NCERT Solutions for Exercise 13.3 Class 9 Maths Chapter 13 - Surface Area and Volumes

# NCERT Solutions for Exercise 13.3 Class 9 Maths Chapter 13 - Surface Area and Volumes

Edited By Vishal kumar | Updated on Oct 13, 2023 11:26 AM IST

## NCERT Solutions for Class 9 Maths Exercise 13.3 Chapter 13 Surface Areas and Volumes- Download Free PDF

NCERT Solutions for Class 9 Maths exercise 13.3 deals with the concept of the right circular cone and it’s surface areas . A three-dimensional shape which narrows smoothly from a flat base to a point is known as cone. Mathematically, there are two types of cones namely right circular cone and oblique cone. In exercise 13.3 Class 9 Maths of NCERT Solution, a type of cone whose axis falls perpendicular on the plane of the base is known as the right circular cone. The distance from the vertex or apex to the point on the outer line of the circular base of the cone is known as slant height which is derived from the Pythagoras Theorem. The formula for calculating the slant height of right circular cone is l2=r2+h2, from the formula l can be calculated. The surface area of a right circular cone is the area covered by the surface of the right circular cone. Surface area can be divided into two categories. They are

• Area of Lateral Surface

• Area of Total Surface

The curved surface area of the right circular cone, also known as the lateral surface area of the right circular cone, is the area covered by the curved surface of the cone. The total surface area of the right circular cone is the area occupied by the complete cone . NCERT solutions for Class 9 Maths chapter 13 exercise 13.3 include eight questions, seven of which are simple and the remaining one may take some time to complete . This Class 9 Maths chapter 13 exercise 13.3 thoroughly explains the concepts of surface area and volume. Along with class 9 maths chapter 13 exercise 13.3 the following exercises are also present.

** As per the CBSE Syllabus for 2023-24, please note that this chapter has been renumbered as Chapter 11.

## Access Surface Area and Volumes Class 9 Chapter 13 Exercise: 13.3

Given,

Base diameter of the cone = $d=10.5\ cm$

Slant height = $l=10\ cm$

We know, Curved surface area of a cone $= \pi r l$

$\therefore$ Required curved surface area of the cone=

$\\ = \frac{22}{7}\times \frac{10.5}{2}\times10 \\ \\ = 165\ cm^2$

Given,

Base diameter of the cone = $d=24\ m$

Slant height = $l=21\ cm$

We know, Total surface area of a cone = Curved surface area + Base area

$= \pi r l + \pi r^2 = \pi r (l + r)$

$\therefore$ Required total surface area of the cone=

$\\ = \frac{22}{7}\times\frac{24}{2}\times(21+12) \\ \\ =\frac{22}{7}\times\frac{24}{2}\times33 \\ = 1244.57 \ m^2$

Given,

The curved surface area of a cone = $\small 308\hspace{1mm}cm^2$

Slant height $= l = 14\ cm$

(i) Let the radius of cone be $r\ cm$

We know, the curved surface area of a cone= $\pi rl$

$\therefore$ $\\ \pi rl = 308 \\ \\ \Rightarrow \frac{22}{7}\times r\times14 = 308 \\ \Rightarrow r = \frac{308}{44} = 7$

Therefore, the radius of the cone is $7\ cm$

Given,

The curved surface area of a cone = $\small 308\hspace{1mm}cm^2$

Slant height $= l = 14\ cm$

The radius of the cone is $r =$ $7\ cm$

(ii) We know, Total surface area of a cone = Curved surface area + Base area

$= \pi r l + \pi r^2$

$\\ = 308+\frac{22}{7}\times 7^2 \\ = 308+154 = 462\ cm^2$

Therefore, the total surface area of the cone is $462\ cm^2$

Given,

Base radius of the conical tent = $r=24\ m$

Height of the conical tent = $h=10\ m$

$\therefore$ Slant height = $l=\sqrt{h^2+r^2}$

$\\ =\sqrt{10^2+24^2} \\ = \sqrt{676} \\ = 26\ m$

Therefore, the slant height of the conical tent is $26\ m$

Given,

Base radius of the conical tent = $r=24\ m$

Height of the conical tent = $h=10\ m$

$\therefore$ Slant height = $l=\sqrt{h^2+r^2} = 26\ m$

We know, Curved surface area of a cone $= \pi r l$

$\therefore$ Curved surface area of the tent

$\\ = \frac{22}{7}\times24\times26 \\ \\ =\frac{13728}{7}\ m^2$

Cost of $1\ m^2$ of canvas = $Rs.\ 70$

$\therefore$ Cost of $\frac{13728}{7}\ m^2$ of canvas =

$Rs.\ (\frac{13728}{7}\times70) = Rs.\ 137280$

Therefore, required cost of canvas to make tent is $Rs.\ 137280$

Given,

Base radius of the conical tent = $r =6\ m$

Height of the tent = $h =8\ m$

We know,

Curved surface area of a cone = $\pi rl = \pi r\sqrt{h^2 + r^2}$

$\therefore$ Area of tarpaulin required = Curved surface area of the tent

$\\ =3.14\times6\times \sqrt{8^2+ 6^2} \\ = 3.14\times6\times 10 \\ = 188.4\ m^2$

Now, let the length of the tarpaulin sheet be $x\ m$

Since $20\ cm$ is wasted, effective length = $x - 20 cm = (x - 0.2)\ m$

Breadth of tarpaulin = $3\ m$

$\\ \therefore [(x - 0.2) \times 3] = 188.4 \\ \Rightarrow x - 0.2 = 62.8 \\ \Rightarrow x = 63\ m$

Therefore, the length of the required tarpaulin sheet will be 63 m.

