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

Assume π = ^{22}/_{7} , unless stated otherwise.

**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}/_{3}πr^{2}h

= ^{1}/_{3} x ^{22}/_{7} x 6^{2} x 7

= 264 cm^{3}

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

Volume of the cone = ^{1}/_{3}πr^{2}h

= ^{1}/_{3} x ^{22}/_{7} x 3.5^{2} x 12

= 154 cm^{3}

**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

h^{2} = l^{2} – r^{2}

= 25^{2} – 7^{2}

= 576

h = 24cm

Volume of conical vessel = ^{1}/_{3}πr^{2}h

= ^{1}/_{3} x ^{22}/_{7} x 7^{2} x 24

= 1232 cm^{3}

The capacity of conical vessel in litres = ^{1232}/_{1000} = 1.232 litres

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

h^{2} = l^{2} – r^{2}

r^{2} = l^{2} – h^{2}

= 13^{2} – 12^{2}

= 25

r = 5

Volume of conical vessel = ^{1}/_{3}πr^{2}h

= ^{1}/_{3} x ^{22}/_{7} x 5^{2} x 12

= 314.29 cm^{3}

The capacity of conical vessel in litres = ^{314.29}/_{1000} = 0.31429 litres

**The height of a cone is 15 cm. If its volume is 1570 cm**^{3}, find the radius of the base.(Use π = 3.14)

Solution:

h = 15 cm

Volume of cone = 1570 cm^{3}

We know, volume of the cone = ^{1}/_{3}πr^{2}h

1570 = ^{1}/_{3} x 3.14 x r^{2} x 15

1570 = 15.7 x r^{2}

r^{2} = ^{1570}/_{15.7} = 100

r = 10 cm

**If the volume of a right circular cone of height 9 cm is 48 π cm**^{3}, find the diameter of its base.

Solution:

h = 9 cm

Volume of the right circular cone = 48π cm^{3}

We know, volume of the cone = ^{1}/_{3}πr^{2}h

48 π =^{1}/_{3} x π x r^{2} x 9

48 = ^{1}/_{3} x r^{2} x 9

r^{2} = ^{48}/_{3} = 16

r = 4 cm

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

**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}/_{3}πr^{2}h

= ^{1}/_{3 }x ^{22}/_{7} x (^{7}/_{4})^{2} x 12

= 38.5 m^{3}

Therefore, the capacity of conical pit in litres = ^{38.5}/_{1000} = 0.0385 litres

**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**

Solution:

(i) Volume of right circular cone = 9856 cm^{3}

d = 28 cm ⇒r = ^{28}/_{2} = 14 cm

Volume of conical pit = ^{1}/_{3}πr^{2}h

9856 = ^{1}/_{3} x ^{22}/_{7} x 14^{2} x h

h = ^{9856x3x7}/_{22×14}^{2} = 48 cm

(ii) h^{2} = l^{2} – r^{2}

48^{2} = l^{2} – 14^{2}

l^{2} = 48^{2} + 14^{2}

l^{2 }= 2500

l = 50 cm

(iii) Curved surface area of the cone = πrl

= ^{22}/_{7} x 14 x 50

= 2200 cm^{2}

**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} πr^{2}h

= ^{1}/_{3} x π x 5^{2} x 12

= 100 π cm^{3}

**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} πr^{2}h

= ^{1}/_{3} x π x 12^{2} x 5

= 240 π cm^{3}

Required ratio = ^{100π}/_{240π}= ^{5}/_{12} = 5 : 12

**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} πr^{2}h

= ^{1}/_{3} x π x (5.25)^{2} x 3

= 86.625 cm^{3}

l = √(h^{2} + r^{2}) = √[3^{2}+(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 m^{2}

= 99.825 m^{2}

Therefore, 99.825m^{2} of canvas is needed.

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