WORKSHEET ON VOLUME OF CONE

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

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

(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 cm³, then find the radius at the other end

(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.

(5)  A right angled triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the fixed side of 12 cm. Find the volume of the solid generated.

(6)  The radius and height of a right circular cone are in the ratio 2:3. Find the slant height if its volume is 100.48 cu.cm (take Π = 3.14)

(7)  The volume of a cone with circular base is 216 Π cu.cm. If the base radius is 9 cm, then find the height of the cone.

Detailed Answer Key

(1)  Solution :

Radius of the cone (r)  =  20 cm

Slant height of the cone (l)  =  29 cm

l²  =  r²+h²

29²  =  20² + h²

841 = 400 + h²

h²  =  841-400

 h²  =  441

 h  =  √(21 21

 h  =  21 cm

Volume of the cone  =  (1/3) Π r² h

=   (1/3) ⋅ (22/7) ⋅ (20)² ⋅ 21

=  8800 cm³ 

Volume of the cone = 8800 cm³

(2)  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) ⋅ 7² ⋅ 12

=  (1/3) ⋅ (22/7) ⋅ 7 ⋅ 7 ⋅ 12

=  616 cm³ 

Volume of the cone  =  616 cm³ 

(3)  Solution :

Volume of the frustum cone  =  (5676/3) cm³

Let r be the required radius

Radius (R)  =  8 cm

height (h)  =  14 cm

(1/3) Π h (R²+r²+R r) = (5676/3)

(1/3) ⋅ (22/7) ⋅ (14) (8²+ r²+8r) = 5676/3

r²+8r+64  =  129

r²+ 8r+64-29  =  0

r²+8r-65  =  0

(r+13) (r-5)  =  0

r  =  -13, r  =  5 cm

So, the required radius = 5 cm

(4)  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 (R²+r²+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 (7²+4.2²+7(4.2))

=  (44/3) (49+29.4+17.64)

=  (44/3) (96.04)

=  (44) (32.013)

=  1408.57 cm³

Volume of the frustum cone =  1408.57 cm³

(5)  Solution :

In any right triangle the longer side must be hypotenuse side. The longer side of the given sides is 13 cm.

So it must be hypotenuse side of the triangle.

From the diagram we know that slant height is 13 cm, radius is 5 cm and height is 12 cm.

l  =  13 cm, r  =  5 cm and h = 12 cm

Volume of cone   =   (1/3)Πr²h

=  (1/3) ⋅ (22/7) ⋅ 5²⋅12

=  (22/7) ⋅ 5 ⋅ 5 ⋅ 4

=  314.29 cm³

Volume of cone  =  314.29 cm³

(6)  Solution :

Radius and height of right circular cone are in the ratio

2 : 3

                                    r : h  =  2 : 3

                                    r/h  =  2/3

                                       r  =  2h/3

Volume of cone  =  100.48 cu.cm

(1/3)Πr² h  =  100.48

(1/3) (3.14) (2h/3)² h  =  100.48

(1/3) (3.14) (4h²/9) h  =  100.48

h³ = 100.48 ⋅ (3/1) ⋅ (1/3.14) ⋅ (9/4)

h³ = 8 ⋅ 27

h  =  6 cm

r  =  4 cm 

l²  =  r²+h²

l²  =  4²+6²

l  =  √52

l  =  2√13 cm

So, slant height of the cone is 2√13 cm.

(7)  Solution :

Volume of the cone  =  216Π cu.cm

Radius of the cone  =  9 cm

(1/3) Π r² h  =  216Π

9² h  =  216 ⋅ 3

h  =  216 ⋅ 3 ⋅ (1/9) (1/9)

h  =  24/3

h  =  8 cm

Required height of cone is 8 cm.

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