Question

In $\Delta PQR,\angle PQR={90}^o$
seg $QS\perp$ hypotenuse $PR$, if $PS=4$
$SR=9$ find $QS$

Answer 1

We have using Pythagorus theorem,

$Q{S}^2=PS.SR$......(1) [ Can be proved by applying Pythagorus theorem on the triangles $\Delta PQR$, $\Delta PQS$, $\Delta QSR$]

Given $PS=4,SR=9$.

Then form (1) we have

$Q{S}^2=4\times 9$

or, $Q{S}^2=36$

or, $QS=6$,