Question

In the adjoining figure O is the center of the circle. $\angle$AOD = ${120}^{\circ }$. If the radius of the circle be 'r', then find the sum of the areas of quadrilaterals AODP and OBQC:

• A. √3/2 r²
• B. 3√3 r²
• C. √3 r²
• D. None of these

Answer 1

The correct option is C

√3 r²

Consider triangle ARO and DRO

AO = DO (radius)

RO is common base

AR = RD (Line joining the center to the chord bisects the chord)

Therefore triangles ARO and DRO are congruent.

$\Rightarrow$ $\angle$AOR = $\angle$DOR

As $\angle$AOD = ${120}^{\circ }$

$\angle$AOR = ${60}^{\circ }$

In triangle ARO

$\angle$AOR + $\angle$ARO + $\angle$RAO = ${180}^{\circ }$

$\Rightarrow$ ${60}^{\circ }$ + ${90}^{\circ }$ + $\angle$RAO = ${180}^{\circ }$

$\angle$RAO = ${30}^{\circ }$

cos $\angle$RAO = $\frac{AR}{AO}$

cos ${30}^{\circ }$ = $\frac{AR}{r}$

AR = ( $\frac{\sqrt{3}}{2}$) r
Area of triangle APO = $\frac{1}{2}$ $\times$ PO $\times$ AR

= $\frac{1}{2}$ $\times$ r $\times$ $\frac{\sqrt{3}}{2}$ r = $\frac{\sqrt{3}}{4}{r}^2$

Required area = 4 $\times$ Area of triangle APO = $\sqrt{3}$ $r^2$