Question

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

Answer 1

ans.: 12 cm2.
first prove that af=fe
then by same base same parallel rule adf's area will be equal to dbf= 3cm2..
now adf will be congruent to cfe..
then in abc triangle apply similarity and area side theorem as i did..
then ar.abc will be 12 cm2..
nkw subtract the area of cfe for ar of abfc amd add area of adf so you will get area of abcd..
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Answer 2

AD = BC
BC = CE
AD = CE
In triangle ADF & CEF,
1. angle DAF = angle FEC (alternate angles)
2. AD = CE (from above)
3. angle ADF = angle FCE (alternate angles)
By ASA citeria, triangle ADF is congurent to triangle CEF.
By CPCT, DF = CF
So, BF is the median of triangle BCD.
So, area of triangle BCD = 2(area of BFD) = 2×3 = 6 cm square
As we know, diagonal in parallelogram divide it into 2 equal areas.
So, area of parallelogram = 2(area of BCD) = 2×6 = 12 cm square...

The area is 12 cm square...........