ABCD is a parallelogram in which BC is produced to E such that CE…

Question

$\text{ABCD}$ is a parallelogram in which $BC$ is produced to $E$ such that $CE = BC$ and $AE$ intersects $CD$ at $F$.

If ar.$(\triangle DFB) = 30 \ cm^2;$ find the area of parallelogram.

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Answer

$BC = CE\dots .....($ given $)$
Also, in parallelogram $\text{ABCD}, BC = AD$
$\Rightarrow AD = CE$
Now, in $\triangle ADF$ and $\triangle ECF,$ We have
$AD = CE$
$\angle ADF = \angle ECF \dots.....($ Alternate angles $)$
$\angle DAF = \angle CEF \dots......($ Alternate angles $)$
$\therefore \triangle ADF \cong \triangle ECF \dots......( \text{ASA}$ Criterion $)$
$\Rightarrow$ Area $( \triangle ADF ) =$ Area $( \triangle ECF )\dots ....(1)$
Also, in $\triangle FBE, FC$ is the median \$dots....( Since $BC = CE )$
$\Rightarrow$ Area $( \triangle BCF ) =$ Area $( \triangle ECF ) \dots.....(2)$
From $(1)$ and $(2)$
Area $( \triangle ADF ) =$ Area $( \triangle BCF )\dots ......(3)$
Again$, \triangle ADF$ and $\triangle BDF$ are on the base $DF$ and between parallels $DF$ and $AB$.
$\Rightarrow $ Area $( \triangle BDF ) =$ Area $( \triangle ADF )\dots ........(4)$
From $(3)$ and $(4),$
Area $( \triangle BDF ) =$ Area $( \triangle BCF ) = 30 \ cm^2$
Area $( \triangle BCD ) =$ Area $( \triangle BDF ) +$ Area $( \triangle BCF ) = 30 + 30 = 60 \ cm^2$
Hence, Area of parallelogram $ABCD = 2 x$ Area $( \triangle BCD ) = 2 \times 60 = 120\ cm^2.$