**Question 1 **:

Radius and slant height of a cone are 20 cm and 29 cm respectively. Find its volume.

**Solution :**

Radius of the cone (r) = 20 cm

Slant height of the cone (l) = 29 cm

l^{2} = r^{2}+h^{2}

29^{2} = 20^{2} + h^{2}

841 = 400 + h^{2}

h^{2 } = 841-400

h^{2} = 441

h = √(21 21

h = 21 cm

Volume of the cone = (1/3) Π r^{2} h

= (1/3) ⋅ (22/7) ⋅ (20)^{2 }⋅ 21

= 8800 cm^{3 }

Volume of the cone = 8800 cm^{3}

**Question 2 :**

The circumference of the base of a 12 m high wooden solid cone is 44 m. Find the volume.

**Solution :**

Circumference of cone = 44 m

Height of the cone (h) = 12 m

2Πr = 44

2^{ }⋅ (22/7) ⋅ r = 44

r = 44 ⋅ (1/2)^{ }⋅ (7/22)

r = 7 cm

Volume of the cone = (1/3) Π r² h

= (1/3) ⋅ (22/7) ⋅ 7^{2 }⋅ 12

= (1/3)^{ }⋅ (22/7)^{ }⋅ 7^{ }⋅ 7 ⋅ 12

= 616 cm^{3 }

Volume of the cone = 616 cm^{3 }

**Question 3 :**

A vessel is in the form of frustum of a cone. Its radius at one end and the height are 8 cm and 14 cm respectively. If its volume is 5676/3 cm^{3}, then find the radius at the other end.

**Solution :**

Volume of the frustum cone = (5676/3) cm^{3}

Let r be the required radius

Radius (R) = 8 cm

height (h) = 14 cm

(1/3) Π h (R^{2}+r^{2}+R r) = (5676/3)

(1/3)^{ }⋅ (22/7)^{ }⋅ (14) (8^{2}+ r^{2}+8r) = 5676/3

r^{2}+8r+64 = 129

r^{2}+ 8r+64-29 = 0

r^{2}+8r-65 = 0

(r+13) (r-5) = 0

r = -13, r = 5 cm

So, the required radius = 5 cm

**Question 4 **:

The perimeter of the ends of a frustum of a cone are 44 cm and 8.4 Π cm. If the depth is 14 cm, then find its volume.

**Solution :**

Perimeter of upper end = 44 cm

Perimeter of lower end = 8.4 Π cm

Height of frustum cone = 14 cm

Now we have to find the volume of frustum cone

Volume of the frustum cone = (1/3) Π h (R^{2}+r^{2}+R r)

2ΠR = 44

2 ⋅ (22/7) ⋅ R = 44

R = 44 ⋅ (1/2) ⋅ (7/22)

R = 2 ⋅ (1/2) ⋅ 7

R = 7

2Πr = 8.4 Π

r = 8.4 Π ⋅ (1/2Π)

r = 4.2

Volume of the frustum cone

= (1/3) ⋅ (22/7) ⋅ 14 (7^{2}+4.2^{2}+7(4.2))

= (44/3) (49+29.4+17.64)

= (44/3) (96.04)

= (44) (32.013)

= 1408.57 cm^{3}

Volume of the frustum cone = 1408.57 cm^{3}

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