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Area Theorems [Proof and Use] — MATHEMATICS STD 9 — Question

ICSE BoardEnglish MediumSTD 9MATHEMATICSArea Theorems [Proof and Use]5 Marks
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.
✓

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 $A B$.
$\Rightarrow$ Area$(\triangle BDF)=$Area$(\triangle ADF )$
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 $\text{ABCD}=2 \times$ Area $(\triangle B C D)=2 \times 60=120 \ cm^2$.

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