Question

An equilateral triangle with side a is revolved about one of its sides as axis. What is the volume of the solid of revolution thus obtained ?

Answer 1

When an equilateral triangle is revolved about one of its sides say BC then a double cone is generated, whose vertical radius = \(\frac{a}{2}\) cm and slant height = a cm

∴ Height of the cone = \(\sqrt{a^2-(\frac{a}{2})^2}\) = \(\sqrt{a^2-(\frac{a}{2})^2}=\sqrt{\frac{3a^2}{4}}=\frac{a}{2}\sqrt3\) cm

Hence, volume of the solid = 2 x \(\frac13\) πr²h = 2 x \(\frac13\) x π x \(\frac{a}{2}\) x \(\big(\frac{\sqrt3a}{2}\big)^2\)

= \(\frac23\) x π x \(\frac{a}{2}\) x \(\frac{3a^2}{4}\) = \(\frac{πa^3}{4}\) cm³.