RD Sharma Solutions for Class 9 Maths Chapter 20–Surface Area and Volume of A Right Circular Cone

Class 9: Maths Chapter 20 solutions. Complete Class 9 Maths Chapter 20 Notes.

RD Sharma Solutions for Class 9 Maths Chapter 20–Surface Area and Volume of A Right Circular Cone

RD Sharma 9th Maths Chapter 20, Class 9 Maths Chapter 20 solutions

Exercise 20.1 Page No: 20.7

Question 1: Find the curved surface area of a cone, if its slant height is 60 cm and the radius of its base is 21 cm.

Solution:

Slant height of cone (l) = 60 cm

Radius of the base of the cone (r) = 21 cm

Now,

Curved surface area of the right circular cone = πrl = 22/7 x 21 x 60 = 3960 cm²

Therefore the curved surface area of the right circular cone is 3960 cm²

Question 2: The radius of a cone is 5cm and vertical height is 12cm. Find the area of the curved surface.

Solution:

Radius of cone (r) = 5 cm

Height of cone (h) = 12 cm

Find Slant Height of cone (l):

We know, l² = r² + h²

l² = 5² +12²

l² = 25 + 144 = 169

Or l = 13 cm

Now,

C.S.A = πrl =3.14 x 5 x 13 = 204.28

Therefore, the curved surface area of the cone is 204.28 cm²

Question 3 : The radius of a cone is 7 cm and area of curved surface is 176 cm² .Find the slant height.

Solution:

Radius of cone(r) = 7 cm

Curved surface area(C.S.A)= 176cm²

We know, C.S.A. = πrl

⇒ πrl = 176

⇒ 22/7 x 7 x l = 176

or l = 8

Therefore, slant height of the cone is 8 cm.

Question 4: The height of a cone 21 cm. Find the area of the base if the slant height is 28 cm.

Solution:

Height of cone(h) = 21 cm

Slant height of cone (l) = 28 cm

We know that, l² = r² + h²

28²=r²+21²

r²=28²−21²

or r= 7√7 cm

Now,

Area of the circular base = πr²

= 22/7 x (7√7 )²

=1078

Therefore, area of the base is 1078 cm².

Question 5: Find the total surface area of a right circular cone with radius 6 cm and height 8 cm.

Solution:

Radius of cone (r) = 6 cm

Height of cone (h) = 8 cm

Total Surface area of the cone (T.S.A)=?

Find slant height of cone:

We know, l² = r² + h²

=6²+8²

= 36 + 64

= 100

or l = 10 cm

Now,

Total Surface area of the cone (T.S.A) = Curved surface area of cone + Area of circular base

= πrl + πr²

= (22/7 x 6 x 10) + (22/7 x 6 x 6)

= 1320/7 + 792/7

= 301.71

Therefore, area of the base is 301.71cm².

Question 6: Find the curved surface area of a cone with base radius 5.25 cm and slant height 10 cm.

Solution:

Base radius of the cone(r) = 5.25 cm

Slant height of the cone(l) = 10 cm

Curved surface area (C.S.A) = πrl

=22/7 x 5.25 x 10

= 165

Therefore, curved surface area of the cone is 165cm².

Question 7: Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Solution:

Diameter of the cone(d)=24 m

So, radius of the cone(r)= diameter/ 2 = 24/2 m = 12m

Slant height of the cone(l) = 21 m

T.S.A = Curved surface area of cone + Area of circular base

= πrl+ πr²

= (22/7 x 12 x 21) + (22/7 x 12 x 12)

= 1244.57

Therefore, total surface area of the cone is 1244.57 m².

Question 8: The area of the curved surface of a cone is 60 π cm². If the slant height of the cone be 8 cm, find the radius of the base.

Solution:

Curved surface area(C.S.A)= 60 π cm²

Slant height of the cone(l) = 8 cm

We know, Curved surface area(C.S.A )=πrl

⇒ πrl = 60 π

⇒ r x 8 = 60

or r = 60/8 = 7.5

Therefore, radius of the base of the cone is 7.5 cm.

Question 9: The curved surface area of a cone is 4070 cm² and diameter is 70 cm .What is its slant height? (Use π =22/7)

Solution:

Diameter of the cone(d) = 70 cm

So, radius of the cone(r)= diameter/2 = 70/2 cm = 35 cm

Curved surface area = 4070 cm²

Now,

We know, Curved surface area = πrl

So, πrl = 4070

By substituting the values, we get

22/7 x 35 x l = 4070

or l = 37

Therefore, slant height of cone is 37 cm.

