Question

In a trapezium ABCD, O is the point of intersection of AC and BD, AB || CD, and $AB=\frac{5}{2}\times CD$. The area of $AOB=100c{m}^2$. The area of $\Delta OCD$ is :

[1 mark]

• A. $16c{m}^2$
• B. $32c{m}^2$
• C. $48c{m}^2$
• D. $64c{m}^2$

Answer 1

The correct option is A $16c{m}^2$

$\angle OAB=\angle OCD\text{(Alternate angle)}$
$\angle OBA=\angle ODC\text{(Alternate angle)}$

$\therefore △AOB\sim △COD\text{(AA criterion)}$

$\Rightarrow \frac{ar\left(AOB\right)}{arCOD}=\frac{A{B}^2}{C{D}^2}=\frac{(\frac{5}{2}CD{)}^2}{C{D}^2}[\because AB=\frac{5}{2}\times CD]$

$=\frac{25C{D}^2}{4C{D}^2}=\frac{25}{4}$

$\Rightarrow ar△COD=\frac{4}{25}ar△AOB$

$=\frac{4}{25}\times 100=16c{m}^2$