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Question
In the given figure AF = BF and DCBF is a parallelogram. If the area of ΔABC is 30 square units, find the area of the parallelogram DCBF.
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Solution
In ΔABC,
AF = FB and EF || BC ...(given)
∴ AE = EC ...(Converse of Midpoint theorem) ...(i)
In ΔAEF and ΔCED,
∠FEA = ∠DEC ...(Vertically opposite angles)
CE = AE ...[From (i)]
∠FAE = ∠DCE ...(Alternate angles)
∴ ΔFAE ≅ ΔCED ...( ASA test of congruency)
⇒ A(ΔAEF) = A(ΔCED) ....(ii)
A(ΔABC)
= A(ΔAEF) + A(EFBC)
= A(ΔCED) + A(EFBC) ....[From (ii)]
∴ A(ΔABC) = A(||gm DCBF)
⇒ A(||gm DCBF) = 30 sq. units.
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