10.281. A quadrilateral with angles of $120, 90, 60$, and $90^{\circ}$ is inscribed in a circle. The area of the quadrilateral is $9 \sqrt{3}$ cm $^{2}$. Find the radius of the circle, given that the diagonals of the quadrilateral are perpendicular to each other.
Solution.

Let $ABCD$ be the quadrilateral mentioned in the problem (Fig. 10.76), $\angle ABC = \angle ADC = 90^\circ$, $\angle BAD = 120^\circ$, $\angle BCD = 60^\circ$. $P$ is the intersection point of $AC$ and $BD$, and $AC \perp BD$. Since $\angle ABC = 90^\circ$, $AC$ is the diameter of the given circle. $BP$ is the height of the right triangle $\triangle ABC$. Therefore, $\angle ABD = \angle ACB$. $\angle ABD$ and $\angle ACD$ are inscribed angles subtending the chord $AD$. Thus, $\angle ABD = \angle ACD$. Consequently, $\angle ACB = \angle ACD = \frac{1}{2} \angle BCD = 30^\circ$. Therefore, $\triangle ABC = \triangle ADC$ (by hypotenuse and acute angle), $S_{\triangle ABC} = \frac{1}{2} S_{ABCD} = \frac{9\sqrt{3}}{2} \text{ cm}^2$. Let $AB = a$. Then $BC = AB \cot \angle ACB = a \sqrt{3}$ and $S_{\triangle ABC} = \frac{1}{2} AB \cdot BC = \frac{a^2 \sqrt{3}}{2}$. Therefore, $\frac{a^2 \sqrt{3}}{2} = \frac{9\sqrt{3}}{2}$, $a = 3$. The desired radius $R = \frac{1}{2} AC = AB = 3$ cm. Answer: 3 cm.

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Fig. 10.77