Let ABCD be a square and let P be point on segment CD such that DP : PC = 1 : 2. Let Q be a point on segment AP such that ∠BQP = 90°. Then the ratio of the area of quadrilateral PQBC to the area of the square ABCD is

Hello student,
Please find the answer to your question below

From the above figure
CP=2a/3 ,PD=a/3
LetPAD=So tan=(1/3) from triangleAPD
NowDAP=QBA=
Required ratio=(area of quadrilateral PQBC)/a²
=(a²-(area of ADP+area ofAQB))/a²
=1-()
=41/60

Another Method:
Let us take side length AB = BC = CD = AD = x
Let DP = a
So PC = 2a
ratio given
DP + PC = 2a + a = 3a = x
a = x/3 ….. (1)
Now in Triangle ADP and Triangle BQA
Angle ADP = BQA = 90
Angle DPA = QAB (Alternate interior angles)
Hence ADP and BQA are similar triangles …. (2)
Hence AP/AD = BQ/AB Basic Proportionality Theorem
But AB = AD = x
So AP = BQ
Traingle BQA and ADP are congruent triangles RHS Congruency … (3)
Area of QBAC = Area of ABCD – (Ar. triangle ABQ + Ar. triangle ADP) from diagram
= x²– 2(Area of triangle ADP) Congruent triangles (3) have same area
= x²- 2(1/2 * a * x)
= x² - ax
= x² - x²/3 as a=x/3 from (1)
= 2x²/3 …. (4)
Ratio of areas = (Area of PQBC)/( Area of ABCD)
= (2x²/3) / (x²)
= 2/3(approximate answer)