Question 10: The radius and slant height of a cone are in the ratio 4:7. If its curved surface area is 792 cm², find its radius. (Use π =22/7)

Solution:

Curved surface area = 792 cm²

The radius and slant height of a cone are in the ratio 4:7 (Given)

Let 4x be the radius and 7x be the height of cone.

Now,

Curved surface area (C.S.A.) = πrl

So, 22/7 x (4x) x (7x) = 792

or x² = 9

or x = 3

Therefore, Radius = 4x = 4(3) cm = 12 cm

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Exercise 20.2 Page No: 20.20

Question 1: Find the volume of the right circular cone with:

(i) Radius 6cm, height 7cm

(ii)Radius 3.5cm, height 12cm

(iii) Height is 21cm and slant height 28cm

Solution:

(i) Radius of cone(r)=6cm

Height of cone(h)=7cm

We know, Volume of a right circular cone = 1/3 πr²h

By substituting the values, we get

= 1/3 x 3.14 x 6² x 7

= 264

Volume of a right circular cone is 264 cm³

(ii) Radius of cone(r)=3.5 cm

Height of cone(h)=12cm

Volume of a right circular cone = 1/3 πr²h

By substituting the values, we get

= 1/3 x 3.14 x 3.5² x 12

=154

Volume of a right circular cone is 154 cm³

(iii) Height of cone(h)=21 cm

Slant height of cone(l) = 28 cm

Find the measure of r:

We know, l² = r² + h²

28² = r² + 21²

or r = 7√7

Now,

Volume of a right circular cone = 1/3 πr²h

By substituting the values, we get

= 1/3 x 3.14 x (7√7)² x 21

=7546

Volume of a right circular cone is 7546 cm³

Question 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.

Solution:

(i) Radius of the cone(r) =7 cm

Slant height of the cone (l) =25 cm

As we know that, l² = r² + h²

25² = 7² + h²

or h = 24

Now, Volume of a right circular cone = = 1/3 πr²h

By substituting the values, we get

= 1/3 x 3.14 x (7)² x 24

= 1232

Volume of a right circular cone is 1232 cm³ or 1.232 litres[1 cm³ = 0.01 liter]

(ii) Height of cone(h)=12 cm

Slant height of cone(l)=13 cm

As we know that, l² = r² + h²

13² = r² + 12²

or r = 5

Now, Volume of a right circular cone = 1/3 πr²h

By substituting the values, we get

= 1/3 x 3.14 x (5)² x 12

= 314.28

Volume of a right circular cone is 314.28 cm³ or 0.314 litres.[1 cm³ = 0.01 liters]

Question 3: Two cones have their heights in the ratio 1:3 and the radii of their bases in the ratio 3:1. Find the ratio of their volumes.

Solution:

Let the heights of the cones be h and 3h and radii of their bases be 3r and r respectively. Then, their volumes are

Volume of first cone (V1) = 1/3 π(3r)²h

Volume of second cone (V2) = 1/3 πr²(3h)

Now, V1/V2 = 3/1

Ratio of two volumes is 3:1.

Question 4: The radius and the height of a right circular cone are in the ratio 5:12. If its volume is 314 cubic meter, find the slant height and the radius. (Use π=3.14).

Solution:

Let us assume the ratio of radius and the height of a right circular cone to be x.

Then, radius be 5x and height be 12x

We know, l² = r² + h²

= (5x) ² + (12x)²

= 25 x² + 144 x²

or l = 13x

Therefore, slant height is 13 m.

Now it is given that volume of cone = 314 m³

⇒1/3πr²h = 314

⇒1/3 x 3.14 x (25x² ) x (12x) = 314

⇒x³=1

or x = 1

So, radius = 5x 1 = 5 m

Therefore ,

Answer: Slant height = 13m

Radius = 5m

Question 5: The radius and height of a right circular cone are in the ratio 5 : 12 and its volume is 2512 cubic cm. Find the slant height and radius of the cone. (Use π=3.14).

Solution:

Let the ratio of radius and height of a right circular cone be y.

Radius of cone(r) = 5y

Height of cone (h) =12y

Now we know, l² = r² + h²

= (5y) ² + (12y)²

= 25 y² + 144 y²

or l = 13y

Now, volume of the cone is given 2512cm³

⇒1/3πr²h=2512

⇒1/3 x 3.14 x (5y)² x 12y = 2512

⇒ y³ = (2512 x 3)/(3.14 x 25 x 12) = 8

or y = 2

Therefore,

Radius of cone = 5y = 5×2 = 10cm

Slant height (l) =13y = 13×2 = 26cm

Question 6: The ratio of volumes of two cones is 4 : 5 and the ratio of the radii of their bases is 2 : 3. Find the ratio of their vertical heights.

Solution:

Let the ratio of the radius be x and ratio of the volume be y.

