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 $\pi =22/7$)

Answer 1

Let $ABC$ be the right angled triangle such that $AB=15cm$ and $AC=20cm$

Using Pythagoras theorem we have
$\Rightarrow {BC}^2={AB}^2+{AC}^2$

$\Rightarrow {BC}^2={15}^2+{20}^2$

$\Rightarrow {BC}^2=225+400=625$

$\Rightarrow BC=25cm$

Let $OB=x$ and $OA=y$

Applying pythagoras theorem in triangles $OAB$ and $OAC$ we have

${AB}^2={OB}^2+{OA}^2;{AC}^2={OA}^2+{OC}^2$

$\Rightarrow {15}^2=x^2+y^2;{20}^2=y^2+{(25-x)}^2$

$\Rightarrow x^2+y^2=225;{(25-x)}^2+y^2=400$

$\{{(25-x)}^2+y^2\}-\{x^2+y^2\}=400-225$

$\Rightarrow {(25-x)}^2-x^2=175\Rightarrow x=9$

Putting $x=9$ in $x^2+y^2=225$ we get
$81+y^2=225\Rightarrow y^2=144\Rightarrow y=12$

Thus we have $OA=12cm$ and $OB=9cm$

Volume of the double cone = volume of cone CAA' + volume of cone BAA'
$=\frac{1}{3}\pi \times {\left(OA\right)}^2\times OC+\frac{1}{3}\pi \times {\left(OA\right)}^2\times OB=\frac{1}{3}\pi \times {12}^2\times 16+\frac{1}{3}\pi \times {12}^2\times 9$

$=\frac{1}{3}\times 3.14\times 144\times 253768{cm}^3$

Surface area of the double cone $=$ curved surface area of cone $CA{A}^{\prime }$ + curved surface area of cone $BA{A}^{\prime }$

$=\pi \times OA\times AC+\pi \times OA\times AB$

$=(\pi \times 12\times 20+\pi \times 12\times 15){cm}^2$

$=420\pi {cm}^2=420\times \pi =420\times 3.14$

$=1318.8{cm}^2$