NCERT solutions for class 9 Maths chapter 13 Surface Areas and Volumes (Exercise

NCERT solutions for class 9 Maths

Surface Areas and Volumes

Chapter 13 Exercise 13.7

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, height 12 cm
Solution:
(i) Radius of the cone (r) =6 cm
Height (h) = 7 cm

Ex 13.7 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 of conical vessel (r) = 7 cm and slant height (l) =25 cm

Thus, the capacity of the conical vessel = 1.232 litres.

(ii) Height of conical vessel (h) = 12 cm and slant height (l) = 13 cm

∴ Capacity of the conical vessel

Thus, the capacity of the conical vessel = $\frac { 11 }{ 35 }$ litres.

Ex 13.7 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:
Volume of the cone = 1570 cm³
Height of the cone (h) = 15 cm
Volume of cone = $\frac { 1 }{ 3 }$ πr²h
⇒1570= $\frac { 1 }{ 3 }$ x 3.14 x r² x 15

⇒1570 = 3.14 x 5 x r²

r² = 1570 / 3.14 x 5 = 100

r = 10 cm

Thus, the radius of the base of cone = 10 cm.

Ex 13.7 Question 4.

If the volume of a right circular cone of height 9 cm is 48 cm³, find the diameter of its base.
Solution:
Volume of the cone = 48 cm³
Height of the cone (h) = 9 cm
Volume of cone = $\frac { 1 }{ 3 }$ πr²h

Diameter = 2 x radius .
∴ Diameter of the base of the cone = (2 x 4)cm = 8 cm

Ex 13.7 Question 5.

A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Solution:
Radius of pit (r) = $\frac { 3.5 }{ 2 }$ m
Depth (h) = 12m
⇒ Volume (capacity) = $\frac { 1 }{ 3 }$ πr²h

Thus, the capacity of the pit = 38.5 kilolitres.

Ex 13.7 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 the cone V = 9856 cm³
Radius of the base (r) = $\frac { 28 }{ 2 }$ = 14 cm

(i) Let the height of the cone be h cm.

Thus, the height of cone = 48 cm.

(ii) Height of cone h = 48 cm and radius r = 14 cm. Let the slant height = l cm.
⇒ l² = r² +h²
⇒ l² = 14² + 48² = 196 + 2304 = 2500
∴ l = 50
Thus, the slant height of cone = 50 cm.

(iii) Slant height of cone l = 50cm and radius r =. 14 cm Curved surface area of a cone = πrl
∴ Curved surface area = $\frac { 22 }{ 7 }$ x 14 x 50 cm²
= 2200 cm²
Thus, the curved surface area of the cone is 2200 cm².

Ex 13.7 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:

If the triangle is revolved about 12 cm side, a cone will be formed. Therefore, the radius of of cone r = 5 cm Height(h) = 12cm and slant height l = 13 cm
∴ Volume of the cone so obtained = $\frac { 1 }{ 3 }$ πr²h
= $\frac { 1 }{ 3 }$ x π x (5)² x 12cm³
= 100 π cm³
Thus, the volume of the cone = 100πcm³.

Ex 13.7 Question 8.

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:
Since the right triangle is revolved about the side 5 cm.
∴ Height of the cone so obtained (h) = 5 cm
Radius of the cone (r) = 12 cm
Volume of solid = $\frac { 1 }{ 3 }$ πr²h = $\frac { 1 }{ 3 }$ x π x 12 x 12 x 5 = 240πcm³
Thus, the volume of solid = 240πcm³

Ex 13.7 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 its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Solution:

Thus, the volume of heap of wheat = 86.625m³
Let, the slant height of heap of. Wheat = l cm

Thus the required area of the canvas to protect wheat = 99.825 m²

NCERT solutions for class 9 Maths chapter 13 Surface Areas and Volumes (Exercise – 13.7)

Published by admin on September 3, 2020September 3, 2020