Question

In the adjoining figure two parallelogram ABCD and AEFB are drawn on opposite sides of AB . Prove that : ar(llgm ABCD) ar(lagn AEFB)= ar(llgm efcd)

Answer 1

ABCD and AEFB are two parallelogram on opposite side of AB,

we have to prove :-

ar(II grm ABCD)+ ar ( AEFB) = ar( EFCD),

First we extend AB , and it intersects ED at point P, and FC at point Q,

Now in EPQF----

EF parallel to PQ,( AB parallel to EF)

EP parallel to FQ,(FC parallel to ED,)

therefore, EPQF is also a parallelogram,

same for PQCD is also a parallelogram,

(PQ parallel to CD,

PD parallel to QC),

we know that, parallelograms on the same base and between same parallel lines are equal in area,

therefore :--

area of AEFB + area of ABCD

= area of EPQF + area of PQCD

= area of EFCD,

proved,

Answer 2

Area of EFCD = Area of ABCD + Area of AEFB

Step-by-step explanation:

Compare Δ ADE & ΔBCF

DE = CE   (opposite sides of parallelogram EFCD)

AD = BC  (opposite sides of parallelogram ABCD)

AE = BF    (opposite sides of parallelogram AFEB)

=> Δ ADE ≅ ΔBCF

=> Area of Δ ADE = Area of ΔBCF

Area of  EFCD = Area of ABCD + Area of AEFB  -  Area of Δ ADE + Area of ΔBCF

=> Area of EFCD = Area of ABCD + Area of AEFB

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