Question

In A PQR, Q = 90°,
QN , PR, QR = b andA (A PQR) = a
then show that
QN=2ab/
$\sqrt{ {b}^{4}+ {4a}^{2} }$
​

Answer 1

Given that

              PQR is a triangle and QM is perpendicular on PR

Also,

                     PR^2 - PQ^2 = QR^2

Now in traingle QMR

                                     QR^2 = QM^2+MR^2

Thus fro  above two equations for QR

 We get

                    PR^2 - PQ^2 = QM^2 + MR^2

                        QM^2 = PR^2 - PQ^2 -MR^2

                         QM^2 = (PM+MR)^2 - PQ^2 - MR^2

                           QM^2 = PM^2 + MR^2 + 2PM*MR - PQ^2 - MR^2

                            QM^2 = PM^2 + 2 PM^MR - PQ^2

                            QM^2 = PQ^2 - QM^2 + 2PM*MR - PQ^2

                           thus,         2QM^2 = 2 PM * MR

Answer 2

Answer:

Given that

              PQR is a triangle and QM is perpendicular on PR

Also,

                     PR^2 - PQ^2 = QR^2

Now in traingle QMR

                                     QR^2 = QM^2+MR^2

Thus fro  above two equations for QR

 We get

                    PR^2 - PQ^2 = QM^2 + MR^2

                        QM^2 = PR^2 - PQ^2 -MR^2

                         QM^2 = (PM+MR)^2 - PQ^2 - MR^2

                           QM^2 = PM^2 + MR^2 + 2PM*MR - PQ^2 - MR^2

                            QM^2 = PM^2 + 2 PM^MR - PQ^2

                            QM^2 = PQ^2 - QM^2 + 2PM*MR - PQ^2

                           thus,         2QM^2 = 2 PM * MR