# From a right circular cylinder with height 10 cm and radius of base 6 cm , a right circular cone of the same height and base is removed . Find the volume of the remaining solid.

Solution : Let $V_{1}$ and $V_{2}$ be the volumes of the right circular cylinder and cone respectively .

Then ,

$V_{1}=\frac{22}{7}\times6 \times6\times 10cm^3$                                 $[Using : V_{1}=\pi r^2h]$

and ,     $V_{2}=\frac{1}{3}\times \frac{22}{7}\times6 \times6 \times10 cm^3$                           $[Using :V_{2}=\frac{1}{3}\pi r^2h]$

$\therefore$           Volume of the remaining solid $=V_{1}-V_{2}$

$\Rightarrow$          Volume of the remaining solid $=(\frac{22}{7}\times6\times 6 \times 10- \frac{1}{3}\times \frac{22}{7}\times 6 \times 6 \times 10)cm^3$

$\Rightarrow$           Volume of the remaining solid $=754.28$$cm^3$

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