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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

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

Grade:10

3 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
9 years ago
Hello student,
Please find the answer to your question below
240-1459_untitled.JPG
From the above figure
CP=2a/3 ,PD=a/3
Let\anglePAD=\phiSo tan\phi=(1/3) from triangle\DeltaAPD
Now\angleDAP=\angleQBA=\phi
Required ratio=(area of quadrilateral PQBC)/a2
=(a2-(area of \DeltaADP+area of\DeltaAQB))/a2
=1-(\frac{1}{6}+\frac{3}{20})
=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)
Varun
18 Points
9 years ago
Ans. 2/3
After drawing, 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 look at Triangle ADP and Triangle BQA
Angle ADP = BQA = 90o
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(Ar. 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
STAN007
11 Points
8 years ago
But sir in both above answer how  AP/AD = BQ/AB is possible. According to basic proportionality theorem AD/BQ=AP/AB

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