KSEEB Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.7

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 cm³.
(ii) r = 3.5 cm, h = 12 cm.
Volume of Cone, V = \(\frac{1}{3} \pi r^{2} h\)

= 11 × 14
∴ V = 154 cm³.

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 cm³
∴ Volume of conical vessel
= \(\frac{1232}{1000}\) = 1.232 litres
(ii) height, h =12 cm, slant height, l = 13cm. r = ?, V = ? .
h² = l² – r²
∴ r² = l² – h²

∴ r = 5 cm.
Volume of vessel, V = \(\frac{1}{3} \pi r^{2} h\)

∴ V = 314.28 cm³.
In litres, = \(\frac{314.28}{1000}\) = 0.31428 litres.

Question 3.
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:
Height of the cone, h = 15 cm.
Volume, V = 1570 cm³, radius, r = ?
Volume of a cone. V = \(\frac{1}{3} \pi r^{2} h\)

r² = 100
∴ r = 10 cm.

Question 4.
If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.
Solution:
Volume of a right circular cone,V= 48π cm³,
height, h = 9 cm.
diameter, d = ?
Volume of cone, V = \(\frac{1}{3} \pi r^{2} h\)

∴ r² = 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 m³
V = 38.5 kilolitres

Question 6.
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:
Volume of Cone, V = 9856 cm³
diameter of the base, d = 28 cm.
(i) h = ?,
(ii) I = ?,
(iii) C.S.A. = ?
(i) V = 9856 cm³.
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) l² = h² + r²

l = 50 cm.

(iii) Curved surface area of the cone, A
A = πrl

= 22 × 100
A= 2200 cm².

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 cm³.

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 cm³.
Volume of Cone in problem 7 = 314.29 cm³
Volume of Cone in problem 8 = 754.28 cm³
∴ 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 m³.
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 m².

KSEEB Solutions for Class 9 Maths