# In the Below Figure, Abcd is Parallelogram. O is Any Point on Ac. Pq || Ab and Lm || Ad. Prove that Ar (||Gm Dlop) = Ar (||Gm Bmoq) - Mathematics

In the below figure, ABCD is parallelogram. O is any point on AC. PQ || AB and LM ||
AD. Prove that ar (||gm DLOP) = ar (||gm BMOQ)

#### Solution

Since, a diagonal of a parallelogram divides it into two triangles of equal area
∴ area (ΔADC) = area (ΔABC)

⇒ area (ΔAPO) + area (11gm DLOP) + area (ΔOLC)

⇒ area ( Δ AOM) + ar (11gm DLOP) + area (ΔOQC) ..... (1)

Since, AO and OC are diagonals of parallelograms AMOP and OQCL respectively.
∴ area   (ΔAPO) = area (ΔAMO)       ........ (2)

And, area  (ΔOLC)  = Area  (ΔOQC)        ......(3)
Subtracting (2) and (3) from (1), we get

Area  (11gm DLOP)  = area ( 11gm BMOQ)

Concept: Corollary: A rectangle and a parallelogram on the same base and between the same parallels are equal in area.
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 14 Areas of Parallelograms and Triangles
Exercise 14.3 | Q 27 | Page 48

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