Question

If ∆ABC is an isosceles triangle with ∠C = 90° and AC = 5 cm, then AB is
$\left[1\right]$
a) 2.5 cm
b) 5 cm
c) 10 cm
d) $5\sqrt{2}$ cm

Answer 1

$A{B}^2=A{C}^2+B{C}^{2}A{B}^2=5^2+5^2...\left(Itisanisoscelestriangle\right)AB=\sqrt{50}\therefore AB=5\sqrt{2}$
Correct Option : d) $5\sqrt{2}$