Question

In PQR, if angle Q= 90 degrees and QS is perpendicular to PR, then find the value of 1/PQ square + 1/QR square ​

Answer 1

Given:  Angle Q= 90 degrees, QS is perpendicular to PR.

To find:  1/PQ square + 1/QR square ​

Solution:

• As we have given that, QS is perpendicular to PR, so triangle QSP and triangle QSR are right angled triangle.
• Now by applying area of triangle in the triangle ABC, we get:
• area of triangle ABC = 1/2 x base x height

             = 1/2 x QR x PQ

• Also area of triangle ABC can be

             = 1/2 x PR x QS

• so equating both, we get:

            QR x PQ = PR x QS

            PR = (QR x PQ )/ QS

• Now in triangle ABC, applying pythagoras theorem, we get:

            PQ² + QR² = PR²

            PQ² + QR² = {(QR x PQ )/ QS }²

            PQ² + QR² = {(QR² x PQ² )/ QS² }

            PQ² + QR² / (QR² x PQ² ) = 1 / QS²

• Separating the terms, we get:

            PQ² / (QR² x PQ² ) +  QR² / (QR² x PQ² ) = 1 / QS²

            1 / QR² + 1 / PQ² = 1 / QS²

Answer:

             the value of 1/PQ square + 1/QR square ​is 1 / QS².

Answer 2

Answer:

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