Q15 of 77 Page 15

ABCD is a parallelogram in which BC is produced to E such that CE=BC. AE intersects CD at F.

(i) Prove that ar(Δ ADF) = ar(Δ ECF)


(ii) If the area of Δ DFB=3 cm2, find the area of ||gm ABCD.

In ADF and ECF

We have,


∠ADF = ∠ECF


AD = EC


And,


∠DFA = ∠CFA


So, by AAS congruence rule,


Δ ADF ≅ Δ ECF


Area (ΔADF) = Area (ΔECF)


DF = CF


BF is a median in ΔBCD


Area (ΔBCD) = 2 Area (ΔBDF)


Area (ΔBCD) = 2 * 3


= 6cm2


Hence, Area of parallelogram ABCD = 2 Area (ΔBCD)


= 2 * 6


= 12 cm2



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13

A point D is taken on the side BC of a Δ ABC such that BD = 2DC. Prove that

ar(Δ ABD) = 2ar(Δ ADC)

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ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that

(i) ar(Δ ADO) = ar(Δ CDO)


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