ABCD is a parallelogram in which BC is produced to E such that CE = BC(Fig. 9.17). AE intersects CD at F.If ar (DFB) = 3 cm2, find the area of the parallelogram ABCD.

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!!!

Hope this helps!!! And draw the diagram before u start with the answer!! That will help u a lot!!! I couldn't provide the diagram here!! :(