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# 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). Show that ar (ABCD) = ar (PBQR). - CBSE Class 9 - Mathematics

ConceptTriangles on the Same Base and Between the Same Parallels

#### 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). Show that

ar (ABCD) = ar (PBQR).

[Hint: Join AC and PQ. Now compare area (ACQ) and area (APQ)]

#### Solution

Let us join AC and PQ.

ΔACQ and ΔAQP are on the same base AQ and between the same parallels AQ and CP.

∴ Area (ΔACQ) = Area (ΔAPQ)

⇒ Area (ΔACQ) − Area (ΔABQ) = Area (ΔAPQ) − Area (ΔABQ)

⇒ Area (ΔABC) = Area (ΔQBP) ... (1)

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

∴ Area (ΔABC) = 1/2Area (ABCD) ... (2)

Area (ΔQBP) = 1/2Area (PBQR) ... (3)

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

1/2Area (ABCD) = 1/2Area (PBQR)

Area (ABCD) = Area (PBQR)

Is there an error in this question or solution?

#### APPEARS IN

NCERT Solution for Mathematics Class 9 (2018 to Current)
Chapter 9: Areas of Parallelograms and Triangles
Ex. 9.30 | Q: 9 | Page no. 163

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Solution 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). Show that ar (ABCD) = ar (PBQR). Concept: Triangles on the Same Base and Between the Same Parallels.
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