Q. 343.6( 19 Votes )

# A right triangle

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,

AC^{2} = AB^{2} + BC^{2}

AC^{2} = (15)^{2} + (20)^{2}

AC^{2} = 225 + 400

AC^{2} = 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,

AD^{2} + BD^{2} = AB^{2}

m^{2} + BD^{2} = 225

BD^{2} = 225 – m^{2} ……….(i)

∆CBD,

BD^{2} + CD^{2} = BC^{2}

BD^{2} + (25 – m)^{2} = (20)^{2}

BD^{2} = (20)^{2} – (25 – m)^{2} …..(ii)

By putting both the equations (i) and (ii) together,

225 – m^{2} = (20)^{2} – (625 – m)^{2}

225 – m^{2} = 400 – (625 + m^{2} – 50m) ……….by (a^{2} – b^{2})

225 – m^{2} = - 225 – m^{2} + 50m

225 – m^{2} + 225 + m^{2} = 50m

450 = 50m

So,

m = = 9 cm

BD^{2} = (15)^{2} – (9)^{2}

BD^{2} = 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 cm^{3} = 3,771.42 cm^{3}

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 cm^{2} = 1320 cm^{2}

So, the volume is 1200π cm^{3} and surface area is 420π cm^{2}, of the double cone so formed.

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