RD Sharma Solutions Class 9 Chapter 20 comes with questions and answers related to cone and its dimensions. Let us discuss some of the important terms used in this chapter.
Total surface area: The total area occupied by the surface including the curved part and the base(s).
Curved surface area: The area occupied by the surface excluding the base(s) is known as curved surface area.
Volume: The space occupied by an object, which is measured in cubic units.
A right circular cone is a geometric figure having a circular base. Consider a right circular cone having ‘r’ as the base area, ‘h’ as the height of a cone and ‘l’ as a slant height. Then,
Total surface area = πr(r + l),
Curved surface area = πrl and
Volume = 1/3πr2h
Check the detailed RD Sharma solutions for class 9 for the chapter “Surface areas and volumes of a right circular cone” given below.
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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
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.
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 Surface Area and Volume of A Right Circular Cone
In the 20th chapter of Class 9 RD Sharma Solutions students will study important concepts listed below:
- Right Circular Cone Introduction
- Some important terms definition – Vertex, Base, Axis, Radius and Slant Height.
- Surface Area of a Right Circular Cone
- Volume of a Right Circular Cone
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