Question

Question 2
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.

Answer 1

In triangle ABC;
$A{C}^2=A{B}^2+B{C}^2$ (using pythagoras theorem)
$A{C}^2=3^2+4^2$
$A{C}^2=9+16=25$
$AC=5cm$
In triangle ABC and triangle BDC;
$\angle ABC=\angle BDC\left(rightangle\right)$
$\angle BAC=\angle DBC$
$Hence,\Delta ABCissimilarto\Delta BDC$
So, we get following equations:
$\frac{AB}{AC}=\frac{BD}{BC}$
$\frac{3}{5}=\frac{BD}{4}$
$BD=\frac{3\times 4}{5}=2.4cm$
In triangle BDC;
$D{C}^2=B{C}^2–B{D}^2$ (using pythagoras theorem)
$Or,D{C}^2=4^2–{2.4}^2$
$Or,D{C}^2=16–5.76=10.24$
$Or,DC=3.2cm$


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
$=\frac{1}{3}\pi r^{2}h$
$=\frac{1}{3}\times 3.14\times {2.4}^2\times 1.8$
$=10.85184c{m}^3$
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 $=19.29216+10.85184=30.144c{m}^3$
Total surface area $=30.144+22.608=52.752c{m}^2$