Question

In a right angled $△ABC$, $\angle B={90}^o$, $AC=17cm$ and $AB=8cm$, find $BC$.

Answer 1

Given, in $△ABC$, $\angle B={90}^o$

$\therefore {AC}^2={AB}^2+{BC}^2$ [$\because$ Pythagoras theorem]

$\Rightarrow {BC}^2={AC}^2-{AB}^2$

${BC}^2={17}^2-8^2$

${BC}^2=289-64=225$

$\therefore BC=\sqrt{225}=15cm$