RD Sharma Class 9 Solutions Surface Area and Volume of A Right Circular Cone
Summary:
 A Right circular cone is a solid generated by revolving a line segment which passes through a fixed point and which makes a constant angle with fixed line. The fixed point is called the vertex of the cone, the fixed line is called the axis of the cone.
 For a circular cone of base radius r, slant height l and height h, we have

 Curved surface area =πrl
 Total surface area = πr(l+r)
 Volume =
RD Sharma Class 9 PDF Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1
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:
Radius of the base of the cone = 21 cm
and slant height (l) = 60 cm
Question 2.
The radius of a cone is 5 cm and vertical height is 12 cm. Find the area of the curved surface.
Solution:
Radius of the base of a cone = 5 cm
Vertical height (h) = 12 cm
Question 3.
The radius of a cone is 7 cm and area of curved surface is 176 cm2. Find the slant height.
Solution:
Curved surface area of a cone = 176 cm^{2}
and radius (r) = 1 cm^{2}
Question 4.
The height of a cone is 21 cm. Find the area of the base if the slant height is 28 cm.
Solution:
Height of the cone (h) = 21 cm
Slant height (l) = 28 cm
∴ l^{2} = r^{2} + h^{2}
⇒ r^{2} = l^{2}– h^{2} = (28)^{2} – (21 )^{2}
⇒ 784 – 441 = 343 …(i)
Now area of base = πr^{2}
= x 343 [From (i)]
= 22 x 49 = 1078 cm^{2}
Question 5.
Find the total surface area of a right circular cone with radius 6 cm and height 8 cm.
Solution:
Radius of base of cone (r) = 6 cm
and height (h) = 8 cm
Question 6.
Find the curved surface area of a cone with base radius 5.25 cm and slant height 10 cm. [NCERT]
Solution:
Radius of base of a cone (r) = 5.25 cm
and slant height (l) = 10 cm
Curved surface area = πrl
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. [NCERT]
Solution:
Slant height of a cone (l) = 21 m
and diameter of its base = 24 m
∴ Radius (r) = = 12 m
Now total surface area = πr(l + r)
Question 8.
The area of the curved surface of a cone is 60π cm^{2}. If the slant height of the cone be 8 cm, find the radius of the base.
Solution:
Curved surface area of a cone = 6071 cm^{2}
Slant height (l) = 8 cm
Question 9.
The curved surface area of a cone is 4070 cm^{2} and its diameter is 70 cm. What is the slant height ? (Use π = 22/7).
Solution:
Surface area of a cone = 4070 cm^{2}
Diameter of its base = 70 cm
Question 10.
The radius and slant height of a cone are in the ratio of 4 : 7. If its curved surface area is 792 cm^{2}, find its radius. (Use π = 22/7)
Solution:
Curved surface area of a cone = 792 cm^{2}
Ratio in radius and slant height = 4:7
Let radius = 4x
Then slant height = 7x
∴ Curved surface area πrl = 792
Question 11.
A Joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps. [NCERT]
Solution:
Radius of the base of a conical cap (r) = 7 cm
and height (h) = 24 cm
Question 12.
Find the ratio of the curved surface areas of two cones if their diameters of the bases are equal and slant heights are in the ratio 4 : 3.
Solution:
Let diameters of each cone = d
Then radius (r) =
Ratio in their slant heights = 4 : 3
Let slant height of first cone = 4x
and height of second cone = 3x
Now curved surface area of the first cone = 2πrh
Question 13.
There are two cones. The curved surface area of one is twice that of the other. The slant height of the later is twice that of the former. Find the ratio of their radii.
Solution:
In two cones, curved surface of the first cone = 2 x curved surface of the second cone
Slant height of the second cone = 2 x slant height of first cone
Let r_{1} and r_{2} be the radii of the two cones
and let height of the first cone = h
Then height of second cone = 2h
∴ Curved surface of the first cone = 2πr_{1}h
Question 14.
The diameters of two cones are equal. If their slant heights are in the ratio 5 : 4, find the ratio of their curved surfaces.
Solution:
Let diameter of one cone = d
and diameter of second cone = d
∴ Radius of the first cone (r) =
and of second cone (r_{2}) =
Ratio in their slant heights = 5:4
Let slant height of the first cone = 5x
Then that of second cone = 4x
Now curved surface of the first cone = 2πrh
Question 15.