Then, Radius of 1st cone (r₁) =2x

Radius of 2nd cone (r₂) =3x

Volume of 1st cone (V₁)= 4y

Volume of 2nd cone (V₂)= 5y

We know formula for volume of a cone = 1/3πr²h

Let h₁ and h₂ be the heights of respective cones.

Therefore, heights are in the ratio of 9 : 5.

Question 7: A cylinder and a cone have equal radii of their bases and equal heights. Show that their volumes are in the ratio 3:1.

Solution:

We are given, a cylinder and a cone are having equal radii of their bases and heights.

Let, radius of the cone = radius of the cylinder = r and

Height of the cone = height of the cylinder = h

Now,

Therefore, ratio of their volumes is 3:1.

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Exercise VSAQs Page No: 20.23

Question 1: The height of a cone is 15 cm. If its volume is 500π cm³, then find the radius of its base.

Solution:

Height of a cone = 15 cm

Volume of cone = 500 π cm³

We know, Volume of cone = 1/3 πr²h

So, 500π = 1/3 π r² x 15

r² = 100

or r = 10

Radius of base is 10 cm.

Question 2: If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.

Solution:

Height of a cone = 9 cm

Volume of cone = 48 π cm³

We know, Volume of cone = 1/3 πr²h

So, 48π = 1/3 π r² x 9

r² = 16

or r = 4

Radius of base r = 4 cm

Therefore, Diameter = 2 Radius = 2 x 4 cm = 8 cm.

Question 3: If the height and slant height of a cone are 21 cm and 28 cm respectively. Find its volume.

Solution:

Height of cone (h) = 21 cm

Slant height of cone (l) = 28 cm

Find radius of cone:

We know, l² = r² + h²

28² = r² + 21²

or r = 7√7 cm

Now,

We know, Volume of cone = 1/3 πr²h

= 1/3 x π x (7√7 )² x 21

= 2401 π

Therefore, Volume of cone is 2401 π cm³.

Question 4: The height of a conical vessel is 3.5 cm. If its capacity is 3.3 litres of milk. Find the diameter of its base.

Solution:

Height of a conical vessel = 3.5 cm and

Capacity of conical vessel is 3.3 litres or 3300 cm³

Now,

We know, Volume of cone = 1/3 πr²h

3300 = 1/3 x 22/7 x r² x 3.5

or r2 = 900

or r = 30

So, radius of cone is 30 cm

Hence, diameter of its base = 2 Radius = 2×30 cm = 60 cm

RD Sharma Solutions for Class 9 Maths Chapter 20: Download PDF

RD Sharma Solutions for Class 9 Maths Chapter 20–Surface Area and Volume of A Right Circular Cone

Download PDF: RD Sharma Solutions for Class 9 Maths Chapter 20–Surface Area and Volume of A Right Circular Cone PDF

Chapterwise RD Sharma Solutions for Class 9 Maths :

• Chapter 1–Number System
• Chapter 2–Exponents of Real Numbers
• Chapter 3–Rationalisation
• Chapter 4–Algebraic Identities
• Chapter 5–Factorization of Algebraic Expressions
• Chapter 6–Factorization Of Polynomials
• Chapter 7–Introduction to Euclid’s Geometry
• Chapter 8–Lines and Angles
• Chapter 9–Triangle and its Angles
• Chapter 10–Congruent Triangles
• Chapter 11–Coordinate Geometry
• Chapter 12–Heron’s Formula
• Chapter 13–Linear Equations in Two Variables
• Chapter 14–Quadrilaterals
• Chapter 15–Area of Parallelograms and Triangles
• Chapter 16–Circles
• Chapter 17–Construction
• Chapter 18–Surface Area and Volume of Cuboid and Cube
• Chapter 19–Surface Area and Volume of A Right Circular Cylinder
• Chapter 20–Surface Area and Volume of A Right Circular Cone
• Chapter 21–Surface Area And Volume Of Sphere
• Chapter 22–Tabular Representation of Statistical Data
• Chapter 23–Graphical Representation of Statistical Data
• Chapter 24–Measure of Central Tendency
• Chapter 25–Probability

About RD Sharma

RD Sharma isn't the kind of author you'd bump into at lit fests. But his bestselling books have helped many CBSE students lose their dread of maths. Sunday Times profiles the tutor turned internet star
He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like 'series solution of linear differential equations'. Meet Dr Ravi Dutt Sharma — mathematics teacher and author of 25 reference books — whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it's only recently that a spoof video turned the tutor into a YouTube star.

R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. "I like to spend all my time thinking and writing about maths problems. I find it relaxing," he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government's Guru Nanak Dev Institute of Technology.

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