Question

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$<b>In the adjacent figure, the height of a solid cylinder is 10 cm and diameter is 7 cm. Two equal conical holes of radius 3 cm and height 4 cm are cut off as shown the figure. Find the volume of the remaining solid.$

Answer 1

Volume of cylinder = pie R^2h

= 22/7× 7/2× 7/2 × 10

= 385

Now 2 conical holes volume

= 2 × 1/3 pie r^2 h

= 2/3× 22/7× 3× 3× 4

=75.42

Now volume of remaining = 385 -75.42= 309.5 cm^3

Answer 2

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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

:)

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