Question

in the given figure QR||AB, RP||BD, CQ = x+2, QA= x, CP=5x +4, PD=3x ​

Answer 1

In △PXY and △PQR,XY is parallel to QR, so corresponding angles are equal.

∠PXY=∠PQR

∠PYX=∠PRQ

Hence, △PXY∼△PQR     [By AAsimilarity criterion]

PQPX=QRXY

⇒1+31=QRXY

⇒41=9XY

⇒XY=2.25 cm

We know that the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

ar(△PQR)ar(△PXY)=(PQPX)2

⇒ar(△PQR)A=161

⇒ar(△PQR)=16A cm2

Now, ar(trapeziumXYRQ)=(16A−A)cm2=15A cm2

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Answer 2

The value of x is 1.

Given:

QR||AB, RP||BD, CQ = x+2, QA= x, CP=5x +4, PD=3x ​

To Find:

The value x.

Solution:

Given that QR||AB and RP||BD

In ΔABC the line QR splits two sides of the triangle and QR is parallel to AB. Similarly, In ΔDBC the line RP splits two sides of the triangle and RP is parallel to BD.

By The fundamental theorem of similarity

$\frac{CQ}{QA} = \frac{CP}{PD}$

$\frac{x+2}{x} = \frac{5x+4}{3x} \\\\3x(x+2) = x(5x+4)\\\\3x^{2} + 6x = 5x^{2} +4x\\\\5x^{2} -3x^{2} = 6x-4x\\\\2x^{2} =2x\\\\x =1$

Therefore, the value of x is 1.

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