Question

In figure , $\angle MNP={90}^o,segNQ\perp segMP,MQ=9$cm ,$QP=4$cm, find $NQ$

Answer 1

Area of $\Delta MPN=\frac{1}{2}\left(MN\right)\left(NP\right)=\frac{1}{2}\left(NQ\right)\left(MP\right)$

$\therefore \left(MN\right)\left(NP\right)=\left(NQ\right)=\left(MP\right)$

$\therefore MP=9+4=13$

$\therefore M{N}^2+N{P}^2=169$

$\therefore M{N}^2=81+N{Q}^2$

$\therefore N{P}^2=16+N{Q}^2$

$\therefore 81+16+2N{Q}^2=169$

$\therefore N{Q}^2=\frac{169-97}{2}$

$\therefore N{Q}^2=36$

$\therefore NQ=6$cm.