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


$\triangle \mathrm{ABC}$ is an isosceles triangle in which $\angle \mathrm{C}=90$. If $\mathrm{AC}=6 \mathrm{~cm}$, then $\mathrm{AB}=$

(a) 62 cm
(b) 6 cm
(c) 26 cm
(d) 42 cm


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


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



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

We got the result as (a)

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