Question

In the given figure ∆ ABC is a right angle triangle,find the value of x​

Answer 1

$\huge\bold{Answer:-}$

$AC^2=AB^2+BC^2$

$\bold\blue{(By\:Pythagoras\:Theorom)}$

$AB^2=AC^2-BC^2$

=169 - 144

=25

$AB^2$ = 25

AB = $\sqrt25$ = 5cm

$\therefore$ The value of x is 5 cm.