"Question 9 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)]
Class 9 - Math - Areas of Parallelograms and Triangles Page 163"
Answers
Triangles on the same base and between the same parallels are equal in area.
Diagonals of a parallelogram divides it into two Triangles of equal areas
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Given:
Two parallelograms ABCD and PBQR.
To show:
ar (ABCD) = ar (PBQR).Proof:
Join AC & PQ .
Now △ACQ & △APQ are on the same base AQ and between the
same parallel lines AQ and CP
ar(△ACQ) = ar(△APQ)
ar(△ACQ) – ar(△ABQ) = ar(△APQ) – ar(△ABQ)
[Subtracting ar(△ABQ) from both sides]
ar(△ABC) = ar(△QBP) — (i)
AC and QP are diagonals ABCD and PBQR.
Thus,
ar(ABC) = 1/2 ar(||gm ABCD) — (ii)
ar(QBP) = 1/2 ar(||gm PBQR) — (iii)
[Diagonals of a parallelogram divides it into two Triangles of equal areas]
From (ii) and (ii),
1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR)
ar(||gm ABCD) = ar(||gm PBQR)
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Hope this will help you...
Answer:
Step-by-step explanation:
