Area of circumscribed circle I'm looking at the solutions for a 1989 ARML contest, and it has this question:

If the area of a rectangle is 1/2 of the area of its circumscribed circle, compute the acute angle between the diagonals of the rectangle to the nearest degree. [We will accept any integer answer that is no more than 4° away from the correct answer to this problem.] [Reminder: Calculators may not be used!]

and the solution is:

Let the rectangle be ABCD, the center of the circle be O, the radius r, and acute angle AOD be θ. Area of $ABCD = (\frac{1}{2}r^2sinθ)⋅4 = \frac{1}{2}πr^2$, so $sinθ=\frac{π}{4}$ which is about 0.7854. Thus θ is greater than 45° but less than (or equal to) 53° [from the angles of a 3-4-5 triangle]. Choose 49° as the wisest answer, since that will cover any integral angle from 45° through 53°! The actual angle is a bit less than 52°, so we will accept any integer angle from 48° through 56°, inclusive.

The part that I'm struggling to understand is how they got the area of ABCD to be $(\frac{1}{2}r^2sinθ)⋅4$. Can someone explain this to me? Thanks in advance. :)
 A: The area of a triangle  $ABC$ with usual notations $(AB=c,BC=a$ and $AC=b) \, ;\,$  has the area : 
$$\operatorname {Area}(\Delta ABC)=\frac{1}{2} ab\sin C =\frac{1}{2} bc\sin A=\frac{1}{2} ac\sin B$$
(Derivation)
They have simply applied this to all the triangles formed by the diagonals.
You can easily notice that -
$$\operatorname {Area}(\Delta AOD)=\operatorname {Area}(\Delta AOB)=\operatorname {Area}(\Delta BOC)=\operatorname {Area}(\Delta COD)$$ and also that -
$$OA=OB=OC=OD=r$$
\begin{align}
\operatorname {Area}(ABCD)&=\operatorname {Area}(\Delta AOD)+\operatorname {Area}(\Delta AOB)+\operatorname {Area}(\Delta BOC)+\operatorname {Area}(\Delta COD)\\
&=4 \cdot \operatorname {Area}(\Delta AOD)\\
&=\frac 12 OA\cdot OD \sin (\angle AOD)\\
&=\frac 12 r^2 \sin \theta
\end{align}
A: 
If the radius of the circumcircle is $R$, the area of the grey rectangle is 
$$ A = \frac{R^2}{2}\left(\sin\theta+\sin(180^\circ-\theta)+\sin\theta+\sin(180^\circ-\theta)\right) = 2R^2 \sin\theta $$
hence $A=\frac{\pi}{2}R^2$ leads to $\theta=\arcsin\frac{\pi}{4}$. By the Shafer-Fink inequality
$$ \forall x\in(0,1),\qquad \arcsin(x)\approx \frac{3x}{2+\sqrt{1-x^2}} $$
hence the amplitude of $\theta$ in degrees is approximately $\frac{540}{8+\sqrt{16-\pi ^2}}\approx 51+\frac{1}{2}$.
A: 
You can divide the rectangle into 8 congruent triangles.
The area of each is $R^2\frac 12 \sin \frac x2 \cos \frac x2$
the area of the rectangle is $4R^2 \sin \frac x2 \cos \frac x2 = 2R^2 \sin x$
A: Drop a perpendicular $P$ from $C$ to the diagonal $BD$. The length of $PC$ is $r\sin\theta$, and therefore the area of $ABCD$ is $BD\cdot PC=2r\cdot r\sin\theta$.
A: $\hspace{5cm}$
$$\begin{align}
A_\text{rectangle}&=\frac 12 A_\text{circle}\\
2r\sin\theta\cdot 2r\cos\theta&=\frac 12 \pi r^2\\
4\sin 2\theta&=\pi\\
2\theta&=\sin^{-1}\frac{\pi}4\\
&=0.45167\\
&=51.76^\circ\approx 52^\circ
\end{align}
$$
