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KVPY2013PartIISolution

KVPY2013PartIISolution contains the solutions from 63 to 65 .

Part II (63-65)

63) Let ABCD be a square and let P be a point on the segment CD such that DP:PC = 1:2 . Let Q be a point on segment AP such that \(\angle BQP =90^{\circ}\). Then the ratio of the area of the quadrilateral PQBC to the area of ABCD is

A) \(\frac{31}{60}\)    B) \(\frac{37}{60}\)  C) \(\frac{39}{60}\  D)\(\frac{41}{60}\)

Solution:

Let ABCD is a square of side 3x.

P is a point on CD such that CP=2x and DP = x .

Q is a point on AP such that \(\angle AQB = 90^{\circ}\).

Let \(\angle DAP = \theta \) then we get \(\angleABQ =\theta\).

from triangle ADP sin\(\theta\)=




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