Curved surface area of a cone is 308 cm^{2} and its slant height is 14 cm. Find the radius of the base and total surface area of the cone. [NCERT]
Solution:
Area of curved surface of a cone = 308 cm^{2}
and slant height (l) = 14 cm
Question 16.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of Rs. 210 per 100 m^{2}. [NCERT]
Solution:
Slant height of a cone (l) = 25 m
and diameter of base = 14 m
Question 17.
A conical tent is 10 m high and the radius of its base is 24 m. Find the slant height of the tent. If the cost of 1 m^{2} canvas is Rs. 70, find the cost of the canvas required to make the tent. [NCERT]
Solution:
Height of conical tent (A) = 10 m
Radius of the base (r) = 24 m
Question 18.
The circumference of the base of a 10 m height conical tent is 44 metres. Calculate the length of canvas used in making the tent if width of canvas is 2 m. (Use π = 22/7)
Solution:
Circumference of the base of a conical tent = 44 m
Question 19.
What length of tarpaulin 3 m wide will be required to make a conical tent of height 8 m and base radius 6 m? Assum that the extra length of material will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use π = 3.14) [NCERT]
Solution:
Height of the conical tent (h) = 8 m
and radius of the base (r) = 6 m
Now curved surface area of the tent = πrl = 3.14 x 6 x 10 = 188.4 m
Width of tarpaulin used = 3 m
∴ Length = 188.4 , 3 = 62.8 m
Extra length required = 20 cm = 0.2 m
∴ Total length of tarpaulin = 62.8 + 0.2 = 63 m
Question 20.
A bus stop is barricated from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹12 per m^{2}, what will be the cost of painting these cones. (Use π = 3.14 and = 1.02) [NCERT]
Solution:
Diameter of the base of tent = 40 cm
∴ Radius of the base of cone (r) =
Question 21.
A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, show that the radius of each is to the height of each as 3 : 4.
Solution:
Let radius of cylinder = r
and radius of cone = r
and let height of cylinder = h
and height of cone = h
Question 22.
A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone is 16 m above the ground. Find the area of the canvas required for the tent.
Solution:
Diameter of the cylindrical portion = 24 m 24
∴ Radius (r) = = 12 m
Height of cylindrical portion = 11 m
and total height = 16 m
∴ Height of conical portion = 16 – 11 = 5 m
∴ Slant height of the conical portion (l)
Question 23.
A circus tent is cylindrical to a height of 3 meters and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent.
Solution:
Diameter of the cylindrical tent = 105 m
Class 9 RD Sharma Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2
Question 1.
Find the volume of a right circular cone with:
(i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm
(iii) height 21 cm and slant height 28 cm. [NCERT]
Solution:
(i) Radius of a cone (r) = 6 cm
and height (h) = 7 cm
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. [NCERT]
Solution:
(i) Radius of the conical vessel (r) = 7 cm
Slant height (h) = 25 cm
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:
Ratio in the heights of two cones =1:3
and ratio in their radii = 3: 1
Let radius of first cone (r_{1}) = x
and of second cone (r_{2}) = 3x
and height of first cone (h_{1}) = 3y
and of second cone (h_{2})
Question 4.
The radius and the height of a right circular cone are in the ratio 5 : 12. If its volume is 314 cubic metre, find the slant height and the radius (Use π = 3.14).
Solution:
Ratio in the radius and height of a cone = 5 : 12
Volume = 314 cm^{3}
Let radius (r) = 5x
and height (h) = 12x
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:
Ratio in the radius and height of a right circular cone = 5 : 12
Volume = 2512 cm^{3}
Let radius (r) = 5x
Then height (h) = 12x
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:
Ratio in volumes of two cones = 4:5
and ratio in radii = 2:3
Let radius of the first cone (r_{1}) = 2x
Then radius of second cone (r_{2}) = 3x
Let h_{1}, h_{2} be their heights respectively
Question 7.
Ratio between their vertical heights = 9:5 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:
Let r be the radius and h be the height of a cylinder and a cone, then
Volume of cylinder = πr^{2}h
Question 8.
If the radius of the base of a cone is halved, keeping the height same, what is the ratio of the volume of the reduced cone to that of the original cone?
Solution:
Let r be the radius and h be the height of the cone, then
Volume = πr^{2}h
By halving the radius and same height the
Question 9.
