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

Volume of cone (V) = (1/3) × πr^{2}h

(i)Given values,Radius of cone (r) = 6 cm

Height of cone (h) = 7 cm

V = (1/3) × (22/7) × 6 × 6 × 7 (using π=22/7)

V = 22 × 6 × 2

V = 264 cm^{3}

(ii)Given values,Radius of cone (r) = 3.5 cm

Height of cone (h) = 12 cm

V = (1/3) × (22/7) × 3.5 × 3.5 × 12 (using π=22/7)

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:**

Volume of cone (V) = (1/3) × πr^{2}h

(i)Given values,Radius of cone (r) = 7 cm

Slant height of cone (l) = 25 cm

h = √(l^{2}– r^{2})h = √(25

^{2 }– 7^{2})h = √576

h = 24 cmV = (1/3) × (22/7) × 7 × 7 × 24 (using π=22/7)

V = 22 × 7 × 8

V = 1232 cm^{3}

(ii)Given values,Height of cone (h) = 12 cm

Slant height of cone (l) = 13 cm

r = √(l^{2}– h^{2})r = √(13

^{2}– 12^{2})r = √25

r = 5 cmV = (1/3) × (22/7) × 5 × 5 × 12 (using π=22/7)

V = 2200/7 cm

^{3}

V = 314.28 cm^{3}

**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) **

^{3}, find the radius of the base. (Use

**Solution:**

Given values,

Height of cone (h) = 15 cm

Volume of cone (V) = 1570 cm

^{3}

V = (1/3) × πr^{2}h1570 = (1/3) × 3.14 × r

^{2 }× 15 (using π=3.14)r

^{2}= 1570 × 3 / (3.14 × 15)r

^{2}= 100r = √100

r = 10 cm

**Question 4. If the volume of a right circular cone of height 9 cm is 48 **π **cm**^{3}, find the diameter of its base.

^{3}, find the diameter of its base.

**Solution:**

Given values,

Height of cone (h) = 9 cm

Volume of cone (V) = 48π cm

^{3}V = (1/3) × πr

^{2}h48 × π = (1/3) × π × r

^{2 ×}9r

^{2}= 48 × 3 / 9 (canceling π from both sides)r

^{2}= 16r = √16

r = 4 cm

Diameter = 2 times radius = 2 × r= 2 × 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) × πr^{2}h= (1/3) × (22/7) × (3.5/2) × (3.5/2) × 12 (taking π=22/7)

= (22 × 3.5 × 3.5 × 12) / (7 × 3 × 2 × 2)

= 38.5 m

^{3}

Capacity of conical pit in kilo liters:

1000 m^{3}= 1 liter38.5 m

^{3}= 1000 × 38.5 liters

= 38,500 liters

= 38.5 kilo liters

**Question 6. The volume of a right circular cone is 9856 cm**^{3}. If the diameter of the base is 28 cm, find

^{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:**

Given values,

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

Volume of cone (V) = 9856 cm

^{3}

(i)Volume of cone = (1/3) × πr^{2}h9856 = (1/3) × (22/7) × 14 × 14 × h (taking π=22/7)

h = (9856 × 3 × 7) / (22 × 14 × 14)

h = 48 cm

(ii)Let slant height = l

l^{2 }= h^{2}+ r^{2}l = √(h

^{2}+ r^{2})l = √(48

^{2}+ 14^{2})l = √(2304 + 196)

l = √2500

l = 50 cm

(iii) curved surface area of the cone = πrl

=π × 14 × 50 cm^{2}= 22/7 × 700 (taking π=22/7)

= 2,200 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:**

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) × πr^{2}h= (1/3) × π × 5 × 5 × 12

= (12 × π × 5 × 5) / 3

V = 100π cm^{3}

**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) × πr

^{2}h= (1/3) × π × 12 × 12 × 5

= (12 × π × 12 × 5) / 3

V = 240π cm^{3}Volume in Question 7 = 100π cm

^{3}Ratio = (Volume in Question 8) / (Volume in Question 7)

= 240π/100π

= 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) × πr

^{2}h= (1/3) × (22/7) × (105/2) × (105/2) × 3 (taking π=22/7)

= (22 × 105 × 105 × 3) / (3 × 20× 20× 7)

V = 86.625 m^{3}

Area of the canvas = surface area of cone = πrl

Slant height (l)=√(h^{2 }+ r^{2})

l =√(3^{2 }+ (10.5/2)^{2)}l = √(9+ (110.25/4))

l = √(146.25/4)

l = √36.56

l = 6.05 m (approx.)

Surface area of cone= π × (105/20) × 6.05 m^{2}= (22/7) × (635.25/20) (taking π=22/7)

= 99.82 m^{2}

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