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Areas Of Parallelograms And Triangles — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsAreas Of Parallelograms And Triangles4 Marks
Question
$\text{ABCD}$ is a parallelogram in which $BC$ is produced to $E$ such that $CE = BC. AE$ intersects $CD$ at $F.$
$i.$ Prove that $\text{ar}(\triangle\text{ADF})=\text{ar}(\triangle\text{ECF}).$
$ii.$ If the area of $\triangle\text{DFB}=3\text{cm}^2,$ find the area of $||^{gm}\text{ABCD}.$
✓

Answer

In triangles $\text{ADF}$ and $\text{ECF}$,
we have $\angle\text{ADF}=\angle\text{ECF}$
$[$Alternate interior angles,Since $AD \| BE]$
$AD = EC [$since $AD = BC = CE]$ And
$\angle\text{DFA}=\angle\text{CFA} [$Vertically opposite angles$]$
So, by $\text{AAS}$ congruence criterion,
we have $\triangle\text{ADF}\cong\triangle\text{ECF}$
$\Rightarrow\text{ar}(\triangle\text{ADF})=\text{ar}(\triangle\text{ECF})$ and $DF = CF$.
Now, $DF = CF$
$\Rightarrow BF$ is a median in $\triangle\text{BCD}.$
$\Rightarrow\text{ar}(\triangle\text{BCD})=2\text{ar}(\triangle\text{BDF})$
$\Rightarrow\text{ar}(\triangle\text{BCD})=2\times3\text{ cm}^2=6\text{ cm}^2$
Hence, area of a parallelogram $=2\text{ar}(\triangle\text{BCD})=2\times6\text{ cm}^2=12\text{ cm}$

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