Question

In the given figure $ PSDA $ is a parallelogram. Points $ Q $ and $ R $ are taken on $ PS $ such that $ PQ = QR = RS $ and $ PA\parallel QB\parallel RC $ . Prove that $ ar(\Delta PQE) = ar(\Delta CFD) $

Answer 1

Hint: Area of congruent triangles is equal. Use the properties of parallelogram to prove that the two triangles given in the question are congruent.

Complete step-by-step answer:
It is given that,
 $ PQ = QR = RS $ and $ PA\parallel QB\parallel RC $
 $ PSDA $ is a parallelogram
 $ \Rightarrow AP||SD $
And $ PS||AD $
We know that parts of parallel lines are also equal.
Therefore, $ RS||CD $
Also, $ RC||SD $ $ (\because AP||SD $ and $ AP||RC) $
Therefore, $ CDSR $ is a parallelogram
 $ \Rightarrow RS = CD $
In $ \Delta PQE $ and $ \Delta DCF $
 $ \angle QPE = \angle FDC $ (Alternate interior angles)
 $ PQ = DC $ $ (\because PQ = RS $ and $ RS = CD) $
 $ \angle PQE = \angle QRF $ (Corresponding angles between two parallel lines)
 $ \angle QRF = \angle FCD $ (Alternate interior angles)
 $ \Rightarrow \angle PQE = \angle FCD $
Therefore, by ASA criteria of congruence of triangles, we get
 $ \Delta PQE \cong \Delta DCF $
We know that the area of two congruent triangles is equal.
Therefore, it is proved that
 $ ar(\Delta PQE) = ar(\Delta CFD) $

Note: From this question, you can observe that it is not necessary to literally calculate the areas to prove them to be equal. We can also use the properties of triangles to prove that their area is equal. To solve this question of the questions of this type, you need to be able to connect the concepts of triangle properties with the properties of parallelogram.