(b) a
In isosceles ΔABC, \(\frac{BC}{x}\) = sin 30º
⇒ BC = \(x\) sin 30º
⇒ BD = 2x sin 30°
and \(\frac{AC}{AB}\) = cos 30º
⇒ AC = \(x\) cos 30°
∴ Area of ΔABC = a = \(\frac{1}{2}\) x BD x AC
= \(\frac{1}{2}\) x 2\(x\) sin 30° x \(x\) cos 30°
= \(x\)2 x \(\frac{1}{2}\)x \(\frac{\sqrt3}{2}\) = \(\frac{\sqrt3}{4}\)\(x\)2 = a (given)
⇒ \(x\)2 = \(\frac{4}{\sqrt3}a\)
Now angle between equal sides = 120°
∴ Area of triangle = \(\frac{1}{2}\) \(x\)2 sin 120°
[Using Area of Δ = \(\frac{1}{2}\) AB AD sin 120°]
= \(\frac{1}{2}\) x \(\bigg(\frac{4a}{\sqrt3}\bigg)^2\) x \(\frac{\sqrt3}{2}\) = a.