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.
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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!! :(