A heap of wheat is in the form of a cone of diameter 9 m and height 3.5 m. Find its volume. How much canvas cloth is required to just cover the heap? (Use π = 3.14). [NCERT]
Solution:
Diameter of conical heap of wheat = 9 m
Question 10.
Find the weight of a solid cone whose base is of diameter 14 cm and vertical height 51 cm, supposing the material of which it is made weighs 10 grams per cubic cm.
Solution:
Diameter of the base of solid cone = 14 cm
and vertical height (h) = 51 cm
Question 11.
A right angled triangle of which the sides containing the right angle are 6.3 cm and 10 cm in length, is made to turn round on the longer side. Find the volume of the solid, thus generated. Also, find its curved surface area.
Solution:
Length of sides of a right angled triangle are 6.3 cm and 10 cm
By turning around the longer side, a cone is formed in which radius (r) = 6.3 cm
and height (h) = 10 cm
Question 12.
Find the volume of the largest right circular cone that can be fitted in a cube whose edge is 14 cm.
Solution:
Side of cube = 14 cm ,
Radius of the largest cone that can be fitted
Question 13.
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. [NCERT]
Solution:
Volume of a right circular cone = 9856 cm^{3}
Diameter of the base = 28 cm
Question 14.
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres? [NCERT]
Solution:
Diameter of the top of the conical pit = 3.5 m
Question 15.
Monica has a piece of Canvas whose area is 551 m^{2}. She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and wastage incurred while cutting, amounts to approximately 1 m^{2}. Find the volume of the tent that can be made with it. [NCERT]
Solution:
Area of Canvas = 551 m^{2}
Area of wastage = 1 m^{2}
Actual area = 551 – 1 = 550 m^{2}
Base radius of the conical tent = 7 m
Let l be the slant height and h be the vertical
RD Sharma Mathematics Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS
Question 1.
The height of a cone is 15 cm. If its volume is 500π cm^{3}, then find the radius of its base.
Solution:
Volume of cone = 500π cm^{3}
and height (h) = 15 cm
Question 2.
If the volume of a right circular cone of height 9 cm is 48π cm^{3}, find the diameter of its base.
Solution:
Volume of a cone = 48π cm^{3}
Height (h) = 9 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 a cone (h) = 21 cm
and slant height (l) = 28 cm
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:
Capacity of conical vessel = 3.3 litres
Volume = 3.3 m^{3}
= 3.3 x 1000 = 3300 cm^{2}
Question 5.
If the radius and slant height of a cone are in the ratio 7 : 13 and its curved surface area is 286 cm2, find its radius.
Solution:
Two ratio in radius and slant height of a cone = 7 : 13
Let radius (r) = 7x
and slant height (1) = 13x
Curved surface area = πrl
Question 6.
Find the area of canvas required for a conical tent of height 24 m and base radius 7 m.
Solution:
Base radius of the closed cone (r) = 7 cm
and vertical height (h) = 24 cm
Question 7.
Find the area of metal sheet required in making a closed hollow cone of base radius 7 cm and height 24 cm. making a closed hollow cone of base radius 7 cm and height 24 cm.
Solution:
Base radius of the closed cone (r) = 7 cm
and vertical height (h) = 24 cm
Question 8.
Find the length of cloth used in making a conical pandal of height 100 m and base radius 240 m, if the cloth is 100π m wide.
Solution:
Height of conical pandal (A) = 100 m
Base radius (r) = 240 m
RD Sharma Class 9 Solution Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS
Question 1.
The number of surfaces of a cone has, is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
Number of surfaces of a cone are 2 (b)
Question 2.
The area of the curved surface of a cone of radius 2r and slant height , is
Solution:
Radius of a cone = 2r
Question 3.
The total surface area of a cone of radius and length 2l, is
Solution:
Question 4.
A solid cylinder is melted and cast into a cone of same radius. The heights of the cone and cylinder are in the ratio
(a) 9 : 1
(b) 1 : 9
(c) 3 : 1
(d) 1 : 3
Solution:
Let r be the radius and h be the height of cylinder, then volume = πr^{2}h
Now volume of cone = πr^{2}h
r is the radius
Question 5.
If the radius of the base of a right circular cone is 3r and its height is equal to the radius of the base, then its volume is
Solution:
Radius of the base of a cone (R) = 3r
and height (H) = 3r
Question 6.
