# RD SHARMA Solutions for Class 9 Maths Chapter 20 - Surface Areas and Volume of A Right Circular Cone

## Chapter 20 - Surface Areas and Volume of A Right Circular Cone Exercise 20.24

The number of surfaces of a cone has, is

(a) 1

(b) 2

(c) 3

(d) 4

A cone has two surfaces as follows: one curved surface and another bottom surface.

Hence, correct option is (b).

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

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

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

## Chapter 20 - Surface Areas and Volume of A Right Circular Cone Exercise 20.25

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

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

A right circular cylinder cylinder and a right circular cone have the same radius and the same volume. 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

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

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

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%

The height of a solid cone is 12 cm and the area of the circular base is 64∏ cm^{2}. A plane parallel to the base of the cone cuts through the cone 9 cm above the vertex of the cone, thee area of the base of the new cone so formed is

(a) 9∏ cm^{2}

(b) 16∏ cm^{2}

(c) 25∏ cm^{2}

(d) 36∏ cm^{2}

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

If h, S and V denote respectively the height, curved surface area and volume of a right circular cone, then 3∏Vh^{3} - S^{2}h^{2 }+ 9V^{2} is equal to

(a) 8

(b) 0

(c) 4∏

(d) 32∏^{2}

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

## Chapter 20 - Surface Areas and Volume of A Right Circular Cone Exercise Ex. 20.1

^{2}and its diameter is 70 cm. What is its slant height? (Use = 22/7).

^{2}and its slant height is 14 cm. Find the radius of the base and total surface area of the cone.

Let radius of circular end of cone be r.

CSA of cone =

Thus, the radius of circular end of the cone is 7 cm.

(ii) Total surface area of cone = CSA of cone + Area of base

=

^{2}.

^{2}canvas is Rs 70, find the cost of the canvas required to make the tent,

What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use = 3.14).

Radius (r) of base of tent = 6 m

Slant height (l) of tent =

CSA of conical tent = = (3.14 6 10) = 188.4

Let length of tarpaulin sheet required be L.

As 20 cm will be wasted so, effective length will be (L - 0.2 m)

Breadth of tarpaulin = 3 m

Area of sheet = CSA of tent

[(L - 0.2 m) 3] m = 188.4

L - 0.2 m = 62.8 m

L = 63 m

Thus, the length of the tarpaulin sheet will be 63 m.

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 to the height of each is 3:4.

## Chapter 20 - Surface Areas and Volume of A Right Circular Cone Exercise Ex. 20.2

(i) radius 6 cm, height 7 cm

(ii) radius 3.5 cm, height 12 cm

(iii) height 21 cm and slant height 28 cm.

Height (h) of cone = 7 cm

Volume of cone

(ii) Radius (r) of cone = 3.5 cm

Height (h) of cone = 12 cm

Volume of cone

(iii)

(i) radius 7 cm, slant height 25 cm

(ii) height 12 cm, slant height 13 cm

Slant height (l) of cone = 25 cm

Height (h) of cone

Volume of cone

Capacity of the conical vessel = litres= 1.232 litres

Slant height (l) of cone = 13 cm

Radius (r) of cone

Volume of cone = 314.28 cm

^{3}

Capacity of the conical vessel = litres = litres.

^{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

Let height of cone be h.

Volume of cone = 9856 cm

^{3}

Thus, the slant height of the cone is 50 cm.

A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

Depth (h) of pit = 12 m

Volume of pit =

^{2}. She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and the wastage incurred while cutting, amounts to approximately 1 m

^{2}, find the volume of the tent that can be made with it.

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