In Delta PQR,mangle Q = 90^{circ},PR = a^{2} | vedclass

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(C) In $\Delta PQR$,since $m\angle Q = 90^{\circ}$,the triangle is a right-angled triangle.By the Pythagoras theorem,$PR^{2} = PQ^{2} + QR^{2}$.Theref

Vedclass · Accepted Answer

$2ab$

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(C) In $\Delta PQR$,since $m\angle Q = 90^{\circ}$,the triangle is a right-angled triangle.By the Pythagoras theorem,$PR^{2} = PQ^{2} + QR^{2}$.Therefore,$PQ^{2} = PR^{2} - QR^{2}$.Substituting the given values:$PQ^{2} = (a^{2} + b^{2})^{2} - (a^{2} - b^{2})^{2}$.Using the algebraic identity $(x + y)^{2} - (x - y)^{2} = 4xy$,where $x = a^{2}$ and $y = b^{2}$:$PQ^{2} = 4(a^{2})(b^{2}) = 4a^{2}b^{2}$.Taking the square root on both sides:$PQ = \sqrt{4a^{2}b^{2}} = 2ab$.

In $\Delta PQR$,$m\angle Q = 90^{\circ}$,$PR = a^{2} + b^{2}$ and $QR = a^{2} - b^{2}$. Then,$PQ = \ldots$ $(a, b \in R^{+}; a > b)$

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