Two Rectangles on a Quarter Circle

Let the vertex of the inscribed rectangle be $(6 \cos t, 6 \sin t )$ , (Note that $t$ is the angle that the diagonal connecting the origin to that vertex, makes with the horizontal axis), then

$\text{Area} = 36 \sin t \cos t$

Point $p$ , where the second rectangle touches the quarter circle, is given by

$p = (0, 6 \sin t) + c ( \sin t, \cos t )$

where $c$ is yet to be determined. Substituting the coordinates of this point into $x^2 + y^2 = 36$ , yields

$c^2 + 6 c \sin 2 t - 18 - 18 \cos 2 t = 0 \hspace{24pt} (1)$

Now, the condition on the areas of the two rectangles, implies,

$6 c = 36 \sin t \cos t$

Hence, $c = 6 \sin t \cos t = 3 \sin 2 t$

Substituting this last equation into Eq. (1) , yields,

$9 \sin^2 2 t + 18 \sin^2 2t - 18 - 18 \cos 2 t = 0$

so that

$3 \sin^2 2 t - 2 - 2 \cos 2 t = 0$

Define $u = 2 t$ , then this becomes,

$3 (1 - \cos^2 u ) - 2 - 2 \cos u = 0$

Re-arranged,

$3 \cos^2 u + 2 \cos u - 1 = 0$

And finally,

$(3 \cos u - 1)(\cos u + 1) = 0$

$\cos u = 1/3$

So that,

$t = \dfrac{1}{2} \cos^{-1} (\dfrac{1}{3})$

Remember, the area $= 6 c = 18 \sin 2 t$

We know that $\cos 2 t = \dfrac{1}{3}$ , so $\sin 2 t = \dfrac{\sqrt{8}}{ 3}$ , from which the area $= 6 \sqrt{8} = \boxed{12 \sqrt{2}} = \boxed{16.97056274847714}$

As labelled in the figure, we note that the area of the rectangle $ABCO$ , $A=\overline{OB}\cdot \overline{BC} = 36\sin \theta \cos \theta$ and the area of rectangle $ADEC$ , $\overline{AC} \cdot \overline{AD} = 6a$ . Therefore, $a = 6 \sin \theta \cos \theta$ .

Applying cosine rule on $\triangle OAD$ , we have:

$\begin{aligned} \overline{OD}^2 & = \overline{AD}^2 + \overline{AO}^2 - 2\overline{AD}\cdot \overline{AO} \cos \angle OAD \\ 6^2 & = a^2 + (6\sin \theta)^2 - 2 a (6\sin \theta) \color{#3D99F6} \cos (180^\circ - \theta) & \small \color{#3D99F6} \text{Note that }\cos (180^\circ - \theta) = - \cos \theta \\ 36 & = a^2 + 36\sin^2 \theta + 12 a \sin \theta \color{#3D99F6} \cos \theta & \small \color{#3D99F6} \text{Recall } a = 6 \sin \theta \cos \theta \\ 36 & = 108 \sin^2 \theta \cos^2 \theta + 36 \sin^2 \theta \\ 1 & = 3 \sin^2 \theta \cos^2 \theta + \sin^2 \theta \end{aligned}$

$\begin{aligned} \implies 3\sin^4 \theta - 4 \sin^2 \theta + 1 & = 0 \\ (3\sin^2 \theta - 1)(\sin^2 \theta - 1) & = 0 \end{aligned}$

$\implies \sin \theta = \frac 1{\sqrt 3}$ , since $\theta < 90^\circ$ and $\cos \theta = \sqrt{1- \frac 13} = \sqrt {\frac 23}$ and the area $A = 36 \sin \theta \cos \theta = 12 \sqrt 2 \approx \boxed{16.971}$ .

Area is exactly $12\sqrt{2}$