If the volumes of two cones are in the ratio 1 : 4 and their diameters are in the ratio 4 : 5, then the ratio of their heights, is
(a) 1 : 5
(b) 5 : 4
(c) 5 : 16
(d) 25 : 64
Solution:
Ratio in the volumes of two cones =1:4
and ratio in their diameter = 4:5
Let h_{1}, h_{2} be their heights
Question 7.
The curved surface area of one cone is twice that of the other while the slant height of the latter is twice that of the former. The ratio of their radii is
(a) 2 : 1
(b) 4 : 1
(c) 8 : 1
(d) 1 : 1
Solution:
Let r be the radius and l be the slant height
∴ Curved surface area of first cone = πr_{1}l_{1}
and let curved surface area of second cone = πr_{2}l_{2}
Question 8.
If the height and radius of a cone of volume V are doubled, then the volume of the cone, is
(a) 3V
(b) 4V
(c) 6V
(d) 8V
Solution:
Let r and h be the radius and height of a cone, then
Question 9.
The ratio of the volume of a right circular cylinder and a right circular cone of the same base and height, is
(a) 1 : 3
(b) 3 : 1
(c) 4 : 3
(d) 3 : 4
Solution:
Let r be the radius and h be the height of a right circular cylinder and a right circular cone, and V_{1} and V_{2} are their volumes, the V_{1} = πr^{2}h and
Question 10.
A right cylinder and a right circular cone have the same radius and same volumes. The ratio of the height of the cylinder to that of the cone is
(a) 3 : 5
(b) 2 : 5
(c) 3 : 1
(d) 1 : 3
Solution:
Let r be the radius of cylinder and cone and volumes are equal
and h_{1}, and h_{2} be their have h_{2} is respectively
Question 11.
The diameters of two cones are equal. If their slant heights are in the ratio 5 : 4, the ratio of their curved surface areas, is
(a) 4 : 5
(b) 25 : 16
(c) 16 : 25
(d) 5 : 4
Solution:
∵ Diameters of two cones are equal
∴ Their radii are also be equal
Let r be their radius of each cone,
and ratio in their slant heights = 5:4
Let slant height of first cone (h_{1}) = 5x
Then height of second cone (h_{2}) = 4x
Question 12.
If the heights of two cones are in the ratio of 1 : 4 and the radii of their bases are in the ratio 4 : 1, then the ratio of their volumes is
(a) 1 : 2
(b) 2 : 3
(c) 3 : 4
(d) 4 : 1
Solution:
Ratio in the heights of two cones =1 : 4
and ratio in their radii of their bases = 4 : 1
Let height of the first cone = x
and height of the second cone = 4x
Radius of the first cone = 4y
and radius of the second cone = y
Question 13.
The slant height of a cone is increased by 10%. If the radius remains the same, the curved surface area is increased by
(a) 10%
(b) 12.1%
(c) 20%
(d) 21%
Solution:
Let r be radius and l be the slant height of a cone, then curved surface area = πrl
If slant height is increased by 10%, then
Question 14.
The height of a solid cone is 12 cm and the area of the circular base is 6471 cm^{2}. A plane parallel to the base of the cone cuts through the cone 9 cm above the vertex of the cone, the area of the base of the new cone so formed is
(a) 9π cm^{2}
(b) 16π cm^{2}
(c) 25π cm^{2}
(d) 36π cm^{2}
Solution:
Height of a solid cone (h) = 12 cm
Area of circular base = 64π cm^{2}
Question 15.
If the base radius and the height of a right circular cone are increased by 20%, then the percentage increase in volume is approximately
(a) 60
(b) 68
(c) 73
(d) 78
Solution:
In first case,
Let r be radius and h be height, in volume
Question 16.
If h, S and V denote respectively the height, curved surface area and volume of a right circular cone, then 3πVh^{3} – S2h^{2} + 9V^{2} is equal to
(a) 8
(b) 0
(c) 4π
(d) 32π^{2}
Solution:
h = height, S = curved surface area
V = volume of a cone
Let r be the radius of the cone, then
Question 17.
If a cone is cut into two parts by a horizontal plane passing through the mid¬point of its axis, the ratio of the volumes of upper and lower part is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 7
(d) 1 : 8
Solution:
∴ ∆PDC ~ ∆PBA (AA axiom)
and O’ is mid point of PO