERT solutions for Class 9 Mathematics. Students can achieve excellent marks in class, board, and competitive exams by using NCERT solutions. By working through the chapter-by-chapter NCERT solutions for Class 9 Mathematics given on the Extramarks learning portal, students can put their mathematical abilities into practice and improve them. These solutions comprise concepts from the exercises included in the NCERT textbooks in accordance with the curriculum.

**CBSE Study Materials for Class 9**

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**CBSE Study Materials**

Students can learn and prepare for the test more effectively with the help of the academic reference materials offered by Extramarks that are made in an exciting, engaging, and student-friendly manner. Subject matter specialists have created the study materials with the most recent CBSE curriculum in mind. These online learning tools, which include the syllabus, books, sample exams, test questions, NCERT solutions, NCERT exemplar solutions, significant questions, and CBSE notes, are beneficial for all students. Students can easily study for their exams by referring to these learning resources.

**Q.1** Find the volume of the right circular cone with

(i) radius 6 cm, height 7 cm

(ii) radius 3.5 cm, height 12 cm

**Ans**

$\begin{array}{l}\text{(i)}\text{Volume}\text{o}\text{f cone}=\frac{1}{3}\pi {r}^{2}h\\ =\frac{1}{3}\times \frac{22}{7}\times 6\times 6\times 7\\ =264\text{\hspace{0.17em}}c{m}^{3}\\ \text{Thus},{\text{the volume of right circular cone is 264 cm}}^{\text{3}}\text{.}\\ \text{(ii)}\text{Volume of cone}=\frac{1}{3}\pi {r}^{2}h\\ =\frac{1}{3}\times \frac{22}{7}\times 3.5\times 3.5\times 12\\ =154\text{\hspace{0.17em}}c{m}^{3}\\ \text{Thus},{\text{the volume of right circular cone is 154 cm}}^{\text{3}}\text{.}\end{array}$

**Q.2** Find the capacity in litres of a conical vessel with

(i) radius 7 cm, slant height 25 cm

(ii) height 12 cm, slant height 13 cm

**Ans**

$\begin{array}{l}\left(\text{i}\right)\text{Radius of cone}=\text{7 cm},\text{Slant height of cone}=\text{25 cm}\\ \text{Height of cone}\left(h\right)=\sqrt{{l}^{2}-{r}^{2}}\\ =\sqrt{{25}^{2}-{7}^{2}}\\ =\sqrt{625-49}\\ =24\text{\hspace{0.17em}}cm\\ \text{Capacity of vessel}=\frac{1}{3}\pi {r}^{2}h\\ =\frac{1}{3}\times \frac{22}{7}\times 7\times 7\times 24\\ =22\times 7\times 8\\ =1232\text{\hspace{0.17em}}c{m}^{3}\\ =\frac{1232}{1000}\text{\hspace{0.17em}}litres[\because 1c{m}^{3}=\frac{1}{1000}\text{\hspace{0.17em}}litres]\\ =1.232\text{\hspace{0.17em}}litres\\ \left(\text{ii}\right)\text{Height of cone}=\text{12 cm},\text{Slant height of cone}=\text{13 cm}\\ \text{Radius of cone}\left(r\right)=\sqrt{{l}^{2}-{h}^{2}}\\ =\sqrt{{13}^{2}-{12}^{2}}\\ =\sqrt{169-144}\\ =5\text{\hspace{0.17em}}cm\\ \text{Capacity of vessel}=\frac{1}{3}\pi {r}^{2}h\\ =\frac{1}{3}\times \frac{22}{7}\times 5\times 5\times 12\\ =\frac{22}{7}\times 25\times 4\\ =\frac{2200}{7}\text{\hspace{0.17em}}c{m}^{3}\\ =\frac{2200}{7\times 1000}\text{\hspace{0.17em}}litres[\because 1c{m}^{3}=\frac{1}{1000}\text{\hspace{0.17em}}litres]\\ =\frac{11}{35}\text{\hspace{0.17em}}\text{\hspace{0.17em}}litres\end{array}$

**Q.3** The height of a cone is 15 cm. If its volume is 1570 cm^{3}, find the radius of the base. (Use π = 3.14)

**Ans**

$\begin{array}{l}\text{\hspace{0.17em} Height of the cone}=15\text{\hspace{0.17em}}cm\\ \text{Volume of the cone}=1570\text{\hspace{0.17em}}c{m}^{3}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{3}\pi {r}^{2}h=1570\text{\hspace{0.17em}}c{m}^{3}\\ \Rightarrow \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{3}\times 3.14\times {r}^{2}\times 15=1570\text{\hspace{0.17em}}c{m}^{3}\\ \Rightarrow \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{r}^{2}=\frac{1570}{3.14\times 5}\\ \Rightarrow \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}r=10\\ \text{Thus},\text{the radius of the base is 10 cm}\text{.}\end{array}$

**Q.4 **If the volume of a right circular cone of height 9 cm is 48 π cm^{3}, find the diameter of its base.

**Ans**

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}Heigh}t\text{of the cone}=9\text{\hspace{0.17em}}cm\\ \text{Volume of the cone}=48\pi \text{\hspace{0.17em}}c{m}^{3}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{3}\pi {r}^{2}h=48\pi \text{\hspace{0.17em}}c{m}^{3}\\ \Rightarrow \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{3}\times {r}^{2}\times 9=48\text{\hspace{0.17em}}c{m}^{3}\\ \Rightarrow \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{r}^{2}=\frac{48}{3}\\ \Rightarrow \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=16\\ \Rightarrow \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}r=4\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}diameter}=2\times 4\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=8\text{\hspace{0.17em}}cm\\ \text{Thus},\text{the diameter of the base is 8 cm}\text{.}\end{array}$

