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