Question

A right-angled triangle whose sides are 15 cm and 20 cm, is made to revolve about its hypotenuse. Find the volume and the surface area of the double cone so formed. [Take $\pi \simeq 3.14]$

• A. 3878 $c{m}^3$,1315.8$c{m}^2$
• B. 3777 $c{m}^3$,1312.8 $c{m}^2$
• C. 3788 $c{m}^3$ ,1310.8 $c{m}^2$
• D. 3768 $c{m}^3$,1318.8 $c{m}^2$

Answer 1

The correct option is D 3768 $c{m}^3$,1318.8 $c{m}^2$

Let ABC be the right triangle, right angled triangle at A, whose sides AB and AC measures 6 cm and 8 cm respectively.
$\therefore$ Hypotenuse $BC=\sqrt{(15{)}^2+(20{)}^2}$
$=\sqrt{225}=25cm$
AO(A'O) is the radius of the common base of the double cone fanned by revolving the triangle around BC.
Area of $\Delta ABC=\frac{1}{2}\times 15\times 20=\frac{1}{2}BC\times AO$.
$\Rightarrow 150=\frac{1}{2}\times 25\times AO$
$\Rightarrow AO=12cm$
Volume of the double cone
$=\frac{1}{3}\pi \times A{O}^2\times BO+\frac{1}{3}\pi \times A{O}^2\times CO$
$=\frac{1}{3}\pi \times A{O}^2(BO+CO)$
$=\frac{1}{3}\pi \times A{O}^2\times BC$
$=\frac{1}{3}\times 3.14\times 12\times 12\times 25=3768c{m}^3$
Surface area of the double cone
$=\pi \times AO\times AB+\pi \times AO\times AC$
$=\pi \times AO(AB+AC)$
$=3.14\times 12\times (15+20)$
$=3.14\times 12\times 35=1318.8c{m}^2$