Question

A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone formed. (Choose value of $\pi$ as found appropriate).

Answer 1

The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure.
Hypotenuse $AC=\sqrt{3^2+4^2}$
$=\sqrt{25}=5cm$
Area of $\Delta ABC=\frac{1}{2}\times AB\times AC$
$\frac{1}{2}\times AC\times OB=\frac{1}{2}\times 4\times 3\frac{1}{2}\times 5\times OB=6OB=\frac{12}{5}=2.4cm$
Volume of double cone =Volume of cone 1 + Volume of cone 2
$=\frac{1}{3}\pi r^2{h}_1+\frac{1}{3}\pi r^2{h}_2=\frac{1}{3}\pi r^2(h_1+h_2)=\frac{1}{3}\pi r^2(OA+OC)=\frac{1}{3}\times 3.14\times \left(2.4{)}^4\right(5)$
$=30.14c{m}^3$
Surface area of the double cone =Surface area of cone 1 +Surface area of cone 2
$=\pi r{l}_1+\pi r{l}_2=\pi r[4+3]=3.14\times 2.4\times 7=52.75c{m}^2$