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The base BC of the triangle ABC is divided at BD so that 3BD=BC.Prove that ar(triangle ABD)=1/2ar(triangle ADC).

In the triangle ABC 

(given) 3BD=BC

 ⇒BD= (1/3) BC

⇒(1/2)BD*AM=(1/3)BC*(1/2)AM  [ multiplying both sides by (1/2)AM ]

⇒Ar(Triangle ABD) = (1/3)Ar ( Triangle ABC)  ...... (1)

Now 

DC= (2/3) BC

(1/2)DC*AM=(2/3)(1/2)BC*AM

Ar (triangle ADC)=(2/3) Ar (Triangle ABC)  ...... (2)

From (1) and (2) we get

3 Ar ( triangle ABD)= (3/2) Ar ( triangle ADC)

⇒Ar ( triangle ABD)= (1/2) Ar ( triangle ADC)

Hence proved.

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Draw AM perpendicular to BC.

Ar.(ABC)= BC*AM/2--------(1)

Ar.(ABD)= BD*AM/2

or Ar.(ABD)= 1/3 (BC*AM/2) [Since BD=1/3rd of BC]

or Ar.(ABD)= Ar.(ABC)/3 [Using (1)]

Ar.(ABC)= 3 *Ar.(ABD)---------(2)

Ar.(ADC)= DC*AM/2

or Ar.(ADC)= 2/3(BC*AM/2) [Since DC= 2/3rd of BC]

or Ar.(ADC)= 2*Ar.(ABC)/3 [Using (1)]

or Ar.(ABC)= 3*Ar.(ADC)/2-----------(3)

From (2) and (3)

3*Ar.(ABD)= 3*Ar.(ADC)/2

or Ar.(ABD)= 1/2Ar.(ADC)

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