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In triangle ABC, D and E are two points ...

In triangle ABC, D and E are two points on the sides AB and AC respectively so that DE II BC and AD/BD = 2/3.Then the area of trapezium DECB / the area of `Delta`ABC is equal to 1) 5/9 2) 21/25 3) 9/5 (4) 21/4

Text Solution

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As, (AD)/(BD)=2/3
So, `AD=2k, BD= 3k& AB=AD+BD=5k`
Since,`/_ADE and /_ABC` are Similar triangles(by using AA)
`(Ar(/_ ADE))/(Ar(/_ ABC))=((AD)/(BD))^2=((2k)/(5k))^2=(4k^2)/(25k^2)` So, `Ar(/_ADE)=4k^2 , Ar(/_ABC)=25k^2`
Hence,`(Ar(trep DECB))/(Ar(ABC))=(25k^2-4k^2)/(25k^2)=21/25`
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