Question

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see the following figure). Find the relation between area (ABCD) and area (PBQR). [3 MARKS]


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Answer 1

Concept : 1 Mark
Application : 1 Mark
Proof : 1 Mark

Let us join AC and PQ.

$△$ACQ and $△$AQP are on the same base AQ and between the same parallels.

$\therefore$ ar ($△$ACQ) = ar ($△$APQ)

$\Rightarrow$ ar ($△$ACQ) − ar ($△$ABQ) = ar ($△$APQ) − ar ($△$ABQ)

$\Rightarrow$ ar ($△$ABC) = ar ($△$QBP) ... (1)

Since AC and PQ are diagonals of parallelograms ABCD and PBQR respectively,

$\therefore$ ar ($△$ABC) = $\frac{1}{2}$ ar (ABCD) ... (2)

ar ($△$QBP) = $\frac{1}{2}$ ar (PBQR) ... (3)

From equations (1), (2), and (3), we obtain

$\frac{1}{2}$ ar (ABCD) = $\frac{1}{2}$ ar (PBQR)

ar (ABCD) = ar (PBQR)