Question

Please solve this?

A right angled triangle whose sides are 15 cm and 20 cm , is made to revolve about its hypotenuse. What will be the volume of the cone so formed?

Answer 1

`V=1200pi " cm"^3`

Explanation:

See Fig 1.
`AC=sqrt(AB^2+BC^2)=sqrt(20^2+15^2)=25`,
Draw a line `BO` perpendicular to `AC`,
Area of `DeltaABC=1/2*20*15=1/2*BO*25`
`=> BO=(20*15)/25=12`
`=> AO=sqrt(20^2-12^2)=16, => OC=25-16=9`
As shown in Fig 2,
after revolving about the hypotenuse `AC`, the sides `AB and BC` become two right circular cones, namely, cone 1 and cone 2, respectively.
volume of a cone is given by `V=1/3pir^2h`
where `r` is the radius of the base and `h` is the height of the cone.
Now you have 2 cones, cone 1 and cone 2.
cone 1 : `h_1=16, r=12`,
cone 2 : `h_2=9, r=12`,
let `V_1, V_2` be the volume of cone 1 and cone 2, respectively.
total volume of the cones `V=V_1+V_2`
`=> V=1/3*pi*12^2*16+1/3*pi*12^2*9`
`= 1/3*pi*12^2*(16+9)=1200pi " cm"^3`