The curved surface area of a right circular cone is \(\sqrt{10}\) times the area of its base. Show that the height of the cone is three times the radius of its base.

Q.The curved surface area of a right circular cone is \(\sqrt{10}\) times the area of its base. Show that the height of the cone is three times the radius of its base.

Let the height of the cone be \(h\) units and the radius be \(r\) units. ∴ According to the question, \[ πrl = \sqrt{10} × πr^2 \] ⇒ \(l = \sqrt{10} r\) ⇒ \(\sqrt{h^2 + r^2} = \sqrt{10} r\) ⇒ \(h^2 + r^2 = 10r^2\) ⇒ \(h^2 = 9r^2\) ⇒ \(\frac{h^2}{r^2} = \frac{9}{1}\) ⇒ \(\frac{h}{r} = \frac{3}{1}\) ⇒ \(h = 3r\) ∴ The height of the cone is three times the radius of its base. (Proved)