Yes. Perfectly right. Kudo

x = \(\frac{2}{2 \sqrt{2}}\)

Apply Pythagoras Theorem in \(ΔABE\), \(BE = 2\sqrt{2}\)
Length of the rectangle \(= BE = 2\sqrt{2}\)

Area of \(ΔABE\) = \(\frac{1}{2}(2)(2) = 2\)
Area of rectangle = \(2\sqrt{2}(DE)\)

Given that Area of \(ΔABE\) = Area of rectangle

\(DE = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}\)

Col. A: \(\frac{BE}{DE}\)
Col. B: \(\frac{4}{1}\)

Col. A: \((2\sqrt{2})(\sqrt{2})\)
Col. B: \(\frac{4}{1}\)

Col. A: \(\frac{4}{1}\)
Col. B: \(\frac{4}{1}\)

Hence, option C