Diagonals $\mathrm{AC}$ and $\mathrm{BD}$ of a trapezium $\mathrm{ABCD}$ with $\mathrm{AB} \| \mathrm{DC}$ intersect each other at $\mathrm{O}$. Prove that ar $(\mathrm{AOD})=\operatorname{ar}(\mathrm{BOC})$.

AcademicMathematicsNCERTClass 9

Complete Python Prime Pack for 2023

9 Courses     2 eBooks

Artificial Intelligence & Machine Learning Prime Pack

6 Courses     1 eBooks

Java Prime Pack 2023

8 Courses     2 eBooks

Given:

Diagonals $\mathrm{AC}$ and $\mathrm{BD}$ of a trapezium $\mathrm{ABCD}$ with $\mathrm{AB} \| \mathrm{DC}$ intersect each other at $\mathrm{O}$.

To do:

We have to prove that ar $(\mathrm{AOD})=\operatorname{ar}(\mathrm{BOC})$.

Solution:

$\triangle ABC$ and $\triangle ABD$ lie on the same base $AB$ and between the parallels $AB$ and $CD$.

This implies,

$ar (\triangle ABD) = ar (\triangle ABC)$

Subtracting $ar (\triangle AOB)$ from both sides, we get,

$ar (\triangle ABD) - ar (\triangle AOB) = ar (\triangle ABC) - ar (\triangle AOB)$

$ar (\triangle AOD) = ar (\triangle BOC)$

Hence proved.

Updated on 10-Oct-2022 13:42:01

Advertisements