Can You Find the Area?

Let the top left corner of the rectangle be $C$ , the segment $AB$ intersect the left edge and right edge of the rectangle at $D$ and $E$ respectively. Due to symmetry, we note that $AD=BE = d = \dfrac {AB-DE}2 = \dfrac {20-x}2$ . And $BD = BE + DE = \dfrac {20-x}2 + x = \dfrac {20+x}2$ . By Pythagorean theorem ,

$\begin{aligned} BD^2+CD^2 & = BC^2 \\ \left(\frac {20+x}2\right)^2 + \left(\frac {\sqrt 3 x}2\right)^2 & = 20^2 \\ \frac {x^2 + 40x+400}4 + \frac {3x^2}4 & = 400 \\ x^2 + 10x+100 & = 400 \\ x^2 + 10x + 25 & = 325 \\ (x+5)^2 & = 325 \\ x + 5 & = 5\sqrt{13} \\ \implies x & = 5\sqrt{13}-5 \end{aligned}$

Then the area of the rectangle $A = \sqrt 3 x \cdot x = \sqrt 3(5\sqrt{13}-5)^2 \approx 293.967882729 \approx \boxed{294}$ .

By symmetry, we can find $d = \cfrac{20 - x}{2}$

By Intersecting chord theorem ,

$( \cfrac{ \sqrt{3} x }{2} )^2 = d \times (d+x +20) \\ 3x^2 = 4\times d \times (d+x +20) \\ 3x^2 = 4\times \cfrac{20 - x}{2} \times ( \cfrac{ 20-x +2x + 40}{2} ) = (20-x)(60+x)\\ \implies 4x^2 + 40x -1200 = 0 \implies x^2 + 10 - 300 = 0 \implies x = -5 + 5\sqrt{13}$

The area of the rectangle

$= \sqrt{3} x^2 = \sqrt{3} [5(\sqrt{13}-1)]^2 \approx \boxed{294}$