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

Students can Download Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.7 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 9 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.7

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 = $$\frac{1}{3}$$ πr2h

(ii) r = 3.5 cm, h = 12 cm.
Volume of Cone, V = $$\frac{1}{3}$$ πr2h

∴ 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

(ii) height, h = 12 cm, slant height, l = 13cm. r = ?, V = ?
h2 = l2 – r2
∴ r2 = l2 – h2

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 = $$\frac{1}{3}$$ πr2h

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 = $$\frac{1}{3}$$ πr2h
∴ 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:

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= $$\frac{1}{3}$$ πr2h
h = $$\frac{2688}{56}$$
h = 48 cm.

(ii) l2 = h2 + r2

l = 50 cm.

(iii) Curved surface area of the cone, A
A = πrl
$$\frac{22}{7}$$ × 14 × 50
= 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
= $$\frac{1}{3}$$ πr2h = $$\frac{1}{3}$$ × π × 52 × 12
= 100π cm3
∴ Volume of cone soformed is 100π 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 = $$\frac{1}{3}$$ πr2h
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:

Volume of the heap of wheat is 86.625m Area of canvas required CSA of cone

∴ 99.825m2 canvas will be required to protect the heap from rain.