NCERT Solutions for Exercise 13.7 Class 9 Maths Chapter 13 - Surface Area and Volumes

# NCERT Solutions for Exercise 13.7 Class 9 Maths Chapter 13 - Surface Area and Volumes

Edited By Vishal kumar | Updated on Oct 17, 2023 09:35 AM IST

## NCERT Solutions for Class 9 Maths Exercise 13.7 Chapter 13 Surface Areas and Volumes- Download Free PDF

The concept of volume of the right circular cone is discussed in NCERT Solutions for Class 9 Maths exercise 13.7. Cone is a three-dimensional form that narrows smoothly from a flat base to a point. There are two types of cones in mathematics: right circular cones and oblique cones. A right circular cone is a form of cone whose axis is perpendicular to the plane of the base, as shown in exercise 13.7 Class 9 Maths. The space or the capacity of the cone is known as the volume of the cone. The Pythagoras Theorem defines slant height as the distance between the vertex or apex and a point on the outer line of the circular base of the cone. The formula l2=r2+h2 is used to calculate the slant height of a right circular cone. The volume of the right circular cone can be estimated using the right circular cone's dimensions, such as its radius and height.

The volume of the right circular cone is the subject of nine questions in NCERT solutions for NCERT book Class 9 Maths chapter 13 exercise 13.7. The 9th class maths exercise 13.7 answers are thoughtfully created by subject experts from Careers360. They are presented in a straightforward and detailed manner, making complex geometry concepts easy to understand. Additionally, these class 9 maths chapter 13 exercise 13.7 solutions are available in PDF format, enabling students to access them offline for completing assignments and homework with ease. The following tasks are included along with NCERT syllabus class 9 Maths chapter 13 exercise 13.7 .

**According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 11.

## Access Surface Area and Volumes Class 9 Maths Chapter 13 Exercise: 13.7

Given,

Radius = $r =6\ cm$

Height = $h =7\ cm$

We know,

Volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ Required volume = $\frac{1}{3}\times\frac{22}{7}\times6^2\times7$

$\\ = 22\times2\times6 \\ = 264\ cm^3$

Given,

Radius = $r =3.5\ cm$

Height = $h =12\ cm$

We know,

Volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ Required volume = $\frac{1}{3}\times\frac{22}{7}\times3.5^2\times12$

$\\ = 22\times0.5\times3.5\times4 \\ = 11\times14 \\ = 154\ cm^3$

Given,

Radius = $r =7\ cm$

Slant height = $l = \sqrt{r^2 + h^2} = 25\ cm$

Height = $h =\sqrt{l^2-r^2} = \sqrt{25^2-7^2}$

$= \sqrt{(25-7)(25+7)} = \sqrt{(18)(32)}$

$= 24\ cm$

We know,
Volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ Volume of the vessel= $\frac{1}{3}\times\frac{22}{7}\times7^2\times24$

$\\ = 22\times7\times8\\ = 154\times8 \\ = 1232\ cm^3$

$\therefore$ Required capacity of the vessel =

$= \frac{1232}{1000} = 1.232\ litres$

Given,

Height = $h =12\ cm$

Slant height = $l = \sqrt{r^2 + h^2} = 13\ cm$

Radius = $r =\sqrt{l^2-h^2} = \sqrt{13^2-12^2}$

$= \sqrt{(13-12)(13+12)} = \sqrt{(1)(25)}$

$= 5\ cm$

We know,
Volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ Volume of the vessel= $\frac{1}{3}\times\frac{22}{7}\times5^2\times12$

$\\ = \frac{22}{7}\times25\times4\\ = \frac{2200}{7}\ cm^3$

$\therefore$ Required capacity of the vessel =

$= \frac{2200}{7\times1000} = \frac{11}{35}\ litres$

Given,

Height of the cone = $h =15\ cm$

Let the radius of the base of the cone be $r\ cm$

We know,
The volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ $\frac{1}{3}\times3.14\times r^2\times15 = 1570$