Given, a conical tomb

The base diameter of the cone = $d =14\ m$

Slant height $= l = 25\ m$

We know, Curved surface area of a cone $= \pi r l$

$\\ = \frac{22}{7}\times\frac{14}{2}\times25 \\ \\ = 22\times25 \\ = 550\ m^2$

Now, Cost of whitewashing per $\small 100\hspace{1mm}m^2$ = $\small Rs.\ 210$

$\therefore$ Cost of whitewashing per $\small 550\hspace{1mm}m^2$ = $\small \\ Rs. (\frac{210}{100}\times550 )$

$\small \\ = Rs.\ (21\times55 ) = Rs.\ 1155$

Therefore, the cost of white-washing its curved surface of the tomb is $\small Rs.\ 1155$ .

Given, a right circular cone cap (which means no base)

Base radius of the cone = $r=7\ cm$

Height $= h = 24\ cm$

$\therefore l = \sqrt{h^2+r^2}$

We know, Curved surface area of a right circular cone $= \pi r l$

$\therefore$ The curved surface area of a cap =

$\\ = \frac{22}{7}\times7\times\sqrt{24^2+7^2} \\ \\ = 22\times\sqrt{625} \\ = 22\times25\ \\ = 550\ cm^2$

$\therefore$ The curved surface area of 10 caps = $550\times10 = 5500\ cm^2$

Therefore, the area of the sheet required for 10 caps = $5500\ cm^2$

Given, hollow cone.

The base diameter of the cone = $d = 40\ cm = 0.4\ m$

Height of the cone = $h = 1\ m$

$\therefore$ Slant height = $l = \sqrt{h^2+r^2}$ $= \sqrt{1^2+0.2^2}$

We know, Curved surface area of a cone = $\pi r l = \pi r\sqrt{h^2+r^2}$

$\therefore$ The curved surface area of 1 cone = $3.14\times0.2\times\sqrt{1.04} = 3.14\times0.2\times1.02$

$= 0.64056\ m^2$

$\therefore$ The curved surface area of 50 cones $= (50\times0.64056)\ m^2$

$= 32.028\ m^2$

Now, the cost of painting $\small 1\ m^2$ area = $\small Rs.\ 12$

$\therefore$ Cost of the painting $32.028\ m^2$ area $= Rs.\ (32.028\times12)$

$= Rs.\ 384.336$

Therefore, the cost of painting 50 such hollow cones is $\dpi{100} Rs.\ 384.34\ (approx)$

## More About NCERT Solutions for Class 9 Maths Exercise 13.3: Surface Area and Volumes – right circular cone

The NCERT solutions for Class 9 Maths exercise 13.3 is mainly focused on the surface area of the right circular cone. In exercise 13.3 Class 9 Maths, The curved surface area of a cone can be calculated by multiplying the area of the sector with radius length . The area of the lateral surface plus the area of the circular base equals the total surface area of a closed right circular cone. V=πr2×h/3 is the volume of a right circular cone, which is one-third of the product of the circular base's area and its height. Surface areas are measured in square units, but the volume of a cube is measured in cubic units. In NCERT solutions for Class 9 Maths exercise 13.3, the formulas for computing surface areas and volume for the correct circular cone are thoroughly explored.

Also Read| Surface Areas And Volumes Class 9 Notes

## Benefits of NCERT Solutions for Class 9 Maths Exercise 13.3 :

• NCERT solutions for Class 9 Maths exercise 13.3 will help us to easily identify the object which resembles the cone shape example: an ice cream cone.

•NCERT book Exercise 13.3 Class 9 Maths, clearly explained the formula for calculating the curved surface area of the right circular cone step by step for our better understanding.

• By completing the NCERT syllabus Class 9 Maths chapter 13 exercise 13.3 exercises, we can build a solid foundation of mathematical knowledge as well as gain confidence in approaching new topics in our higher classes.

## Key Features of 9th Class Maths Exercise 13.3 Answers

1. Step-by-Step Solutions: The NCERT Solutions for class 9 maths chapter 13 exercise 13.3 provide detailed, step-by-step explanations for each problem, simplifying complex calculations and enhancing students' understanding.

2. Practical Application: This exercise 13.3 class 9 maths challenges students with problems that require the calculation of surface areas and volumes of intricate 3D shapes, allowing them to apply their knowledge to real-world scenarios.

3. Clear and Understandable Language: Class 9 maths ex 13.3 solutions are written in clear and understandable language to ensure that students can grasp the concepts with ease.

4. PDF Format: Students can access the ex 13.3 class 9 solutions in PDF format for free download, enabling offline use and flexible learning.

Also See:

## NCERT Solutions of Class 10 Subject Wise

1. According to NCERT solutions for Class 9 Maths chapter 13 exercise 13.3 , define the right circular cone .

According to NCERT solutions for Class 9 Maths chapter 13 exercise 13.3 , A type of cone whose axis falls perpendicular on the plane of the base is known as the right circular cone.

2. The point formed at the end of the cone is known as _______

The point formed at the end of the cone is known as the apex .

3. How many apexes does the cone have?

A cone has only one apex .

4. The total surface area of the cone is _________

The total surface area of the cone is πr(l + r)

5. The number of surfaces in the right cone is ______

In the right cone, there are two surfaces. There are two types of surfaces:  base and slanted surfaces .

6. How is the total surface area of the right circular cone determined from the lateral surface area and the area of the circle?

The total surface area of a closed right circular cone is computed by adding the area of the lateral surface and the area of the circular base.

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