**Q.5 **A conical pit of top diameter 3.5 m is 12 m deep of height. What is its capacity in kilolitres?

**Ans**

$\begin{array}{l}\text{Diameter of conical pit}=3.5\text{}m\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}Radius of conical pit}=\frac{3.5}{2}\text{\hspace{0.17em}}m\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}Depth of conical pit}=12\text{}m\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}Volume of conical pit}=\frac{1}{3}\pi {r}^{2}h\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{3}\times \frac{22}{7}\times \frac{3.5}{2}\times \frac{3.5}{2}\times 12\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=11\times 3.5=38.5\text{\hspace{0.17em}}{m}^{3}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=38.5\text{\hspace{0.17em}kilolitres\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}[\because 1\text{\hspace{0.17em}}{m}^{3}=1\text{\hspace{0.17em}kilolitre}]\\ \text{Thus, the capacity of conical pit is}38.5\text{\hspace{0.17em}kilolitres}.\end{array}$

**Q.6 **The volume of a right circular cone is 9856 cm^{3}. If the diameter of the base is 28 cm, find:

(i) height of the cone

(ii) slant height of the cone

(iii) curved surface area of the cone.

**Ans**

$\begin{array}{c}\text{Diameter of a right circular cone}=28\text{\hspace{0.17em}}cm\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}Radius of a right circular cone}=14\text{\hspace{0.17em}}cm\\ \text{(i)}\text{Volume}\text{of a right circular cone}=9856\text{\hspace{0.17em}}c{m}^{3}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{3}\times \frac{22}{7}\times 14\times 14\times h=9856\text{\hspace{0.17em}}c{m}^{3}\\ h=\frac{9856\times 3}{22\times 2\times 14}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=48\text{\hspace{0.17em}}cm\\ \left(ii\right)\text{Slant}\text{\hspace{0.17em}}\text{height of the cone}\left(l\right)=\sqrt{{r}^{2}+{h}^{2}}\\ =\sqrt{{14}^{2}+{48}^{2}}\\ =\sqrt{2500}\\ =50\text{\hspace{0.17em}}cm\\ \left(iii\right)\text{Curved surface area of cone}=\pi rl\\ =\frac{22}{7}\times 14\times 50\\ =2200{\text{cm}}^{\text{2}}\end{array}$

Thus, the curved surface area of cone is 2200 cm^{2}.

**Q.7 **A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained..

**Ans**

$\begin{array}{l}\mathrm{In}\text{\hspace{0.17em}}\mathrm{\Delta ABC},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{let}\text{}\mathrm{BC}=5\mathrm{cm},\text{}\mathrm{AB}=12\text{}\mathrm{cm}\text{}\mathrm{and}\text{}\mathrm{AC}=13\text{}\mathrm{c}\text{m}\\ \text{When}\mathrm{\Delta ABC}\text{is revolved around AB, a cone is obtained,}\\ \text{whose radius is 5 cm, height is 12 cm and slant height is 13 cm.}\\ \text{So, Volume of cone}=\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{3}\mathrm{\pi}{\left(5\right)}^{2}\times 12\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=100\mathrm{\pi}\text{\hspace{0.17em}}{\mathrm{cm}}^{2}\\ \text{Thus},\text{the volume of cone is 100}\mathrm{\pi}{\text{cm}}^{\text{2}}\text{.}\end{array}$

**Q.8 **If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8**.**

**Ans**

$\begin{array}{l}\text{When}\Delta \text{ABC is revolved around BC, then radius of cone is}\\ \text{12 cm, height is 5 cm and slant height is 13 cm}\text{.}\\ \therefore \text{Volume of cone}=\frac{1}{3}\pi {r}^{2}h\\ =\frac{1}{3}\pi {\left(12\right)}^{2}\times 5\\ =\frac{1}{3}\pi \times 12\times 12\times 5\\ =240\text{\hspace{0.17em}}\pi \text{\hspace{0.17em}}c{m}^{3}\\ \text{Ratio of volumes of the two solids obtained on revolving triangle}\\ =\frac{100\pi}{240\pi}\\ =5:12\end{array}$

**Q.9 **A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

**Ans**

$\begin{array}{l}\text{Diameter of conical heap}=10.5\text{\hspace{0.17em}}m\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}Radius of conical heap}=\frac{10.5}{2}\text{\hspace{0.17em}}m\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}Height of conical heap}=3\text{\hspace{0.17em}}m\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}Volume of conical heap}=\frac{1}{3}\pi {r}^{2}h\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{3}\times \frac{22}{7}\times \frac{10.5}{2}\times \frac{10.5}{2}\times 3\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{22}{\overline{)7}}\times \frac{\overline{)105}15}{20}\times \frac{105}{20}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{11}{10}\times 15\times \frac{21}{4}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=86.625\text{\hspace{0.17em}}c{m}^{3}\\ \text{Slant height of conical heap}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{{3}^{2}+{\left(\frac{10.5}{2}\right)}^{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{9+\frac{110.25}{4}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{\frac{36+110.25}{4}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{\frac{146.25}{4}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{12.1}{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=6.05\text{\hspace{0.17em}}m\\ \mathrm{Curved}\text{surface area of conical heap}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\pi rl\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{22}{7}\times \frac{10.5}{2}\times 6.05\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=99.825\text{\hspace{0.17em}}{m}^{2}\\ \text{Thus},\text{the area of canvas required is 99}{\text{.825 m}}^{\text{2}}\text{.}\end{array}$

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