Q The side AB of a IIgm ABCD is produced to any point P A line through A - Maths - Areas of Parallelograms and Triangles

Question

Q The side AB of a IIgm ABCD is produced to any point P A line
through A and parallel to CP meets CB produced at Q and then IIgm
PBQR is completed Show that ar(IIgm ABCD) = ar(IIgm PBQR)

Lalit Mehra · Accepted Answer

Given, ABCD is a parallelogram and AC is the diagonal on the parallelogram.

We know that, the diagonal of a parallelogram divides it into two triangle having equal areas.

$∴ area (Δ ABC) = \frac{1}{2} area (| |^{gm} ABCD) ... (1)$

Now, PBQR is a parallelogram and PQ is the diagonal of the parallelogram.

$∴ area (Δ PBQ) = \frac{1}{2} area (| |^{gm} PBQR) ... (2)$

ΔAQC and ΔAQP are on the same base AQ and between the same parallel AQ and CP.

∴ area (ΔACQ) = area (ΔAQP) ( Triangles on the same base and between the same parallels have equal area)

⇒ area (ΔACQ) – (ΔABQ) = area (ΔAQP) – area (ΔABQ)

⇒ area (ΔABC) = area (ΔPBQ)

$\Rightarrow \frac{1}{2} area (| |^{gm} ABCD) = \frac{1}{2} area (| |^{gm} PBQR) [U \sin g (1) & (2)] \Rightarrow area (| |^{gm} ABCD) = area (| |^{gm} PBQR)$

67

Q. The side AB of a IIgm ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then IIgm PBQR is completed.Show that ar(IIgm ABCD) = ar(IIgm PBQR)