Question

In $△PQR,m\angle Q={90}^o$ and $\overline{QM}$ is an altitiude; $M\in \overline{PR}$. If $QM=12,PR=26$; then find $PM$ and $RM$; If $PM<RM$; then find $PQ$ and $QR$.

Answer 1

In $△PQR,m\angle Q={90}^o$ and $\overline{QM}$ is an altitude; $M\in \overline{PR}$
$\therefore$ $P-M-R$, ${QM}^2=PM.MR$
Suppose $PM=a$
$\therefore$ $RM=PR-PM=26-a$
Now, ${QM}^2=PM.RM$
$\Rightarrow$ ${12}^2=a(26-a)$
$\Rightarrow$ $144=26a-a^2$
$\Rightarrow$ $a^2-26a+144=0$
$\Rightarrow$ $(a-8)(a-18)=0$
$\Rightarrow$ $a=8$ or $a=18$
So $PM=8$ and $RM=18$ or
$PM=18$ and $RM=8$

If $PM<RM$, then $PM=8$ and $RM=18$
${PQ}^2=PM.PR=8\times 26=16\times 13$
$\Rightarrow$ $PQ=4\sqrt{13}$
${QR}^2=RM.PR=18\times 26=36\times 13$
$QR=6\sqrt{13}$