# A right triangle

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

CD = (25 – m) cm

By using the Pythagoras theorem in ∆ABD and ∆CBD;

∆ABD,

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

(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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