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 .