A right triangle having sides 15 cm and 20 cm is made to revolve about its hypotenuse. Find the Volume and Surface Area of the double cone so formed. (Use π=3.14).
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let the hypotenuse be h
h=√15×15+20×20
=√ 225+400
=√625
= 25 cm
by applying Pythagoras theorem
x2+y2=15×15 and (25-x)^2+y2= 400
{(25-x)+y2}-{x2+y2}=400-225
x=9. ......(by solving)
put x=9 in x2+y2=225
81+y2=225
y2= 144
y= 12cm
now put the formula of cone =1/3πr2h+1/3πr2H
1/3π×144(16+9)
=1/3×3.14×144×25
=3768 cm3
TSA of dual cone = πrl+πrL
=3.14×12(20+15)
=3.14×420
=1318.8cm2
h=√15×15+20×20
=√ 225+400
=√625
= 25 cm
by applying Pythagoras theorem
x2+y2=15×15 and (25-x)^2+y2= 400
{(25-x)+y2}-{x2+y2}=400-225
x=9. ......(by solving)
put x=9 in x2+y2=225
81+y2=225
y2= 144
y= 12cm
now put the formula of cone =1/3πr2h+1/3πr2H
1/3π×144(16+9)
=1/3×3.14×144×25
=3768 cm3
TSA of dual cone = πrl+πrL
=3.14×12(20+15)
=3.14×420
=1318.8cm2
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