Question

A right triangle With sides 3cm,4cm and 5cm is revolved around its hypotenuse. Find volume of double cone thus generated.

Answer 1

area of triangle ABC = [tex] \frac{1}{2}*base *height [/tex]
                              = $\frac{1}{2}*4cm *3cm$
                              = 6cm^2
again ara of  triangle = $\frac{1}{2} * AC * BO$
                           [tex]6cm = \frac{1}{2} *5*BO [/tex]
            therefore BO = $6cm* \frac{2}{5}$
                               = 12/5
area of both cones = $\frac{1}{3} \pi r^2 AO + \frac{1}{3} \pi r^2 CO$
                           = $\frac{1}{3} \pi r^2( AO + CO )$
                           = $\frac{1}{3} * \frac{22}{7} * \frac{12}{5} \frac{12}{5} *AC$
                            = $\frac{1}{3} * \frac{22}{7} * \frac{12}{5} \frac{12}{5} *5$
                            = $30.17 cm^3$

Answer 2

By using Pythagoras theorem.

AB² + AC² = BC²
9 + 16 = BC²
5 cm = BC

Now let OB be x cm 
Then OC = BC - x = 5 - x cm

Now in triangle AOB.
OB² + OA² = AB²
OA² + x² = 9
OA² = 9 - x² --------------------------------------1

Similarly in triangle AOC
OA² + OC² = AC²
OA² = 16 - ( 5 - x )² ---------------------------2

From 1 and 2

9 - x²  = 16 - ( 5 - x )²
We will get 
OB =  x = 9/5

And OC = ( 5 - 9/5 ) = 16/5

From 1 we will get ,
OA² = 9 - (9/5)²  = 144/25
OA = 12/5

Now ,

Volume of double cone = volume of smaller cone + volume of bigger cone
                                  = 1/3 π (OA)² × OB + 1/3 π (OA)² × OC
                                  = 1/3 π (OA)² × (OB + OC)
                                  = 1/3 π (OA)² × (BC)
                                  = 1/3 × 22/7 × (12/5)² × 5
                                  = 1056/35
                                  = 30.17 cm³