Question

A right tringle having sides 15cm and 20cmis made torevolve about its hypotenuse.find the volume and surface area of the double cone so formed (use π=3.14

Answer 1

SEE THE ATTACHED IMAGE

BD is perpendicular to AC

BD is the radius of the double cone
Using Pythagoras theorem

AC2 =15² + 20²

AC2 = 225 + 400

AC2 = 625

AC = 25

If AD = x cm

CD = 25 – x

Using Pythagoras theorem in triangle ABD

x² + BD² = 15² ------ (i)

BD² = 20² – (25 – x)² ------- (ii)

Solving (i) and (ii)

x² + 20² – (25 – x)²  = 15²

x² + 400 – (625 – 50x + x²) = 15²

x² + 400 –  625 + 50x – x²² = 225

 50x = 450

x = 9 cm

BD² = 20² – (25 – x)²

BD² = 20² – (25 – 9)² 

BD² = 400 – 256 = 144

BD = 12

Volume of the double cone = upper volume + lower volume

Volume = 1/3 pi (BD)2 x(AD + CD)

Volume = 1/3 x 3.14 x 12 x 12 x 25

             = 1200 x 3.14= 3768 cm³


Surface Area

Surface area of double cone = upper S.A. + lower S.A.

S.A. = 3.14 x 12 x 15 + 3.14 x 12 x 20

= 1318.8 cm²