CAT 2016 Quantitative Ability

Let’s draw AF perpendicular to BC.
In ΔADE, as AD = AE
∠AED = ∠ADE = 60°
⇒ ΔDAE = 60°
Hence, we can conclude that ΔADE is an equilateral triangle.

⇒$\mathit{A}\mathit{F}=\frac{\sqrt{3}}{2}\times 4$
= $2\sqrt{3}$cm
and DF = FE = 2 cm
In ΔAFB, using Pythagoras’ Theorem,
$\mathit{A}\mathit{B}=\sqrt{\mathit{A}{\mathit{F}}^2+\mathit{B}{\mathit{F}}^2}$$=\sqrt{{\left(2\sqrt{3}\right)}^2+4^2}$ $=2\sqrt{7}\mathit{c}\mathit{m}$