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Question

ABCD is a parallelogram in which BC is produced to E such that CE=BC.AE intersects CD at G. If ar(DFD)=3 cm2 find the area of the parallelogram(ABCD).

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Solution

Given: ABCD is a parallelogram, BC is produced to E such that CE= BC , ar(BDF) = 3cm2!

To find: ar(paralelogram ABCD)!

Solution: In triangle ADF and ECF, we have,

angle ADF = angle ECF (Alternate interior angles, Ad is parallel to BC,therfore AC is parallel to CE)

AD= CE (Since AD =BC and BC=CE)

angle DFA = angle CFE ( AAS congruence rule)

Therefore, ar(ADF) = ar(ECF)

Also, DF= CF( By CPCT)

Therfore, BF is the median in triangle BCD.

So, ar(BCD) = 2 ar(BDF)

ar( BCD) = 2(3 cm2) = 6cm2

We know that BD is a diagonal of the paralelogram ABCD,

Therefore, ar( ABCD) = 2 ar(BCD) ( Since, A diagonal of a parallelogram bisects it into 2 congruent triangles)

= 2 (6cm2) = 12 cm2

Hence, the area of parallelogram is 12 cm2!!!


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