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Question

In a triangle ABC, points D,E and F are taken on the sides BC,CA and AB respectively, such that BDDC=CEEA=AFFB=n. Then Area of DEF is equal to

A
n2n+12(n+1)2Area(ABC)
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B
n2n+1(n+1)2Area(ABC)
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C
n2+n+12(n1)2Area(ABC)
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D
n2+n+1(n1)2Area(ABC)
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Solution

The correct option is B n2n+1(n+1)2Area(ABC)
Let A is the origin and the position vector of points B and C are b and c respectively.
As BDDC=CEEA=AFFB=n
By using section formula :
Position vector of D,E,F are nc+bn+1,cn+1,nbn+1 are respectively.
So, ED=ADAE=(n1)c+bn+1 and
EF=nbcn+1
Now, vector area of triangle DEF=12(EF×ED)
=12(n+1)2[(nbc)×{(n1)c+b}]=12(n+1)2[(n2n)b×c+b×c]=n2n+12(n+1)2(b×c)=n2n+1(n+1)2Area(ABC)

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