$\\ \Rightarrow 3.14\times r^2\times5 = 1570 \\ \Rightarrow r^2 = \frac{1570}{15.7} \\ \Rightarrow r^2 = 100 \\ \Rightarrow r = 10\ cm$

Given,

Height of the cone = $h =9\ cm$

Let the radius of the base of the cone be $r\ cm$

We know,
The volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ $\frac{1}{3}\times\pi\times r^2\times9 = 48\pi$

$\\ \Rightarrow 3r^2 = 48 \\ \Rightarrow r^2 = 16\\ \Rightarrow r = 4\ cm$

Therefore the diameter of the right circular cone is $8\ cm$

Given,

Depth of the conical pit = $h =12\ m$

The top radius of the conical pit = $r = \frac{3.5}{2}\ m$

We know,
The volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ The volume of the conical pit =

$= \frac{1}{3}\times\frac{22}{7}\times \left (\frac{3.5}{2} \right )^2\times12$

$\\ = \frac{1}{3}\times\frac{22}{7}\times \frac{3.5\times 3.5}{4}\times12 \\ \\ = 22\times 0.5\times 3.5 \\ = 38.5\ m^3$

Now, $1\ m^3 = 1\ kilolitre$

$\therefore$ The capacity of the pit = $38.5\ kilolitre$

Given, a right circular cone.

The radius of the base of the cone = $r = \frac{28}{2} = 14\ cm$

The volume of the cone = $\small 9856\hspace{1mm}cm^3$

(i) Let the height of the cone be $h\ m$

We know,
The volume of a right circular cone = $\frac{1}{3}\pi r^2 h$

$\therefore$ $\frac{1}{3}\times\frac{22}{7}\times(14)^2\times h = 9856$

$\\ \Rightarrow \frac{1}{3}\times\frac{22}{7}\times14\times14\times h = 9856 \\ \Rightarrow \frac{1}{3}\times22\times2\times14\times h = 9856 \\ \Rightarrow h = \frac{9856\times3}{22\times2\times14} \\ \\ \Rightarrow h =48\ cm$

Therefore, the height of the cone is $48\ cm$

Given, a right circular cone.

The volume of the cone = $\small 9856\hspace{1mm}cm^3$

The radius of the base of the cone = $r = \frac{28}{2} = 14\ cm$

And the height of the cone = $h = 48\ cm$

(ii) We know, Slant height, $l = \sqrt{r^2+h^2}$

$\\ \Rightarrow l = \sqrt{14^2+48^2} \\ \Rightarrow l = \sqrt{196+2304} = \sqrt{2500} \\ \Rightarrow l = 50\ cm$

Therefore, the slant height of the cone is $50\ cm$ .

Given, a right circular cone.

The radius of the base of the cone = $r = \frac{28}{2} = 14\ cm$

And Slant height of the cone = $l = 50\ cm$

(iii) We know,

The curved surface area of a cone = $\pi r l$

$\therefore$ Required curved surface area= $\frac{22}{7}\times14\times50$

$\\ = 22\times2\times50 \\ = 2200\ cm^2$

When a right-angled triangle is revolved about the perpendicular side, a cone is formed whose,

Height of the cone = Length of the axis= $h = 12\ cm$

Base radius of the cone = $r = 5\ cm$

And, Slant height of the cone = $l = 13\ cm$

We know,

The volume of a cone = $\frac{1}{3}\pi r^2 h$

The required volume of the cone formed = $\frac{1}{3}\times\pi\times5^2\times12$

$\\ = \pi\times25\times4 \\ = 100\pi\ cm^3$

Therefore, the volume of the solid cone obtained is $100\pi\ cm^3$

When a right-angled triangle is revolved about the perpendicular side, a cone is formed whose,

Height of the cone = Length of the axis= $h = 5\ cm$

Base radius of the cone = $r = 12\ cm$

And, Slant height of the cone = $l = 13\ cm$

We know,

The volume of a cone = $\frac{1}{3}\pi r^2 h$

The required volume of the cone formed = $\frac{1}{3}\times\frac{22}{7}\times12^2\times5$

