In the given figure, ΔABC is an isosceles triangle, in which AB = AC, AD ⊥ BC, BC = 6 cm and AD = 4 cm. The length of AB is:

7 cm 6 cm 5 cm 4 cm

5 cm

BC = 6 cm BD = \(\frac{1}{2}\) x BC = \(\frac{1}{2}\) x 6 = 3 cm AD = 4 cm In right angled triangle \( {AB }^{2 } \) = \( {AD }^{2 } \) + \( {BD }^{2 } \) = \( {AB }^{2 } \) = \( {4 }^{2 } \) + \( {3 }^{2 } \) = 16 + 9 = 25 = AB = \(\sqrt { 25}\) = 5 cm