Question

from a solid cylinder of height 14cm and base diameter 7cm two equal conical holes each of radius 2.1cm and height 4cm are hollowed out find the volum of remaining solid​

Answer 1

HEYA!

Volume of cylinder= π×r×r×h

=22/7×3.5×3.5×14

=539 sqcm

Volume of conical holes=1/3π×r×r×h

=1/3×22/7×2.1×2.1×4

=18.48 sqcm

There are two conical holes so

2×18.48

=36.96 sqcm

So remaining volume= 539 - 36.96

=502.04 sqcm

HOPE IT WORKS!!!

Answer 2

Answer:

502.04

Step-by-step explanation:

Volume of cylinder= π(r^2)(h)

=22/7(7/2)(7/2)(14)

=539 cm^3.

Volume of conical holes= 2[(1/3)(πr^2)(h)

=2[(1/3)(22/7)(2.1)(2.1)(4)

=2[18.48]

=36.96

Volume of remaining solid= Volume of cylinder- Volume of cone

=539 cm^3- 36.96cm^3

Volume of remaining solid=502.04 cm^3.