In the adjacent figure, the height of a solid cylinder is 10 cm and diameter is 7cm. Two equal conical holes of radius 3cm and height 4 cm are cut off as shown the figure. Find the volume of the remaining solid.
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Hi ,
i ) Dimensions of the solid cylinder :
Height ( H ) = 10 cm
Diameter ( D ) = 7cm
Radius ( R ) = D/2 = 7/2 = 3.5 cm
Volume of cylinder ( V1) = πR²H
= ( 22/7 ) × ( 3.5 ) × ( 3.5 ) × 10
= 385 cm³ -----( 1 )
ii ) Dimensions of the cone :
height ( h ) = 4 cm
radius ( r ) = 3 cm
Volume of cone ( V2 ) = ( πr²h)/3
= ( 22 × 3 × 3 × 4 )/7
= 37.71 cm³ ------( 2 )
According to the problem given ,
Volume of Remaining solid
= V1 - 2 × V2
= 385 cm³ - 2 × 37.71 cm³
=385 - 75.43
= 309.57 cm³
Therefore ,
Volume of the remaining solid
= 309.57 cm³
I hope this helps you.
: )
i ) Dimensions of the solid cylinder :
Height ( H ) = 10 cm
Diameter ( D ) = 7cm
Radius ( R ) = D/2 = 7/2 = 3.5 cm
Volume of cylinder ( V1) = πR²H
= ( 22/7 ) × ( 3.5 ) × ( 3.5 ) × 10
= 385 cm³ -----( 1 )
ii ) Dimensions of the cone :
height ( h ) = 4 cm
radius ( r ) = 3 cm
Volume of cone ( V2 ) = ( πr²h)/3
= ( 22 × 3 × 3 × 4 )/7
= 37.71 cm³ ------( 2 )
According to the problem given ,
Volume of Remaining solid
= V1 - 2 × V2
= 385 cm³ - 2 × 37.71 cm³
=385 - 75.43
= 309.57 cm³
Therefore ,
Volume of the remaining solid
= 309.57 cm³
I hope this helps you.
: )
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8
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