(b) Equilateral, \(2\sqrt3a^2\) sq. units
Let A(a, a), B(–a, –a) and C \((-\sqrt3a,\sqrt3a)\) be the vertices of ΔABC. Then,
AB = \(\sqrt{(a+a)^2+(a+a)^2}\) = \(\sqrt{4a^2+4a^2}\) = \(\sqrt{8a^2}\) = \(2\sqrt2a\)
BC = \(\sqrt{(-a+\sqrt3a)^2+(-a-\sqrt3a)^2}\)
= \(\sqrt{a^2-2\sqrt3a+3a^2+a^2+2\sqrt3a+3a^2}\) = \(\sqrt{8a^2}\) = \(2\sqrt2a\)
AC = \(\sqrt{(-a+\sqrt3a)^2+(-a-\sqrt3a)^2}\)
= \(\sqrt{a^2+2\sqrt3a+3a^2+a^2-2\sqrt3a+3a^2}\) = \(\sqrt{8a^2}\) = \(2\sqrt2a\)
∵ AB = BC = AC, ΔABC is equilateral.
Area = \(\frac{\sqrt3}{4}\) (side)2 = \(\frac{\sqrt3}{4}\) x (\(2\sqrt2a\))2
= \(\frac{\sqrt3}{4}\) x 8a2 = \(2\sqrt3a^2\).