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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. - Mathematics

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

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 14 Areas of Parallelograms and Triangles
Exercise 14.3 | Q 13 | Page 46
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