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Areas Of Parallelograms And Triangles — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsAreas Of Parallelograms And Triangles5 Marks
Question
$\text{ABCD}$ is a parallelogram whose diagonals intersect at $O$ .If $P$ is any point on $BO,$ prove that:
$i. \text{ar}(\triangle\text{ADO})=\text{ar}(\triangle\text{CDO})$
$ii. \text{ar}(\triangle\text{ABP})=\text{ar}(\triangle\text{CBP})$
✓

Answer

Given that $\text{ABCD}$ is the parallelogram To Prove:
$i. \text{ar}(\triangle\text{ADO})=\text{ar}(\triangle\text{CDO}).$
$ii. \text{ar}(\triangle\text{ABP})=\text{ar}(\triangle\text{CBP}).$
Proof: we know that diagonals of parallelogram bisect each other
$\therefore AO = OC$ and $BO = OD$
$i.$ In $\triangle\text{DAC},$ since $DO$ is a median.
Then $\text{ar}(\triangle\text{ADO})=\text{ar}(\triangle\text{CDO}).$
$ii.$ In $\triangle\text{BAC},$ since $BO$ is a median.
Then $\text{ar}(\triangle\text{BAO})=\text{ar}(\triangle\text{BCO})\ ....(1)$
In $\triangle\text{PAC},$ since $PO$ is a median.
Then $\text{ar}(\triangle\text{PAO})=\text{ar}(\triangle\text{PCO})\ .....(2)$
Subtract equation $2$ from $1.$
$\Rightarrow\text{ar}(\triangle\text{BAO})−\text{ar}(\triangle\text{PAO})\\ \ =\text{ar}(\triangle\text{BCO})−\text{ar}(\triangle\text{PCO})$
$\Rightarrow\text{ar}(\triangle\text{ABP})=2 \text{ar}(\triangle\text{CBP}).$

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