Question

In a trapezium $ABCD,AB$ is parallel to $DC$ and $AB=2DC$. If $AC$ and $BD$ meets at $O$, then area of $△AOB$ is equal to

• A. The area of $△COD$
• B. Twice the area of $△COD$
• C. Thrice the area of $△COD$
• D. Four times the area of $△COD$

Answer 1

The correct option is D Four times the area of $△COD$

$△AOB\sim △DOC$
$\therefore \frac{Area△AOB}{Area△DOC}=\frac{(AB{)}^2}{(DC{)}^2}$
$=\frac{4D{C}^2}{D{C}^2}=4$.
$\Rightarrow Area△AOB=4.Area△DOC$.