## Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.7

(Assume π = \(\frac{22}{7}\)unless stated otherwise)

Question 1.

Find the volume of the right circular cone with

(i) radius 6 cm, height 7 cm.

(ii) radius 3.5 cm, slant height 12 cm.

Solution:

(i) Volume of Cone, V = \(\frac{1}{3} \pi r^{2} h\)

∴ V = 264 cm^{3}.

(ii) r = 3.5 cm, h = 12 cm.

Volume of Cone, V = \(\frac{1}{3} \pi r^{2} h\)

= 11 × 14

∴ V = 154 cm^{3}.

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.

Solution:

(i) radius, r = 7 cm, slant height, l = 25 cm

Volume of conical vessel, V = \(\frac{1}{3} \pi r^{2} h\)

∴ h = 24 cm.

V = \(\frac{1}{3} \pi r^{2} h\)

= 22 × 7 × 8

∴ V = 1232 cm^{3}

∴ Volume of conical vessel

= \(\frac{1232}{1000}\) = 1.232 litres

(ii) height, h =12 cm, slant height, l = 13cm. r = ?, V = ? .

h^{2} = l^{2} – r^{2}

∴ r^{2} = l^{2} – h^{2}

∴ r = 5 cm.

Volume of vessel, V = \(\frac{1}{3} \pi r^{2} h\)

∴ V = 314.28 cm^{3}.

In litres, = \(\frac{314.28}{1000}\) = 0.31428 litres.

Question 3.

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:

Height of the cone, h = 15 cm.

Volume, V = 1570 cm^{3}, radius, r = ?

Volume of a cone. V = \(\frac{1}{3} \pi r^{2} h\)

r^{2} = 100

∴ r = 10 cm.

Question 4.

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

Solution:

Volume of a right circular cone,V= 48π cm^{3},

height, h = 9 cm.

diameter, d = ?

Volume of cone, V = \(\frac{1}{3} \pi r^{2} h\)

∴ r^{2} = 16

∴ r = 4 cm

∴ diameter, d = 2r

= 2 × 4

∴ d = 8 cm.

Question 5.

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

Solution:

d = 3.5m., h = 12 m., V = ?

Volume of Conical pit, V

= 22 × 0.5 × 3.5

∴ V = 38.5 m^{3}

V = 38.5 kilolitres

Question 6.

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:

Volume of Cone, V = 9856 cm^{3}

diameter of the base, d = 28 cm.

(i) h = ?,

(ii) I = ?,

(iii) C.S.A. = ?

(i) V = 9856 cm^{3}.

d = 28 cm. ∴ r = \(\frac{28}{2}\) = 14 cm.

Volume of cone, V= \(\frac{1}{3} \pi r^{2} h\)

h = \(\frac{2688}{56}\)

h = 48 cm.

(ii) l^{2} = h^{2} + r^{2}

l = 50 cm.

(iii) Curved surface area of the cone, A

A = πrl

= 22 × 100

A= 2200 cm^{2}.

Question 7.

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:

Cone is formed ∆ABC is revolved about the side 12 cm.

r = 5 cm,

h = 12 cm,

V = ?

Volume of Cone, V

∴ V = 314.29 cm^{3}.

Question 8.

If the triangle ABC in 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:

Cone is formed when ∆ABC is revolved about 5 cm.

r = 12 cm,

h = 5 cm.

Volume of Cone, V

v = \(\frac{1}{3} \pi r^{2} h\)

V = 754.28 cm^{3}.

Volume of Cone in problem 7 = 314.29 cm^{3}

Volume of Cone in problem 8 = 754.28 cm^{3}

∴ ratio of both Volumes = \(\frac{314.29}{754.28}\)

= \(\frac{5}{2}\) = 5 : 12

Question 9.

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

Solution:

Diameter of heap of wheat in the form of Cone,

diameter, d = 10.5 m = \(\frac{21}{2}\)

height, h = 3 m. V = ?

Volume of Canvas Cover, V = \(\frac{1}{3} \pi r^{2} h\)

= 86.625 m^{3}.

Area of Canvas required to cover a heap of wheat, C.S.A.

C.S.A.= πrl

∴ Area of Canvas, A = πrl

∴ A = 99.83 m^{2}.