A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. Find the volume and the curved surface of the solid so formed.

Answer : Volume $=301.44\; cm^{3}$ and Curved surface area $=188.57\; cm^{2}$

A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm.

So on revolving we get a right circular cone as shown in the figure.

Radius = 6 cm, Height = 8 cm, Slant Height = 10 cm

Now, we know that Volume of cone$=\frac{1}{3}\pi r^{2}h$                    (where r is radius and h is height)

$V=\frac{1}{3}\times 3.14 \times \left ( 6 \right )^{2}\times 8$

$=\frac{1}{3}\times 3.14 \times 36 \times 8$

$=\frac{1}{3}\times 12 \times 8$

$=301.44\; cm^{3}$

Curved surface area $=\pi r l$                        (where r is radius and l is slant height)

$=\frac{22}{7}\times 6\times 10$

$=\frac{1320}{7}$

$=188.57\; cm^{2}$

Hence Volume $=301.44\; cm^{3}$ and curved surface area $=188.57\; cm^{2}$

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