ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects
CD at F.
(i) Prove that ar (ΔADF) = ar (ΔECF)
(ii) If the area of ΔDFB = 3 cm2, find the area of ||gm ABCD.
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Solution
In triangles and , ADF ECF we have
∠ADF = ∠ECF [Alternative interior angles, Since || AD BE]
AD = EC [ Since AD = BC = CE ]
And ∠DFA = ∠CFA [vertically opposite angles]
So, by AAS congruence criterion, we have
ΔADF ≅ ECF
⇒ area (ΔADF) = area (ΔECF) and DF = CF.
Now, DF = CF
⇒ BF is a madian in ΔBCD
⇒ area (ΔBCD) =2ar (ΔBDF)
⇒ area (ΔBCD) = 2× 3 cm2 =6cm2
Hence, ar (||gm ABCD) = 2ar (ΔBCD) 2 × 6cm2
=12cm2
Concept: Concept of Area
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