Question

8.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 area of the solid so formed(pi=3.14) *

Answer 1

hope it helps you..........

Answer 2

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

Then solid formed is a cone whose :

• Height of a cone (h) = 8 cm

• Radius of a cone (r) = 6 cm

• Slant height of a cone (l) = 10 cm

We know,

$\boxed{\bold{\sf{Volume \:of \:a \:cone = \dfrac{1}{3}\pi r^2 h}}}$

So,

$\implies\sf Volume= \dfrac{1}{3} \times \dfrac{22}{7} \times 6 \times 6 \times 8$

$\implies\sf Volume = \dfrac{6336}{21}$

$\implies\sf Volume= 301.7 \:cm^3$

$\boxed{\bold{\sf{Curved\:Surface\:Area \:of \:a \:cone =\pi rl}}}$

So,

$\sf\implies CSA=\dfrac{22}{7}\times 6\times10$

$\sf\implies CSA=\dfrac{1320}{7}$

$\sf\implies CSA=188.5\:cm^2$

Hence,

$\huge{\boxed{\sf Volume= 301.7 \:cm^3}}}$

$\huge{\boxed{\sf CSA= 188.5 \:cm^2}}}$