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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 â–³DEF 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 n2−n+1(n+1)2⋅Area(△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 n→c+→bn+1,→cn+1,n→bn+1 are respectively. So, −−→ED=−−→AD−−−→AE=(n−1)→c+→bn+1 and −−→EF=n→b−→cn+1 Now, vector area of triangle DEF=12(−−→EF×−−→ED) =12(n+1)2[(n→b−→c)×{(n−1)→c+→b}]=12(n+1)2[(n2−n)→b×→c+→b×→c]=n2−n+12(n+1)2(→b×→c)=n2−n+1(n+1)2⋅Area(△ABC)

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