(b) \(\frac{q^2}{4}\) (2√3 + 1).
AC = q, ∠ABC = 90º
⇒ q = \(\sqrt{AB^2+BC^2}\)
⇒ q = \(\sqrt{2x^2}\)
⇒ q2 = 2x2 ⇒ \(x\) = \(\frac{q}{\sqrt2}\)
∴ Area of the re-entrant hexagon = Sum of areas of (ΔABC + ΔADC + ΔBFC + ΔAEB)
= \(\frac{1}{2}\)x \(\frac{q}{\sqrt2}\) x \(\frac{q}{\sqrt2}\) + \(\frac{\sqrt3}{4}\)q2 + \(\frac{\sqrt3}{4}\)\(\big(\frac{q}{\sqrt2}\big)^2\) + \(\frac{\sqrt3}{4}\)\(\big(\frac{q}{\sqrt2}\big)^2\)
= \(\frac{q^2}{4}\) + \(\frac{\sqrt3}{4}\)q2 + \(\frac{\sqrt3}{8}\)q2 + \(\frac{\sqrt3q^2}{8}\) = \(\frac{q^2}{4}\) (2√3 + 1).