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# Class 9 NCERT Solutions- Chapter 13 Surface Areas And Volumes – Exercise 13.7

• Last Updated : 04 Mar, 2021

### (ii) radius 3.5 cm, height 12 cm

Solution:

Volume of cone (V) = (1/3) Ã— Ï€r2h

(i) Given values,

Radius of cone (r) = 6 cm

Height of cone (h) = 7 cm

V = (1/3) Ã— (22/7) Ã— 6 Ã— 6 Ã— 7             (using Ï€=22/7)

V = 22 Ã— 6 Ã— 2

V = 264 cm3

(ii) Given values,

Radius of cone (r) = 3.5 cm

Height of cone (h) = 12 cm

V = (1/3) Ã— (22/7) Ã— 3.5 Ã— 3.5 Ã— 12           (using Ï€=22/7)

V = 154 cm3

### (ii) height 12 cm, slant height 13 cm

Solution:

Volume of cone (V) = (1/3) Ã— Ï€r2h

(i) Given values,

Radius of cone (r) = 7 cm

Slant height of cone (l) = 25 cm

h = âˆš(l2 – r2)

h = âˆš(252 – 72)

h = âˆš576

h = 24 cm

V = (1/3) Ã— (22/7) Ã— 7 Ã— 7 Ã— 24                         (using Ï€=22/7)

V = 22 Ã— 7 Ã— 8

V = 1232 cm3

(ii) Given values,

Height of cone (h) = 12 cm

Slant height of cone (l) = 13 cm

r = âˆš(l2 – h2)

r = âˆš(132 – 122)

r = âˆš25

r = 5 cm

V = (1/3) Ã— (22/7) Ã— 5 Ã— 5 Ã— 12                            (using Ï€=22/7)

V = 2200/7 cm3

V = 314.28 cm3

### Question 3. The height of a cone is 15 cm. If its volume is 1570 cm3, find the radius of the base. (Use Ï€ = 3.14)

Solution:

Given values,

Height of cone (h) = 15 cm

Volume of cone (V) = 1570 cm3

V = (1/3) Ã—  Ï€r2h

1570 = (1/3) Ã— 3.14 Ã— r2 Ã— 15                         (using Ï€=3.14)

r2 = 1570 Ã— 3 / (3.14 Ã— 15)

r2 = 100

r = âˆš100

r = 10 cm

### Question 4. If the volume of a right circular cone of height 9 cm is 48 Ï€ cm3, find the diameter of its base.

Solution:

Given values,

Height of cone (h) = 9 cm

Volume of cone (V) = 48Ï€ cm3

V = (1/3) Ã— Ï€r2h

48 Ã— Ï€ = (1/3) Ã— Ï€ Ã— r2 Ã— 9

r2 = 48 Ã— 3 / 9          (canceling Ï€ from both sides)

r2 = 16

r = âˆš16

r = 4 cm

Diameter = 2 times radius = 2 Ã— r

= 2 Ã— 4

= 8 cm

### Question 5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kiloliters?

Solution:

Given values,

Radius of cone (r) = 3.5/2 m

Height of cone (h) = 12 m

Volume of cone = (1/3) Ã— Ï€r2h

= (1/3) Ã— (22/7) Ã— (3.5/2) Ã— (3.5/2) Ã— 12                (taking Ï€=22/7)

= (22 Ã— 3.5 Ã— 3.5 Ã— 12) / (7 Ã— 3 Ã— 2 Ã— 2)

= 38.5 m3

Capacity of conical pit in kilo liters:

1000 m3 = 1 liter

38.5 m3 = 1000 Ã— 38.5 liters

= 38,500 liters

= 38.5 kilo liters

### (iii) curved surface area of the cone

Solution:

Given values,

Radius of cone (r) = 28/2 = 14 cm

Volume of cone (V) = 9856 cm3

(i) Volume of cone = (1/3) Ã— Ï€r2h

9856 = (1/3) Ã— (22/7) Ã— 14 Ã— 14 Ã— h                   (taking Ï€=22/7)

h = (9856 Ã— 3 Ã— 7) / (22 Ã— 14 Ã— 14)

h = 48 cm

(ii) Let slant height = l

l2 = h2 + r2

l = âˆš(h2 + r2)

l = âˆš(482 + 142)

l = âˆš(2304 + 196)

l = âˆš2500

l = 50 cm

(iii) curved surface area of the cone = Ï€rl

= Ï€ Ã— 14 Ã— 50 cm2

= 22/7 Ã— 700                                   (taking Ï€=22/7)

= 2,200 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:

Here, after revolving the triangle about the side 12 cm, we get

Radius of cone (r) =5 cm

Height of cone (h) = 12 cm

Volume of cone = (1/3) Ã— Ï€r2h

= (1/3) Ã— Ï€ Ã— 5 Ã— 5 Ã— 12

= (12 Ã— Ï€ Ã— 5 Ã— 5) / 3

V = 100Ï€ cm3

### 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:

Here, after revolving the triangle about the side 5 cm, we get

Radius of cone (r) =12 cm

Height of cone (h) = 5 cm

Volume of cone = (1/3) Ã— Ï€r2h

= (1/3) Ã— Ï€ Ã— 12 Ã— 12 Ã— 5

= (12 Ã— Ï€ Ã— 12 Ã— 5) / 3

V = 240Ï€ cm3

Volume in Question 7 = 100Ï€ cm3

Ratio = (Volume in Question 8) / (Volume in Question 7)

= 240Ï€/100Ï€

= 12/5

Hence, the ratio obtained = 12 : 5

### 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:

Given values,

Radius of cone (r) = 10.5/2 = 105/20 m

Height of cone (h) = 3 m

Volume of cone = (1/3) Ã— Ï€r2h

= (1/3) Ã— (22/7) Ã— (105/2) Ã— (105/2) Ã— 3                (taking Ï€=22/7)

= (22 Ã— 105 Ã— 105 Ã— 3) / (3 Ã— 20Ã— 20Ã— 7)

V = 86.625 m3

Area of the canvas = surface area of cone = Ï€rl

Slant height (l) = âˆš(h2 + r2)

l = âˆš(32 + (10.5/2)2)

l = âˆš(9+ (110.25/4))

l = âˆš(146.25/4)

l = âˆš36.56

l = 6.05 m (approx.)

Surface area of cone = Ï€ Ã— (105/20) Ã— 6.05 m2

= (22/7) Ã— (635.25/20)                                         (taking Ï€=22/7)

= 99.82 m2

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