Answer :

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.

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