The area of the equilateral triangle = (1/2) 4 sin120 = 2(√3)/ 2 = √3
The radius of both circles is given by √[ 1^2 + (1/√3)^2 ] = √[1 + 1/3] = 2/ √3
So....the diagonal of the rectangle = 2 ( 2/ √3 ) = 4 / √3
And, using the Pythagorean Thorem, "y" in the rectangular figure = √ [( 4/ √3) ^2 - x^2 ] = √[16/3 - x^2]
So......since the area of the triangle and the rectangle is equal, we have...
y * x = √3
√[16/3 - x^2] * x = √3 square both sides
(16/3)x^2 - x^4 = 3 multiply through by 3
16x^2 - 3x^4 = 9 rearrange
3x^4 - 16x^2 + 9 = 0 let m= x^2 .... so we have
3m^2 - 16m + 9 = 0
Using the quadratic formula [and an assist from the on-site solver], we have
$${\mathtt{3}}\left[{{m}}^{{\mathtt{2}}}\right]{\mathtt{\,-\,}}{\mathtt{16}}{\mathtt{\,\times\,}}{\mathtt{m}}{\mathtt{\,\small\textbf+\,}}{\mathtt{9}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{m}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{37}}}}{\mathtt{\,-\,}}{\mathtt{8}}\right)}{{\mathtt{3}}}}\\
{\mathtt{m}} = {\frac{\left({\sqrt{{\mathtt{37}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}}\right)}{{\mathtt{3}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{m}} = {\mathtt{0.639\: \!079\: \!156\: \!567\: \!260\: \!1}}\\
{\mathtt{m}} = {\mathtt{4.694\: \!254\: \!176\: \!766\: \!073\: \!2}}\\
\end{array} \right\}$$
So x = about .7994 or x = about 2.1666
If x = .799, y = √[16/3 - .7994^2] = .2166 ...but this is impossible....since y is smaller than x
If x = 2.1666, y = √[16/3 - 2.1666^2] = about . 7994
So....x =about .2166 and y = about .7994
