Question

In $△ABC,m\angle B={90}^o,\overline{BM}\perp \overline{AC},M\in AC$. If $AM-MC=7$ and ${AB}^2-{BC}^2=175$, then find $AC$.

Answer 1

In $△ABC,m\angle B={90}^o$
$\overline{BM}\perp \overline{AC},M\in AC$

In $△AMB$ and $△BMC$, $\angle AMB={90}^o,\angle BMC={90}^o$
$\therefore {AB}^2={AM}^2+{BM}^2$ and
${BC}^2={BM}^2+{CM}^2$
Then
${AB}^2-{BC}^2={AM}^2-{CM}^2$
$\Rightarrow$ $175=(AM+CM)(AM-CM)$
$\Rightarrow$ $175=AC(AM-CM)$
$\Rightarrow$ $175=AC\left(7\right)$
$\Rightarrow$ $AC=25$