Question

Given:- $AB||DC$, $AD=AC$. Which of the following is true about the figure?

• A. $ar\left(\Delta AOD\right)=ar\left(\Delta BOC\right)$
• B. $ar\left(\Delta AOB\right)=ar\left(\Delta COD\right)$
• C. $ar\left(\Delta ADC\right)=ar\left(\Delta ABC\right)$
• D. $ar\left(\Delta AOB\right)=\frac{1}{4}ar\left(quadABCD\right)$

Answer 1

The correct option is A $ar\left(\Delta AOD\right)=ar\left(\Delta BOC\right)$

A trapezium $ABCD$ where $AB\parallel DC$ and diagonals $AC$ and $BD$ intersect each other at $O$.

$\Rightarrow$ $△DAB$ and $△CAB$ lie on the same base $DC$ and between the same parallels $AB$ and $CD$

$\therefore$ $ar\left(△DAB\right)=ar\left(△CAB\right)$ [ Triangles with same base and between same parallels are equal in area ]

Subtracting $ar\left(△AOB\right)$ both sides

$\Rightarrow$ $ar\left(△DAB\right)-ar\left(△AOB\right)=ar\left(△CAB\right)-ar\left(△AOB\right)$

$\Rightarrow$ $ar\left(△AOD\right)=ar\left(△BOC\right)$