$\\ = \pi\times4\times60 \\ = 240\pi\ cm^3$

Now, Ratio of the volumes of the two solids = $\\ = \frac{100\pi}{240\pi}$

$\\ = \frac{5}{12}$

Therefore, the required ratio is $5:12$

Given,

Height of the conical heap = $h = 3\ m$

Base radius of the cone = $r = \frac{10.5}{2}\ m$

We know,

The volume of a cone = $\frac{1}{3}\pi r^2 h$

The required volume of the cone formed = $\frac{1}{3}\times\frac{22}{7}\times\left (\frac{10.5}{2} \right )^2\times3$

$\\ = 22\times\frac{1.5\times10.5}{4} \\ = 86.625\ m^3$

Now,

The slant height of the cone = $l = \sqrt{r^2+h^2}$

$\\ \Rightarrow l = \sqrt{3^2+5.25^2} = \sqrt{9+27.5625} \approx 6.05$

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

The required area of the canvas to cover the heap = $\frac{22}{7}\times\frac{10.5}{2}\times6.05$

$= 99.825\ m^2$

## More About NCERT Solutions for Class 9 Maths Exercise 13.7

The volume of a cone formula is one-third the product of the area of the circular base and the height of the cone. To calculate the volume of a right circular cone, we must first look at its parameters, such as the radius of the circular base and the height of the cone. Then we must determine whether or not all of the dimensions, such as radius and height, are in the same units, or convert them to the same ones. After we've made all of the measurements' units the same, we'll need to calculate the area of the circular base. After that, we must multiply the area of the circular base and height by one-third. As a result, the volume of the right circular cone is derived, which should be expressed in cubic units.

Also Read| Surface Areas And Volumes Class 9 Notes

## Benefits of NCERT Solutions for Class 9 Maths Exercise 13.7

• The solved example before the exercise 13.7 Class 10 Maths and the NCERT solutions for Class 9 Maths exercise 13.7 are crucial since they cover questions about the volume of the right circular cone.

• If students can answer all of the questions in exercise 13.7 Class 10 Maths they will be able to grasp the hit-or-miss cone concept as well as the notion of finding the proper circular cone as described in Class 9 Maths chapter 13 exercise 13.7

• For Class 9 final exams, students may receive either short or lengthy answer questions from the types discussed inclass 9 maths ex 13.7.

## Key Features of NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volume Exercise 13.7

1. Comprehensive Coverage: Exercise 13.7 class 9 maths covers a wide range of problems related to surface areas and volumes, ensuring students have a thorough understanding of the concepts.

2. Expert-Crafted Solutions: The 9th class maths exercise 13.7 answers are expertly crafted to provide step-by-step explanations, making it easier for students to grasp complex topics.

3. Alignment with CBSE Syllabus: The class 9 maths ex 13.7 solutions adhere to the CBSE syllabus, making them an ideal resource for homework, assignments, and exam preparation.

4. PDF Format: Class 9 ex 13.7 solutions are available in PDF format, allowing students to access and download them for offline use, ensuring accessibility and convenience.

Also See:

## NCERT Solutions of Class 10 Subject Wise

1. The slant height of a right circular cone is _________

The slant height of a right circular cone is l^2=r^2+h^2.

2. The volume of the right circular cone is ________

The volume of the right circular cone is pi r^2 h/3

3. The cube's volume is calculated using _______ units.

Cubic units are used to measure the volume of the cube

4. Funnel is a _______ a)Cylinder b)Cone c)Sphere

Funnel is a cone. Option (b) cone.

5. Define slant height.

The Pythagoras Theorem defines slant height as the distance between the vertex or apex and a point on the outer line of the circular base of the cone.

6. According to NCERT solutions for Class 9 Maths chapter 13 exercise 13.7 , what are the types of cones?

Mathematically, There are two types of cones in ,

• right circular cones

•  oblique cones.

7. According to NCERT solutions for Class 9 Maths chapter 13 exercise 13.4 , whether the volume of a regular cone or right circular cone and the oblique cone is the same?

Yes, The formula for the volume of a regular cone or right circular cone and the oblique cone is the same.

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