Question

A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed.

In triangle ABC;

Answer 1

AC = AB + BC
Or, AC = 3 + 4
Or, AC = 9 + 16 = 25
Or, AC = 5 cm
In triangle ABC and triangle BDC;
$\angle ABC=\angle BDC$ (right angle)
$\angle BAC=\angle DBC$
Hence. $\Delta ABC\sim \Delta BDC$
So, we get following equations:
$\frac{AB}{AC}=\frac{BC}{BC}$
Or, $\frac{3}{5}=\frac{BD}{4}$
Or, $BD=\frac{3\times 4}{5}=2.4$
In triangle BDC;
DC = BC – BD
Or, DC = 4 – 2.4
Or, DC = 16 – 5.76 = 10.24
Or, DC = 3.2

From above calculations, we get following measurements for the double cone formed:
Upper Cone: r = 2.4 cm, l = 3 cm, h = 1.8 cm
Volume of cone
Lower Cone: r = 2.4 cm, l = 4 cm, h = 3.2 cm
Volume of cone
$=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\times 3.14\times {2.4}^2\times 1.8=10.85184c{m}^3$
Curved surface area of cone
$=\pi rl=3.14\times 2.4\times 3=22.608c{m}^2$
Lower Cone: r = 2.4 cm, l = 4 cm, h = 3.2 cm
Volume of cone
$=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\times 3.14\times {2.4}^2\times 3.2=19.2916c{m}^3$
Curved surface area of cone
$=\pi rl=3.14\times 2.4\times 4=30.144c{m}^2$
Total volume = 19.29216 + 10.85184 = 30.144 cm
Total surface area = 30.144 + 22.608 = 52.752 cm