Question

A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained.

Hint:

r = 5 cm
l = 13 cm
h = 12 cm
Volume = $\frac 13 \pi r^2 h$
= $\frac 13 \pi \times 5^2 \times 12$
= $100\pi$ cm³

r = 12 cm
l = 13 cm
h = 5 cm
Volume = $\frac 13 \pi r^2 h$
= $\frac 13 \pi \times {12}^2 \times 5$
= $240\pi$ cm³

Ratio = $\frac {100\pi}{240\pi}$ = $\frac {10}{24}$ =$\frac {5}{12}$= $5:12$

Answer 1

When right-angled $∆$ ABC is revolved about its side 5 cm, a cone will be formed having radius (r) as 12 cm, height (h) as 5 cm, and slant height (l) as 13 cm. 

Volume of cone= $\frac{1}{3}\pi$r²h
= ($\frac{1}{3}$) × $\pi$× 12cm × 12cm × 5cm
= 240$\pi$ cm³
Volume of the cone = 100$\pi$ cm

Required ratio = 100$\pi$ : 240$\pi$ 
= 5 :12

Therefore, the volume of the cone so formed is 240$\pi$ cm³ .