ABCD is a parallelogram in which BC is produced to P such that CP=BC,as shown in the adjoining figure.AP intersects CD at M.If ar(DMB)=7 $c m^{2}$ ,find the area of parallelogram ABCD.

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Solution

ANSWER:
In △MDA and △MCP,
∠DMA = ∠CMP
(Vertically opposite angles)
∠MDA = ∠MCP
(Alternate interior angles)
AD = CP
(Since AD = CB and CB = CP)
So, △MDA ≅ △MCP
(ASA congruency)
⇒DM = MC
(CPCT)
⇒M is the midpoint of DC
⇒BM is the median of △BDC
⇒ar(△BMC) = ar(△DMB) = $7 c m^{2}$
ar(△BMC) + ar(△DMB) = ar(△DBC) = 7 + 7 = $14 c m^{2}$
Area of parallelogram ABCD = 2 × ar(△DBC) = 2 × 14 = $28 c m^{2}$

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