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?

1 Answer
Dec 29, 2017

V=1200pi " cm"^3

Explanation:

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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
enter image source here