∆ABC is an isosceles triangle in which ∠C = 90. If AC = 6 cm, then AB =

Given: In an isosceles $\triangle \mathrm{ABC}, \angle C=90^{\circ}, \mathrm{AC}=6 \mathrm{~cm}$.

To find: AB

In an isosceles $\triangle \mathrm{ABC}, \angle C=90^{\circ}$.

Therefore, BC = AC = 6 cm

Applying Pythagoras theorem in ΔABC, we get

$\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{BC}^{2}$

$\mathrm{AB}^{2}=6^{2}+6^{2}(\mathrm{AC}=\mathrm{BC}$, sides of isosceles triangle $)$

$\mathrm{AB}^{2}=36+36$

$\mathrm{AB}^{2}=72$

$\mathrm{AB}=6 \sqrt{2} \mathrm{~cm}$

We got the result as (a)