Question

A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 5 cm. Find the volume of the solid so obtained. If it is now revolved about the side 12 cm, then what would be the ratio of the volumes of the two solids obtained in two cases ?

Answer 1

Step-by-step explanation:

Left hypotenuse=13cm

Given triangle recolves around 12cm

Radius =5cm

Height =12cm

Volume =1/3PIxr^2xH

=1/3PIx5x5x12

Volume =100PI cm ^3 [nearly]

Case II (Second case)

R= 2cm

H= 5cm

Vol. = 1/3xPIx12x12x5

Vol. =240PIcm^3

Therefore, the ratio = 100/240

5:12

Answer 2

$\bold{Given,}$$~~~~$

$~~$A right triangle ABC with sides 5cm, 12cm and 13cm is revolved about the side 5cm.

$\rm\color{pink}{Case ~1:}$

$\star$ r = 12 cm

$\star$ h = 5 cm

$\star$ l = 13cm

Now, volume of the cone = πr2h/3

= {π(12)2 * 5}/3

= {π *12 * 12 * 5}/3

= π *12 * 4 * 5

= 240π

$\rm\color{pink}{Case 2:}$

$\star$ r = 5 cm

$\star$ h = 12 cm

$\star$ l = 13cm

$25\times 4$

$\underline\bold{100π}$

$\bold{Now,}$

$\bold{required ~ratio}$

=$\huge\frac{100π}{240π}$

=$\huge \frac{100}{240}$

= $\huge\frac{10}{24}$

= $\huge\frac{5}{12}$

$~~~~~~~~~~~~~~~~~~~~~$=$\underline\color{green}{ 5 : 12}$

Thankyou :)