RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1

RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1

Other Exercises

Question 1.
Find the surface area of a sphere of radius.
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
Solution:
In a sphere,
(i) Radius (r) = 10.5 cm
Surface area = 4πr2
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 1.1
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 1.2

Question 2.
Find the surface area of a sphere of diameter
(i) 14 cm
(ii) 21 cm
(iii) 3.5 cm
Solution:
(i) Diameter of a sphere = 14 cm
Radius (r) = \(\frac { 14 }{ 2 }\) = 7 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 2.1

Question 3.
Find the total surface area of a hemisphere and a solid hemisphere each of radius 10 cm. [Use π = 3.14]
Solution:
(i) Radius of hemisphere = 10 cm
∴ Total surface area of hemisphere = 2πr2
= 2 x 3.14 x 10 x 10 cm2
= 628 cm2
(ii) Total surface area of solid hemisphere
= 3πr2 = 3 x 3.14 x 10 x 10 cm2
= 942 cm2

Question 4.
The surface area of a sphere in 5544 cm2, find the diameter.
Solution:
Let r be the radius of a sphere, then Surface area = 4πr2
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 4.1

Question 5.
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹4 per 100 cm2. [NCERT]
Solution:
Inner diameter of a hemispherical bowl = 10.5 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 5.1

Question 6.
The dome of a building is in the form of a hemisphere. Its radius is 63 dm. Find the cost of painting it at the rate of ₹2 per sq. m.
Solution:
Radius of dome (hemispherical) = 63 dm
Area of curved surface
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 6.1

Question 7.
Assuming the earth to be a sphere of radius 6370 km, how many square kilometres is area of the land, if three-fourth of the earth’s surface is covered by water?
Solution:
Radius of earth (sphere) = 6370 km
Water on the earth = \(\frac { 3 }{ 4 }\) % total area
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 7.1

Question 8.
A cylinder of same height and radius is placed on the top of a hemisphere. Find the curved surface area of the shape if the length of the shape be 7 cm.
Solution:
Total height of the so formed shape = 7 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 8.1

Question 9.
The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Solution:
Diameter of moon = \(\frac { 1 }{ 4 }\) of diameter of earth
Let radius of earth = r km
Then radius of moon = \(\frac { 1 }{ 4 }\) r km
Now surface area of earth = 4πr2
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 9.1

Question 10.
A hemi-spherical dome of a building needs to be painted. If the circumference of the base of the dome is 17.6 m, find the cost of painting it, given the cost of painting is ₹5 per 100 cm2. [NCERT]
Solution:
Circumference of the base of dome (r) = 17.6 m
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 10.1

Question 11.
A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 16 cm and its height is 15 cm. Find the cost of painting the toy at ₹7 per 100 cm2.
Solution:
Diameter of toy = 16 cm
Radius (r) = \(\frac { 16 }{ 2 }\) = 8 cm
Height of conical part (h) = 15 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 11.1
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 11.2

Question 12.
A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of ₹10 per m2.
Solution:
Diameter of the tank = 1.4 m
∴ Radius (r) = \(\frac { 1.4 }{ 2 }\) m = 0.7 m
and height of cylindrical portion = 8m
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 12.1
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 12.2

Question 13.
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in the figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2. [NCERT]
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 13.1
Solution:
Diameter of each spheres = 21 cm
∴ Radius (R) = \(\frac { 21 }{ 2 }\) cm
Radius of each cylinder (r) = 1.5 cm
and height (h) = 7 cm
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 13.2
Now surface area of one sphere = 4πR2
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 13.3
RD Sharma Class 9 Solutions Chapter 21 Surface Areas and Volume of a Sphere Ex 21.1 13.4

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RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS

RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS

Other Exercises

Mark the correct alternative in each of the following:
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 \(\frac { 1 }{ 2 }\), is
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q2.1
Solution:
Radius of a cone = 2r
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q2.2

Question 3.
The total surface area of a cone of radius \(\frac { r }{ 2 }\) and length 2l, is
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q3.1
Solution:
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q3.2

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 = πr2h
Now volume of cone = πr2h
r is the radius
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q4.1

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q5.1
Solution:
Radius of the base of a cone (R) = 3r
and height (H) = 3r
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q5.2

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 h1, h2 be their heights
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q6.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q6.2

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 = πr1l1
and let curved surface area of second cone = πr2l2
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q7.1

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q8.1

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 V1 and V2 are their volumes, the V1 =  πr2h and
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q9.1

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 h1, and h2 be their have h2 is respectively
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q10.1

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 (h1) = 5x
Then height of second cone (h2) = 4x
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q11.1

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q12.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q12.2

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q13.1

Question 14.
The height of a solid cone is 12 cm and the area of the circular base is 6471 cm2. 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 cm2
(b) 16π cm2
(c) 25π cm2
(d) 36π cm2
Solution:
Height of a solid cone (h) = 12 cm
Area of circular base = 64π cm2
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q14.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q14.2

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q15.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q15.2

