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

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

### Solution 15

(i)    Slant height of cone = 14 cm
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
=

Thus, the total surface area of the cone is 462 .

### Solution 19

Height (h) of conical tent = 8 m
Radius (r) of base of tent = 6 m
Slant height (l) of tent =
CSA of conical tent =  = (3.14  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.

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

### Solution 1

(i)    Radius (r) of cone = 6 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)

### Solution 2

(i)    Radius (r) of cone = 7 cm
Slant height (l) of cone = 25 cm
Height (h) of cone
Volume of cone
Capacity of the conical vessel =  litres= 1.232 litres
(ii)    Height (h) of cone = 12 cm
Slant height (l) of cone = 13 cm

Volume of cone   = 314.28 cm3
Capacity of the conical vessel = litres =  litres.

### Solution 13

(i)    Radius of cone =   =14 cm
Let height of cone be h.
Volume of cone = 9856 cm3

h = 48 cm
Thus, the height of the cone is 48 cm.

(ii)   Slant height (l) of cone

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

(iii)    CSA of cone = rl =

### Solution 14

Depth (h) of pit = 12 m
Volume of pit =
Capacity of the pit = (38.5  1) kilolitres = 38.5 kilolitres

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

### Solution 1

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

Hence, correct option is (b).