UPSC 2014 CDS II Math Paper

Let the legs of the right triangle be 3k and 4k.
∴(3k2)+(4k2)=252
⇒25k2=625
⇒k2=25
⇒k=5
So, the sides of the right triangle are 15 cm, 20 cm and 25 cm.Let AE = x cm
∴BE=(25−x)cm
In right ΔACE,CE2=225−x2
In right ΔBCE,CE2=400−(25−x2)
∴225−x2=400−(25−x)2
⇒625+x2−50x−x2=175
⇒50x=450
⇒x=9cm
∴AE=9cm,BE=25−9=16cm and CE=√225−81=√144
=12cm
Volume of the double cone = Volume of cone ACD + Volume of cone BCD
=‌‌

1

3

π×CE2×(AE+BE)
=‌‌

1

3

π×CE2×AB
=‌‌

1

3

×3.14×144×25
=3768cm3