Surface Area and Volumes – Exercise 13.7 – Class IX

For previous exercise – Surface Area and Volumes – Exercise 13.6 – Class IX


Assume π = 22/7 , unless stated otherwise.

  1. Find the volume of the right circular cone with

(i) radius 6 cm, height 7 cm

(ii) radius 3.5 cm, height 12 cm

Solution:

(i) r = 6 cm; h = 7cm

Volume of the cone = 1/3πr2h

= 1/3 x 22/7 x 62 x 7

= 264 cm3

(ii) r = 3.5 cm ; h = 12 cm

Volume of the cone = 1/3πr2h

= 1/3 x 22/7 x 3.52 x 12

= 154 cm3


  1. 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) r = 7cm , l = 25 cm

h2 = l2 – r2

=  252 – 72

= 576

h = 24cm

Volume of conical vessel = 1/3πr2h

= 1/3 x 22/7 x 72 x 24

= 1232 cm3

The capacity of conical vessel in litres = 1232/1000 = 1.232 litres

 

(ii)h = 12 cm , l = 13 cm

h2 = l2 – r2

r2 = l2 – h2

= 132 – 122

= 25

r = 5

Volume of conical vessel = 1/3πr2h

= 1/3 x 22/7 x 52 x 12

= 314.29 cm3

The capacity of conical vessel in litres = 314.29/1000 = 0.31429 litres


  1. The height of a cone is 15 cm. If its volume is 1570 cm3, find the radius of the base.(Use π = 3.14)

Solution:

h = 15 cm

Volume of cone = 1570 cm3

We know, volume of the cone = 1/3πr2h

1570 = 1/3 x 3.14  x r2 x 15

1570 = 15.7 x r2

r2 = 1570/15.7 = 100

r = 10 cm


  1. If the volume of a right circular cone of height 9 cm is 48 π cm3, find the diameter of its base.

Solution:

h = 9 cm

Volume of the right circular cone = 48π cm3

We know, volume of the cone = 1/3πr2h

48 π =1/3 x π x  r2 x 9

48 = 1/3 x r2 x 9

r2 = 48/3 = 16

r = 4 cm

Therefore, diameter of right circular cone, d = 2r = 2 x 4 = 8 cm


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

Solution:

d = 3.5 m ⇒ r = d/2 = 3.5/2 = 7/4 m

h = 12 m

Volume of conical pit = 1/3πr2h

= 1/3 x 22/7 x (7/4)2 x 12

= 38.5 m3

Therefore, the capacity of conical pit in litres = 38.5/1000 = 0.0385 litres


  1. 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

Solution:

(i) Volume of right circular cone = 9856 cm3

d = 28 cm ⇒r = 28/2 = 14 cm

Volume of conical pit = 1/3πr2h

9856 = 1/3 x 22/7 x 142 x h

h = 9856x3x7/22×142 = 48 cm

(ii) h2 = l2 – r2

482 = l2 – 142

l2 = 482 + 142

l2 = 2500

l = 50 cm

(iii) Curved surface area of the cone = πrl

= 22/7 x 14 x 50

= 2200 cm2


  1. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

Solution:

r = 5 cm, h = 12 cm , l = 13 cm

Volume of the cone = 1/3 πr2h

= 1/3 x π x 52 x 12

= 100 π cm3


  1. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

Solution:

r = 12 cm, h = 5 cm

Volume of the cone = 1/3 πr2h

= 1/3 x π x 122 x 5

= 240 π cm3

Required ratio = 100π/240π= 5/12 = 5 : 12


  1. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Solution:

d = 10.5 m

r =  d/2 = 10.5/2 = 5.25 m

h = 3 m

Volume of the cone = 1/3 πr2h

= 1/3 x π x (5.25)2 x 3

= 86.625 cm3

l = √(h2 + r2) = √[32+(5.25)2] = √[9+27.5625] = √(36.5625) = 6.05 m

Curved surface area of the cone = πrl

= 22/7 x 5.25 x 6.05 m2

= 99.825 m2

Therefore, 99.825m2 of canvas is needed.


Next exercise – Surface Area and Volumes – Exercise 13.8 – Class IX