Question 16.
If h, S and V denote respectively the height, curved surface area and volume of a right circular cone, then 3πVh3 – S2h2 + 9V2 is equal to
(a) 8
(b) 0
(c) 4π
(d) 32π2
Solution:
h = height, S = curved surface area
V = volume of a cone
Let r be the radius of the cone, then
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q16.1

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q17.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone MCQS Q17.2

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RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS

RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS

Other Exercises

Question 1.
The height of a cone is 15 cm. If its volume is 500π cm3, then find the radius of its base.
Solution:
Volume of cone = 500π cm3
and height (h) = 15 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q1.1

Question 2.
If the volume of a right circular cone of height 9 cm is 48π cm3, find the diameter of its base.
Solution:
Volume of a cone = 48π cm3
Height (h) = 9 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q2.1

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q3.1

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 m3
= 3.3 x 1000 = 3300 cm2
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q4.1

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q5.1

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q6.1

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q7.1

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 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone VSAQS Q8.1

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RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2

RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2

Other Exercises

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q1.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q1.2

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q2.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q2.2

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 (r1) = x
and of second cone (r2) = 3x
and height of first cone (h1) = 3y
and of second cone (h2)
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q3.1

Question 4.
The radius and the height of a right circular cone are in the ratio 5 : 12. If its volume is 314 cubic 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 cm3
Let radius (r) = 5x
and height (h) = 12x
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q4.1

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 cm3
Let radius (r) = 5x
Then height (h) = 12x
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q5.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q5.2

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 (r1) = 2x
Then radius of second cone (r2) = 3x
Let h1, h2 be their heights respectively
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q6.1

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 = πr2h
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q7.1

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 = \(\frac { 1 }{ 3 }\) πr2h
By halving the radius and same height the
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q8.1

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q9.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q9.2

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q10.1

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q11.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q11.2

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q12.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q12.2

Question 13.
The volume of a right circular cone is 9856 cm3. 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 cm3
Diameter of the base = 28 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q13.1

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q14.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q14.2

Question 15.
Monica has a piece of Canvas whose area is 551 m2. 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 m2. Find the volume of the tent that can be made with it. [NCERT]
Solution:
Area of Canvas = 551 m2
Area of wastage = 1 m2
Actual area = 551 – 1 = 550 m2
Base radius of the conical tent = 7 m
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q15.1
Let l be the slant height and h be the vertical
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.2 Q15.2

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RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1

RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1

Other Exercises

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 1.1

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 2.1

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 cm2
and radius (r) = 1 cm2
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 3.1

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
∴ l2 = r2 + h2
⇒ r2 = l2– h2 = (28)2 – (21 )2
⇒ 784 – 441 = 343 …(i)
Now area of base = πr2
= \(\frac { 22 }{ 7 }\) x 343 [From (i)]
= 22 x 49 = 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 base of cone (r) = 6 cm
and height (h) = 8 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 5.1

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 6.1

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) = \(\frac { 24 }{ 2 }\) = 12 m
Now total surface area = πr(l + r)
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 7.1

Question 8.
The area of the curved surface of a cone is 60π cm2. If the slant height of the cone be 8 cm, find the radius of the base.
Solution:
Curved surface area of a cone = 6071 cm2
Slant height (l) = 8 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 8.1

Question 9.
The curved surface area of a cone is 4070 cm2 and its diameter is 70 cm. What is the slant height ? (Use π = 22/7).
Solution:
Surface area of a cone = 4070 cm2
Diameter of its base = 70 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 9.1

Question 10.
The radius and slant height of a cone are in the ratio of 4 : 7. If its curved surface area is 792 cm2, find its radius. (Use π = 22/7)
Solution:
Curved surface area of a cone = 792 cm2
Ratio in radius and slant height = 4:7
Let radius = 4x
Then slant height = 7x
∴ Curved surface area πrl = 792
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 10.1

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 11.1

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) = \(\frac { d }{ 2 }\)
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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 12.1

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 r1 and r2 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 = 2πr1h
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 13.1

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) = \(\frac { d }{ 2 }\)
and of second cone (r2) = \(\frac { d }{ 2 }\)
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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 14.1

Question 15.
Curved surface area of a cone is 308 cm2 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 cm2
and slant height (l) = 14 cm
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 15.1

Question 16.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2. [NCERT]
Solution:
Slant height of a cone (l) = 25 m
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 16.1

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 m2 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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 17.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 17.2

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 18.1

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 19.1
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 card-board. 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 m2, what will be the cost of painting these cones. (Use π = 3.14 and \(\sqrt { 1.04 } \) = 1.02) [NCERT]
Solution:
Diameter of the base of tent = 40 cm
∴ Radius of the base of cone (r) = \(\frac { 40 }{ 2 }\)
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 20.1

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 21.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 21.2

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) = \(\frac { 24 }{ 2 }\) = 12 m
Height of cylindrical portion = 11 m
and total height = 16 m
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 22.1
∴ Height of conical portion = 16 – 11 = 5 m
∴ Slant height of the conical portion (l)
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 22.2

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
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 23.1
RD Sharma Class 9 Solutions Chapter 20 Surface Areas and Volume of A Right Circular Cone Ex 20.1 23.2

 

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