Question
$ABCD$ is a parallelogram in which $BC$ is produced to $E$ such that $CE = BC$ (Fig.) $AE$ intersects $CD$ at $F.$ If ar $(DFB) = 3\ cm^2$ , find the area of the parallelogram $ABCD$.
✓
Answer
Given, $ABCD$ is a parallelogram and $CE = BC$
i.e., $C$ is the mid-point of $BE.$
Also, $\text{ar}(\triangle\text{DFB})=3\text{cm}^2$
Now, $\triangle\text{ADF}$ and $\triangle\text{DFB}$
are on the same base DF and between parallel $CD$ and $AB.$
Then $\text{ar}(\triangle\text{ADF})=\text{ar}(\triangle\text{DFB})=3\text{cm}^2\ ...(\text{i})$ In
$\triangle\text{ADF}$ by the converse of of mid point theorem,
$EF = AF [$since, $C$ is mid-point of $BE] ...(ii)$ In $\triangle\text{ADF}$
and $\triangle\text{ECF},$
$\angle\text{AFD}=\angle\text{CFG}$ [vertically opposite angle]
$AF = EF \angle\text{DAF}=\angle\text{CEF} [$since $BE || AD$ and $AE$ is transversal,
then alternate interior angles are equal]
$\triangle\text{ADF}\cong\triangle\text{ECF} [$by $ASA$ congurance rule$]$
Then, $\text{ar}(\triangle\text{ADF})=\text{ar}(\triangle\text{CFE})$ [since, congurant figure have equal area]
$\therefore\text{ar}(\triangle\text{CEF})=\text{ar}(\triangle\text{ADF})=3\text{cm}^2 [$From EQ. $(i)] ...(iii)$
Now, In $\triangle\text{BFE}, C$ is the mid point of $BE$ then $CF$ is median of
$\triangle\text{BFE},$
$\therefore\text{ar}(\triangle\text{CEF})=\text{ar}(\triangle\text{BFC})$ [since, median of a triangle divides it into two triangle of equal area]
$\Rightarrow\text{ar}(\triangle\text{CEF})=\text{ar}(\triangle\text{BFC})$ [since, median of a triangle divides it into two triangles of equal area]
$\Rightarrow\text{ar}(\triangle\text{BFC})=3\text{cm}^2$
Now, $\text{ar}(\triangle\text{BDC})=\text{ar}(\triangle\text{CFB})+\text{ar}(\triangle\text{BFC})$
$=3+3=6\ \text{cm}^2 [$from $EQS. (i)$ and $(iv)]$
We know that, diagonal of a parallelogram divides it into two congurant triangle of equal area.
$\therefore$ Area of parallelogram $ABCD = 2 \times $ Area of $\triangle\text{BDC}$
$= 2 \times 6 = 12\ cm^2$
Hence, the area of parallelogram $ABCD$ is $12\ cm^2.$
i.e., $C$ is the mid-point of $BE.$
Also, $\text{ar}(\triangle\text{DFB})=3\text{cm}^2$
Now, $\triangle\text{ADF}$ and $\triangle\text{DFB}$
are on the same base DF and between parallel $CD$ and $AB.$
Then $\text{ar}(\triangle\text{ADF})=\text{ar}(\triangle\text{DFB})=3\text{cm}^2\ ...(\text{i})$ In
$\triangle\text{ADF}$ by the converse of of mid point theorem,
$EF = AF [$since, $C$ is mid-point of $BE] ...(ii)$ In $\triangle\text{ADF}$
and $\triangle\text{ECF},$
$\angle\text{AFD}=\angle\text{CFG}$ [vertically opposite angle]
$AF = EF \angle\text{DAF}=\angle\text{CEF} [$since $BE || AD$ and $AE$ is transversal,
then alternate interior angles are equal]
$\triangle\text{ADF}\cong\triangle\text{ECF} [$by $ASA$ congurance rule$]$
Then, $\text{ar}(\triangle\text{ADF})=\text{ar}(\triangle\text{CFE})$ [since, congurant figure have equal area]
$\therefore\text{ar}(\triangle\text{CEF})=\text{ar}(\triangle\text{ADF})=3\text{cm}^2 [$From EQ. $(i)] ...(iii)$
Now, In $\triangle\text{BFE}, C$ is the mid point of $BE$ then $CF$ is median of
$\triangle\text{BFE},$
$\therefore\text{ar}(\triangle\text{CEF})=\text{ar}(\triangle\text{BFC})$ [since, median of a triangle divides it into two triangle of equal area]
$\Rightarrow\text{ar}(\triangle\text{CEF})=\text{ar}(\triangle\text{BFC})$ [since, median of a triangle divides it into two triangles of equal area]
$\Rightarrow\text{ar}(\triangle\text{BFC})=3\text{cm}^2$
Now, $\text{ar}(\triangle\text{BDC})=\text{ar}(\triangle\text{CFB})+\text{ar}(\triangle\text{BFC})$
$=3+3=6\ \text{cm}^2 [$from $EQS. (i)$ and $(iv)]$
We know that, diagonal of a parallelogram divides it into two congurant triangle of equal area.
$\therefore$ Area of parallelogram $ABCD = 2 \times $ Area of $\triangle\text{BDC}$
$= 2 \times 6 = 12\ cm^2$
Hence, the area of parallelogram $ABCD$ is $12\ cm^2.$
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