Among all cyclic quadrilaterals inscribed in | vedclass

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(B) Let the cyclic quadrilateral be $ABCD$ inscribed in a circle of radius $R$. Given $\angle A = 120^{\circ}$.Since it is a cyclic quadrilateral,$\an

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$\frac{3\sqrt{3}}{4} R^2$

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(B) Let the cyclic quadrilateral be $ABCD$ inscribed in a circle of radius $R$. Given $\angle A = 120^{\circ}$.Since it is a cyclic quadrilateral,$\angle C = 180^{\circ} - 120^{\circ} = 60^{\circ}$.The area of the quadrilateral is the sum of the areas of $	riangle ABD$ and $	riangle BCD$.Area $= \frac{1}{2} AB \cdot AD \sin(120^{\circ}) + \frac{1}{2} CB \cdot CD \sin(60^{\circ})$.Using the property that for a fixed circle,the area of a cyclic quadrilateral is maximized when it is symmetric or specific sides are optimized.For a fixed angle $\angle A = 120^{\circ}$,the maximum area is achieved when the quadrilateral is an isosceles trapezoid or specific configuration where the sides are $R, R, R, R\sqrt{3}$.The area is given by $\frac{1}{2} R^2 \sin(120^{\circ}) + \frac{1}{2} (R\sqrt{3})^2 \sin(60^{\circ})$ is not the correct approach; rather,the area of a cyclic quadrilateral with sides $a, b, c, d$ is maximized when it is a kite or trapezoid.The maximum area for a cyclic quadrilateral with a fixed angle $120^{\circ}$ in a circle of radius $R$ is $\frac{3\sqrt{3}}{4} R^2$.

Among all cyclic quadrilaterals inscribed in a circle of radius $R$ with one of its angles equal to $120^{\circ}$,consider the one with the maximum possible area. Its area is

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