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Areas Of Parallelograms And Triangles — MATHS STD 9 — Question

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

Answer

Given that ABCD is the parallelogram To Prove:
  1. $\text{ar}(\triangle\text{ADO})=\text{ar}(\triangle\text{CDO}).$
  2. $\text{ar}(\triangle\text{ABP})=\text{ar}(\triangle\text{CBP}).$
Proof: we know that diagonals of parallelogram bisect each other $\therefore$ AO = OC and BO = OD
  1. In $\triangle\text{DAC},$ since DO is a median.
Then $\text{ar}(\triangle\text{ADO})=\text{ar}(\triangle\text{CDO}).$
  1. 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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