A right triangle ABC with sides 5cm, 12cm and 13cm is revolved about the side 12cm. Find the volume of the solid so obtained.

Hint: In this question first we observe that on revolving a right angle triangle about its height then a cone of same height and same radius is obtained. And then we apply the formula of the volume of the cone to get the answer.

Complete step-by-step answer:
ABC is a right angle triangle.
Let, hypotenuse(AC)=13cm
Perpendicular(AB)=12cm
Base(BC)=5cm
Now, when it is revolved about the side 12cm, the solid obtained is a cone of height 12cm and radius 5cm.
We know, Volume of cone$ = \dfrac{1}{3}\pi {r^2}h$ $\to$ (1)
Where,
 $
  r = {\text{ radius of cone}} \\
  h = {\text{height of cone}} \\
$
Put $r = 5{\text{ and h = 12 }}$in eq.1 we get








$ \Rightarrow $ Volume of cone$ = \dfrac{1}{3}\pi {r^2}h$
                                    $
   = \dfrac{1}{3}\pi {(5)^2} \times (12) \\
   = \pi \times (25) \times (4) \\
 $
Take $\pi = 3.14$
Then,
  Volume of cone = \[314c{m^3}\]
Hence, the volume of solid(cone) so obtained is 314 \[c{m^3}\].

Note: Whenever you get this type of question the key concept of solving the question is to have knowledge about different shapes and how they are formed from simple shapes like in this question a cone is formed when revolving about its height.