Question

A right triangle whose sides are 15 cm and 20 cm (other than hypotenuse), is made to revolve about its hypotenuse. Find the 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 ABB' = 20 cm
and slant height of cone BCB' = 15 cm

In $△ABC,\angle B={90}^o$

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

Let 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,

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

Subtitute the value in (i),

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

Rdaius of each cone, r = 12 cm

Surface area of double cone = surface area of ABB' + surface area of CBB'

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