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Areas Of Parallelograms And Triangles — Maths STD 9 — Question

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

Answer

In triangles ADF and 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 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
⇒ 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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