Area of a Rectangle

This my solution using basic geometry and algebra.
the intersection point of circles be X.(the intersection bw areas). all angles are in radians so pi conviniently cancels.
(area of B)/2=sector - triangle=.
(angle<XSR)/2-(sin<XSR)(cos<XSR)/2.
similarly.
(area of A/2)=(1-cos<PSX)(sin<PSX)-(<PSX/2-cos<PSX.sin<PSX/2).
<PSX=Pi/2-<XSR.
replacing to bring everything in terms of <XSR and inserting area of A=2B we have. LET <XSR=Y.
Y-(sinY)(cosY)=(cosY)(sinY)/2+(1-sinY)(cosY)-pi/4+Y/2 eq1.
notice that 2cosY is the answer.
Solving eq 1 for 2cosY (we'll need a calculator for this one) we get 1.6856.

Let $S$ be the origin and $SR = a$ lie on the positive $x$ -axis. Since the height of rectangle $PQRS$ is $1$ its area will then just be $a$ .

The circle on the left then has equation $x^{2} + y^{2} = 1$ and the circle on the right has equation $(x - a)^{2} + y^{2} = 1$ . The uppermost point of intersection of the two circles lies at $(\frac{a}{2}, \sqrt{1 - (\frac{a}{2})^{2}})$ .

The area of region $A$ will then be

$2*\displaystyle\int_{0}^{\frac{a}{2}} (1 - \sqrt{1 - x^{2}}) dx$ ,

and the area of region $B$ will be

$2*\displaystyle\int_{\frac{a}{2}}^{1} \sqrt{1 - x^{2}} dx$ .

So for the area of $B$ to be one-half the area of $A$ we must have

$2*\displaystyle\int_{\frac{a}{2}}^{1} \sqrt{1 - x^{2}} dx = \displaystyle\int_{0}^{\frac{a}{2}} (1 - \sqrt{1 - x^{2}}) dx$

$\Longrightarrow \sin^{-1}(1) - \sin^{-1}(\frac{a}{2}) - (\frac{a}{2})\sqrt{1 - \frac{a^{2}}{4}} = \frac{a}{2} - (\frac{1}{2})\sin^{-1}(\frac{a}{2}) - (\frac{a}{4})\sqrt{1 - \frac{a^{2}}{4}}$

$\Longrightarrow a + \sin^{-1}(\frac{a}{2}) + (\frac{a}{2})\sqrt{1 - \frac{a^{2}}{4}} = \pi$ .

At this point I resorted to an application of WolframAlpha, which gave a solution of $a = \boxed{1.686}$ to $3$ decimal places.

Note that in general, if the area of region $A$ is $n$ times that of region $B$ , (where $n \ge 0$ is a real number), then the area $a$ of rectangle $PQRS$ is the solution to the equation

$a + (n - 1)\arcsin(\frac{a}{2}) + (n - 1)(\frac{a}{2})\sqrt{1 - \frac{a^{2}}{4}} = \dfrac{n\pi}{2}$ .

Note that $\displaystyle\int \sqrt{1 - x^{2}} dx = (\frac{1}{2})(\sin^{-1}(x) + x\sqrt{1 - x^{2}}) + C$ .