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Volume of a frustum= \[\dfrac{1}{3}\pi h\left( {{r^2} + rR + {R^2}} \right)\]

Given that,

Width of each ring =3cm

Radius of uppermost ring (r) =4cm

Radius of lowermost ring (R) =10cm

Volume of a frustum= \[\dfrac{1}{3}\pi h\left( {{r^2} + rR + {R^2}} \right)\]

\[

= \dfrac{1}{3} \times \dfrac{{22}}{7} \times 21\left( {{4^2} + 4 \times 10 + {{10}^2}} \right) \\

= 22\left( {16 + 40 + 100} \right) \\

= 22 \times 156 \\

= 3432c{m^3} \\

\]

The only point where a student can get confused is slant height and normal height of frustum. Here they have asked for volume and for that we need normal height h. This height is the perpendicular distance between the cut plane and the base.

Slant height is written as l.