__Exercise 13.7 Page No: 233__

__Exercise 13.7 Page No: 233__

**1. Find the volume of the right circular cone with**

**(i) radius 6cm, height 7 cm (ii) radius 3.5 cm, height 12 cm (Assume π = 22/7)**

**Solution:**

Volume of cone = (1/3) πr^{2}h cube units

Where r be radius and h be the height of the cone

(i) Radius of cone, r = 6 cm

Height of cone, h = 7cm

Say, V be the volume of the cone, we have

V = (1/3)×(22/7)×36×7

= (12×22)

= 264

The volume of the cone is 264 cm^{3}.

(ii) Radius of cone, r = 3.5cm

Height of cone, h = 12cm

Volume of cone = (1/3)×(22/7)×3.5^{2}×7 = 154

Hence,

The volume of the cone is 154 cm^{3}.

**2. Find the capacity in litres of a conical vessel with**

**(i) radius 7cm, slant height 25 cm (ii) height 12 cm, slant height 12 cm**

**(Assume π = 22/7)**

**Solution:**

(i) Radius of cone, r =7 cm

Slant height of cone, l = 25 cm

or h = 24

Height of the cone is 24 cm

Now,

Volume of cone, V = (1/3) πr^{2}h (formula)

V = (1/3)×(22/7) ×7^{2}×24)

= (154×8)

= 1232

So, the volume of the vessel is 1232 cm^{3}

Therefore, capacity of the conical vessel = (1232/1000) liters (because 1L = 1000 cm^{3})

= 1.232 Liters.

(ii) Height of cone, h = 12 cm

Slant height of cone, l = 13 cm

r = 5

Hence, the radius of cone is 5 cm.

Now, Volume of cone, V = (1/3)πr^{2}h

V = (1/3)×(22/7)×52×12 cm^{3}

= 2200/7

Volume of cone is 2200/7 cm^{3}

Now, Capacity of the conical vessel= 2200/7000 litres (1L = 1000 cm^{3})

= 11/35 litres

**3. The height of a cone is 15cm. If its volume is 1570cm ^{3}, find the diameter of its base. (Use π = 3.14)**

**Solution:**

Height of the cone, h = 15 cm

Volume of cone =1570 cm^{3}

Let r be the radius of the cone

As we know: Volume of cone, V = (1/3) πr^{2}h

So, (1/3) πr^{2}h = 1570

(1/3)×3.14×r^{2 }×15 = 1570

r^{2} = 100

r = 10

Radius of the base of cone 10 cm.

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

**Solution:**

Height of cone, h = 9cm

Volume of cone =48π cm^{3}

Let r be the radius of the cone.

As we know: Volume of cone, V = (1/3) πr^{2}h

So, 1/3 π r^{2}(9) = 48 π

r^{2} = 16

r = 4

Radius of cone is 4 cm.

So, diameter = 2×Radius = 8

Thus, diameter of base is 8cm.

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

**(Assume π = 22/7)**

**Solution:**

Diameter of conical pit = 3.5 m

Radius of conical pit, r = diameter/ 2 = (3.5/2)m = 1.75m

Height of pit, h = Depth of pit = 12m

Volume of cone, V = (1/3) πr^{2}h

V = (1/3)×(22/7) ×(1.75)^{2}×12 = 38.5

Volume of cone is 38.5 m^{3}

Hence, capacity of the pit = (38.5×1) kiloliters = 38.5 kiloliters.

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

**(i) height of the cone**

**(ii) slant height of the cone**

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

**(Assume π = 22/7)**

**Solution:**

Volume of a right circular cone = 9856 cm^{3}

Diameter of the base = 28 cm

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

Let the height of the cone be h

Volume of cone, V = (1/3) πr^{2}h

(1/3) πr^{2}h = 9856

(1/3)×(22/7) ×14×14×h = 9856

h = 48

The height of the cone is 48 cm.

Slant height of the cone is 50 cm.

(iii) curved surface area of cone = πrl

= (22/7)×14×50

= 2200

curved surface area of the cone is 2200 cm^{2}.

**7. A right triangle ABC with sides 5cm, 12cm and 13cm is revolved about the side 12 cm. Find the volume of the solid so obtained.**

**Solution:**

Height (h)= 12 cm

Radius (r) = 5 cm, and

Slant height (l) = 13 cm

Volume of cone, V = (1/3) πr^{2}h

V = (1/3)×π×5^{2}×12)

= 100π

Volume of the cone so formed is 100π cm^{3}.

**8. If the triangle ABC in the Question 7 is revolved about the side 5cm, then find the volume of the solids so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.**

**Solution:**

A right-angled ΔABC is revolved about its side 5cm, a cone will be formed of radius as 12 cm, height as 5 cm, and slant height as 13 cm.

Volume of cone = (1/3) πr^{2}h; where r is the radius and h be the height of cone

= (1/3)×π×12×12×5

= 240 π

The volume of the cones of formed is 240π cm^{3}.

So, required ratio = (result of question 7) / (result of question 8) = (100π)/(240π) = 5/12 = 5:12.

**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.**

**(Assume π = 22/7)**

**Solution:**

Radius (r) of heap = (10.5/2) m = 5.25

Height (h) of heap = 3m

Volume of heap = (1/3)πr^{2}h

= (1/3)×(22/7)×5.25×5.25×3

= 86.625

The volume of the heap of wheat is 86.625 m^{3}.

Again,

= (22/7)×5.25×6.05

= 99.825

Therefore, the area of the canvas is 99.825 m^{2}.