A right triangle whose sides are 15 cm and 20 cm is made to revolve about its hypoteneous.Find the volume and the surface area of the double cone so formed. (Take pie = 3.14

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

AC

^{2}= AB^{2}+ BC^{2}= (15 cm)^{2}+ (20 cm)^{2}= 225 cm^{2}+ 400 cm^{2}= 625 cm^{2}= (25 cm)2⇒ AC = 25 cm

Let AD =

*x*cm∴CD = (25 –

*x*) cmUsing Pythagoras theorem in ∆ABD and ∆CBD

AD

^{2}+BD^{2}= AB^{2}and BD^{2}+ CD^{2}= BC^{2}⇒

*x*^{2}+ BD^{2}= 152 and BD^{2}+ (25 –*x*)^{2}= 20^{2}⇒ BD

^{2}= 152 – x^{2}and BD^{2}= 20^{2}– (25 – x)^{2}⇒ 152 –

*x*^{2}= 202 – (25 –*x*)^{2}⇒ 225 –

*x*^{2}= 400 – (625 +*x*^{2}– 50*x*)⇒ 225 –

*x*^{2}= – 225 –*x*^{2}+ 50*x*⇒ 50

*x*= 450⇒

*x*= 9⇒ BD

^{2}= 15^{2}– 9^{2}= 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 cm

^{2}

Hope! This will help you.