Last Activity: 10 Years ago
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 triangle
APD
NowDAP=
QBA=
Required ratio=(area of quadrilateral PQBC)/a2
=(a2-(area of ADP+area of
AQB))/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)
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