A right triangle whose sides are 15 cm and 20 cm (other than hypotenuse), is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate)

#### Solution

We have,

In ΔABC, ∠B = 90° , AB = l_{1 }=15 cm and BC = l_{2 }= 20 cm

Let OD = OB = r , AO = h_{1 }and CO = h_{2 }

Using Pythagoras therom,

`"AC" = sqrt("AB"^2 + "BC"^2)`

`=sqrt(15^2 + 20^2)`

`= sqrt(225 + 400)`

`= sqrt(625)`

⇒ h = 25 cm

As, ar(ΔABC)` = 1/2xxACxxBO=1/2xxABxxBC `

⇒ AC × BO = AB × BC

⇒ 25r = 15 × 20

`rArr r = (15xx20)/25`

⇒ r = 12 cm

Now,

Volume of the double cone so formed = Volume of cone 1 +Volume of cone 2

`= 1/3 pir^2h_1 + 1/3pir^2h_2`

`= 1/3 pir^2 ("h"_1 + "h"_2)`

`= 1/3pi"r"^2"h"`

`= 1/3xx3.14xx12xx12xx25`

= 3768 cm^{3 }

Also,

surace area of the solid so formed = CAS of cone 1 + CSA of cone 2

= πrl_{1} + πrl_{2}

= πr ( l_{1} + l_{2} )

`= 22/7xx12xx(15+20)`

`=22/7xx12xx35`

= 1320 cm^{2 }