VOLUME OF 3D SHAPES EXAMPLES

Volume  =  Πr2h	Volume  =  Πh(R2 - r2)
Volume  =  (1/3) Πr2h	Volume  =  (4/3) Πr3 Volume of hollow sphere   =  (4/3) Π(R3 - r3)
Volume  =  (2/3) Πr3	Volume = (2/3)Π(R3 - r3)
	Volume = (1/3)Πh(R2 + r2 + Rr)

Example 1 :

A 14 m deep well with inner diameter 10 m is dug and the earth taken out is evenly spread all around the well to form an embankment of width 5 m. Find the height of the embankment.

Solution :

Given radius of the well = 5 m

Height of the well = 14 m

Width of the embankment = 5 m

Radius of the embankment = 5 + 5 = 10 m

Let h be the height of the embankment

Hence the volume of the embankment = volume of the well 

That is,  π(R² – r²)h = πr²h

 (10² – 5²)h = 5²(14)

(100 - 25)h  =  25(14)

h  =  25(14)/75

h  =  14/3

h = 4.67 m

Hence the height of the embankment is 4.67 m.

Example 2 :

A cylindrical glass with diameter 20 cm has water to a height of 9 cm. A small cylindrical metal of radius 5 cm and height 4 cm is immersed it completely. Calculate the raise of the water in the glass?

Solution :

Measurement of glass :

Radius of glass  =  10 cm

height of glass  =  h

Measurement of small metal :

Radius of metal  =  5 cm

height of glass  =  4 cm

The quantity of water raised in the cylindrical glass  =  volume of small cylindrical metal

 πr²h  =   πr²h

 10² h  =  5² (4)

  h  =  100/100

  h  =  1 cm

Example 3 :

If the circumference of a conical wooden piece is 484 cm then find its volume when its height is 105 cm.

Solution :

Circumference of piece  =  484 cm

2 π r  =  484

2 (22/7) r  =  484

r  =  484 (7/22) (1/2)

  r  =  77

Volume of cone  =  (1/3) πr²h

=  (1/3)⋅ (22/7) ⋅ 77² ⋅ (105)

=  652190 cm³

Example 4 :

A conical container is fully filled with petrol. The radius is 10m and the height is 15 m. If the container can release the petrol through its bottom at the rate of 25 cu.meter per minute, in how many minutes the container will be emptied. Round off your answer to the nearest minute.

Solution :

Volume of petrol in the container  =  (1/3) πr²h

Radius  =  10 m, height  =  15 m

Releasing rate  =  25 cu.meter

Number of minutes  =  (1/3) π10²(15) / 25

  =  (1/3) (22/7) 10²(15) / 25

  =  62.85

  =  63 minutes

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