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, $AB=17.327$
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