In triangle ABC, if a point D divides BC in the ratio 2 : 5, show that - area (triangle ABD) : area (triangle ACD) = 2 : 5.

Answer :

In triangle ABC, if a point D divides BC in the ratio 2 : 5, , So

BD  =  2x And DC  =  5 x
And
Height of triangle  =  AE (  as AE is perpendicular on BC by construction )

And we know

Area of triangle  = 12 ×Base × Height

So ,
Area of ABD  = 12 ×BD × AE  = 12 ×2x × AE = 2  x AE2

And

Area of ACD  = 12 ×DC × AE  = 12 ×5x × AE = 5  x AE2

So,

Area of ABD  :  Area of ACD  = 2  x AE2 : 5  x AE2  , So

Area of ABD  :  Area of ACD =  2 :  5                                                           ( Hence proved )

  • 0
What are you looking for?