A right triangle whose sides are 15cm & 20cm is made to revolve about its hypotenuse, find the volume and the S.A. of the double cone so formed.
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Answer:
heya friend .♠
Here is ur answer.....✌
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Let ∆ AOB be the given right triangle in which <AOB is rotated about the hypotenuse AB ,the two cones generated are AO'O And BOO'..
Clearly ,OA=15cm, OB=20cm.
in right -angled ∆ AOB, we have
AB=√15^2+20^2
AB=225+400
AB=√625
AB=25.
let OP =x cm
then
1/2*OA*OB=1/2*AB*OP
=>1/2*15*20=1/2*25*OP
=>150=25x
=>x=6cm
in right -angled ∆APO,we have
AP=√(OA)^2-(OP)^2
=√(15)^2-(6)^2
=√225-36
=√289
AP=17cm
BP=AB-AP=(25-17)=8cm
now ,•°•radius of each of the cones AO'O and BO'O,r =OP=6cm
Height of the cone AO'O ,H=AP=17cm
Height of the cone BOO' h=8cm.
volume of the double cone
=volume of the cone AO'O+bole of the cone BOO'H
=1/3πr^2h+1/3πr^2H
=1/3*3.14*(6)^2*8+ 1/3*(3.14*(6)^2*17
=>1/3*3.14(288+612)
=>1/3*3.14*900
=>942cm^3 Ans..
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hope it help you.
@SNEHA[DORY]☺☺
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