Question

A right triangle whose sides are 15 cm and 20 cm (other than hypotenuse), is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of $\pi$ as found appropriate)

Answer 1

When a right angled triangle revolves around its hypotenuse, a double cone is formed.

Slant height of cone ABB' = 20 cm
and slant height of cone BCB' = 15 cm

In $△ABC,\angle B={90}^{\circ }$

$AC=\sqrt{{20}^2+15}=\sqrt{400+225}=\sqrt{625}=25cm$

In $△ABC,AO=xcm$

$B{O}^2=A{B}^2-A{O}^2=400-x^2-----\left(i\right)$

In $△BOC,OC=AC-AO=25-x\Rightarrow B{O}^2=B{C}^2-O{C}^2=225-(25-x{)}^2-----(ii)$

From (i) and (ii), we get

$400-x^2=225-(25-x{)}^{2}400-x^2=225-(625-50x+x^2)400-x^2=225-625+50x-x^{2}50x=800x=\frac{800}{50}=16$

Substitute the value in (i), we get

$B{O}^2=400-(16{)}^2=400-256=144B{O}^2=144\Rightarrow BO=\sqrt{144}=12cm$

Radius of each cone, r = 12 cm

The volume of double cone = Volume of ABB' + Volume of BCB'

$=\frac{1}{3}\pi B{O}^2\times AO+\frac{1}{3}\pi B{O}^2\times CO$

$=\frac{1}{3}\times \frac{22}{7}\times B{O}^2[AO+CO]$

$=\frac{1}{3}\times {12}^2\times 25\times \frac{22}{7}$

$=12\times 4\times 25\times \frac{22}{7}=3771.42c{m}^3$

Surface area of the double cone = CSA of ABB' + CSA of BCB'

$=\pi r\times AB+\pi r\times BC=\pi r(AB+BC)=3.14\times 12\times (20+15)=3.14\times 12\times 35=1318.8c{m}^2$