A right triangle whose sides are 15 cm and 20 cm (Other than hypotenuse) , is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate)
Let name the triangle, ABC
Given,
Sides of the triangle are 15 and 20 cm,
BD ⟘ AC.
In the given case BD is the radius of the double cone generated by triangle by revolving.
Now by Pythagoras theorem,
AC2 = AB2 + BC2
AC2 = (15)2 + (20)2
AC2 = 225 + 400
AC2 = 625 = (25)2
AC = 25 cm
Let AD be m cm;
∴ CD = (25 – m) cm
By using the Pythagoras theorem in ∆ABD and ∆CBD;
∆ABD,
AD2 + BD2 = AB2
m2 + BD2 = 225
BD2 = 225 – m2 ……….(i)
∆CBD,
BD2 + CD2 = BC2
BD2 + (25 – m)2 = (20)2
BD2 = (20)2 – (25 – m)2 …..(ii)
By putting both the equations (i) and (ii) together,
225 – m2 = (20)2 – (625 – m)2
225 – m2 = 400 – (625 + m2 – 50m) ……….by (a2 – b2)
225 – m2 = - 225 – m2 + 50m
225 – m2 + 225 + m2 = 50m
450 = 50m
So,
m = = 9 cm
BD2 = (15)2 – (9)2
BD2 = 225 – 81 = 144 cm
BD = 12 cm
Radius of the generated double cone = 12 cm
Now,
Volume of the cone generated = Volume of the upper cone + Volume of the lower cone
(AD + CD)
(12)2× (25)
= 1200π cm3 = 3,771.42 cm3
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 × 12cm × 15 cm + π × 12 cm × 20 cm
= 420π cm2 = 1320 cm2
So, the volume is 1200π cm3 and surface area is 420π cm2, of the double cone so formed.