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प्रश्न
O is any point on the diagonal PR of a parallelogram PQRS (Figure). Prove that ar (PSO) = ar (PQO).
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उत्तर
Given: In a parallelogram PQRS, O is any point on the diagonal PR.
To prove: ar (ΔPSO) = ar (ΔPQO)
Construction: Join SQ which intersect PR at B.
Proof: We know that, diagonals of a parallelogram bisect each other, so B is the mid-point of SQ.
Here, PB is a median of ΔQPS and we know that, a median of a triangle divides it into two triangles of equal area.
∴ ar (ΔBPQ) = ar (ΔBPS) ...(i)
Also, OB is the median of ΔOSQ.
∴ ar (ΔOBQ) = ar (ΔOBS) ...(ii)
On adding equations (i) and (ii), we get
ar (ΔBPQ) + ar (ΔOBQ) = ar (ΔBPS) + ar (ΔOBS)
⇒ ar (ΔPQO) = ar (ΔPSO)
Hence proved.
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संबंधित प्रश्न
In the given figure, ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).
In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram.
[Hint: From D and B, draw perpendiculars to AC.]
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[Hint : From A and C, draw perpendiculars to BD.]
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