In Delta PQR, m angle Q = 90^{circ}. If PR = | vedclass

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(A) In $\Delta PQR$,since $m \angle Q = 90^{\circ}$,the triangle is a right-angled triangle.According to the Pythagoras theorem,the square of the hypo

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$15$

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(A) In $\Delta PQR$,since $m \angle Q = 90^{\circ}$,the triangle is a right-angled triangle.According to the Pythagoras theorem,the square of the hypotenuse is equal to the sum of the squares of the other two sides.Here,$PR$ is the hypotenuse (the side opposite to the $90^{\circ}$ angle).Thus,$PR^{2} = PQ^{2} + QR^{2}$.Substituting the given values: $17^{2} = 8^{2} + QR^{2}$.$289 = 64 + QR^{2}$.$QR^{2} = 289 - 64$.$QR^{2} = 225$.Taking the square root on both sides,$QR = \sqrt{225} = 15$.

In $\Delta PQR, m \angle Q = 90^{\circ}$. If $PR = 17$ and $PQ = 8$,then $QR = \ldots$

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