KVPY SA Exam 2012 Question Paper

Question : 9 of 80

Marks: +1, -0

In the figure, AHKF, FKDE and HBCK are unit squares; AD and BF intersect in X. Then the ratio of the areas of triangles AXF and ABF is

Solution:

Equation of BF
y = 2x
equation of AD
x + 2y = 4

x(4/5,8/5)

{\ar(\Δ\AFX)}/{\ar(\Δ\ABF)}

={1/2×1×(2-8/5)}/{1/2×1×2}=1/5

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