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Question
ABCD is a parallelogram in which BC is produced to E such that CE = BC (Figure). AE intersects CD at F. If ar (DFB) = 3 cm2, find the area of the parallelogram ABCD.
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Solution
Given: ABCD is a parallelogram in which BC is produced to E such that CE = BC. C is the mid-point BE and ar (ΔDFB) = 3 cm2.
In triangle, ADF and triangle EFC,
∠DAF = ∠CEF ...[Alternate interior angle]
AD = CE ...[AD = BC = CE]
∠ADF = ∠FCE ...[Alternate interior angle]
So, ΔADF ≅ ΔECF ...[By SAS rule of congruence]
Now, ΔADF ≅ ΔECF ...[By SAS rule of congruence]
DF = CF ...[CPCT]
As BF is the median of triangle BCD.
ar (ΔBDF) = `1/2` ar (BCD) ...(i) [Median divides a triangle into two triangle of equal area]
As we know that a triangle and parallelogram are on the same base and between the same parallels then area of the triangles is equal to half the area of the parallelogram.
ar (ΔBCD) = `1/2` ar (ABCD) ...(ii)
ar (ΔBDF) = `1/2 {1/2 "ar (ABCD)"}` ...[By equation (i)]
3 cm2 = `1/4` ar (ABCD)
ar (ABCD) = 12 cm2
Hence, the area of the parallelogram is 12 cm2.
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