question of the day
A right triangle whose sides are 15 cm and 20 cm is made to revolve about its
hypotenuse. Find the volume and surface area of the double cone so formed.
Hi Prakhar!
Consider the following right angled triangle ABC is rotated through its hypotenuse AC
BD ⊥ AC. In this case BD is the radius of the double cone generated.
Using Pythagoras theorem for ∆ABC it is obtained
AC2 = AB2 + BC2 = (15 cm)2 + (20 cm)2 = 225 cm2 + 400 cm2 = 625 cm2 = (25 cm)2
⇒ AC = 25 cm
Let AD = x cm
∴ CD = (25 – x) cm
Using Pythagoras theorem in ∆ABD and ∆CBD
AD2 +BD2 = AB2 and BD2 + CD2 = BC2
⇒ x2 + BD2 = 152 and BD2 + (25 – x)2 = 202
⇒ BD2 = 152 – x2 and BD2 = 202 – (25 – x)2
⇒ 152 – x2 = 202 – (25 – x)2
⇒ 225 – x2 = 400 – (625 + x2 – 50x)
⇒ 225 – x2 = – 225 – x2 + 50x
⇒ 50x = 450
⇒ x = 9
⇒ BD2 = 152 – 92 = 225 – 81 = 144
⇒ BD = 12 cm
Surface area of the double cone formed
= L.S.A of upper cone + L.S.A of the lower cone
= Π (BD) × (AB) + Π (BD) × BC
= Π × 12 cm × 15 cm + Π × 12 cm × 20 cm
= 420 Π cm2
Hope! This will help you.
Cheers!