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MATHMAA

# 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$$\theta$$=