In $ \triangle \mathrm{ABC}, \angle \mathrm{C}=90^{\circ}, \mathrm{AB}=12.5 $ and $ \mathrm{BC}=12 $. Find $ \mathrm{AC} $.

Given:

In \( \triangle \mathrm{ABC}, \angle \mathrm{C}=90^{\circ}, \mathrm{AB}=12.5 \) and \( \mathrm{BC}=12 \).

To do:

We have to find \( \mathrm{AC} \).

Solution:  

In triangle ABC, by Pythagoras theorem,

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

$=(12)^2+(12.5)^2$

$=144+156.25$

$=300.25$

$\Rightarrow AB=\sqrt{300.25}=17.327$

Therefore, $AB=17.327$