Area of triangle AEF = 6sq. cm area of triangle = \(\frac{1}{2}\)bh (Line DX makes right angles with the parallel lines) 6 = \(\frac{1}{2}\) x 3 x h 6 = 3h 3h = \(\frac{12}{3}\) h = 4 = DX From D, C x D, CX\(^2\) = 5\(^2\) - 4\(^2\) = 25 - 16 = 9 Cx = 3. From angle B x D, Bx = 6(i.e. 3 + 3) BD\(^2\) = 4\(^2\) + 6\(^2\) = 16 + 36 = 52 BD = \(\sqrt{4 \times 13}\) = 2\(\sqrt{13}\)