Question

A right triangle, whose sides are $15cm$ and $20cm$ is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed $(Use\pi =3.14)$.

Answer 1

$Hypotenuse=$ sum of square of sides by phythagorus theorem.

$\therefore hypotenuse=\sqrt{{15}^2+{20}^2}=25cm$

By phythagorus theorem,

$A{E}^2+B{E}^2={15}^2\to \left(1\right)C{E}^2+B{E}^2={20}^2\to \left(2\right)AE+CE=25\to \left(3\right)$

Equation $\left(1\right)-\left(2\right),A{E}^2-C{E}^2=-175\to \left(4\right)$

Solving $\left(3\right)-\left(4\right),AE=9cm,CE=16cm,B{E}^2=144c{m}^2$

$\therefore$ Volume of total solid $=$ Volume of cone formed by ABED+Volume of cone formed by CBED

$=\frac{1}{3}\pi \times B{E}^2\times AE+\frac{1}{3}\pi \times B{E}^2\times CE=\frac{1}{3}\pi B{E}^2\times (AE+CE)=\frac{1}{3}\pi \times 144\times 25=3768unit{s}^3$

Volume of double cone formed is $3768c{m}^3$

Surface area of double cone$=CurvedsurfaceareaofABED+CurvedsurfaceareaofCBED=\pi \times BE\times BA+\pi \times BE\times BC=\pi \times BE\times (BA+BC)=3.14\times \sqrt{144}\times (15+20)=1318.8c{m}^2$

Surface area of required double cone is $1318.8c{m}^2$