Question

In the given figure, $O$ is the center of the circle. $OD=4\text{units}$ and $DC=10\text{units}.$ $OD$ bisects the cord $AB.$ If area of $△AOD$ is $10{\text{unit}}^2$, find the area of the quadrilateral $AOBC.$

• A. $70{\text{units}}^2$
• B. $35{\text{units}}^2$

Answer 1

The correct option is A $70{\text{units}}^2$

In the given figure, OD bisects the chord AB and also passes through the center of the circle.
Hence, OD is perpendicular to AB.

Area of $△AOD$
$=\frac{1}{2}\times \text{base}\times \text{height}$
$\Rightarrow 10{\text{unit}}^2=\frac{1}{2}\times AD\times OD$
$\Rightarrow 10{\text{unit}}^2=\frac{1}{2}\times AD\times 4\text{units}$
$\Rightarrow AD=5\text{units}$

As OC is a straight line, $\angle ADC=\angle ODA={90}^o$

$\therefore$ Area of $△ADC$
$=\frac{1}{2}\times \text{base}\times \text{height}$
$=\frac{1}{2}\times 5\text{units}\times 10\text{units}=25{\text{units}}^2$

$△ADC\cong △DBC$
(by SAS property)

Also, $△AOD\cong △ODB$
(by SAS property)

Area of the quadrilateral $AOBC$
$=(△AOD+△ODB)+(△ADC+△DBC)$
$=2\times △AOD+2\times △ADC$
$=2\times 10{\text{units}}^2+2\times 25{\text{units}}^2$
$=70{\text{units}}^2$