In Delta PQR,mangle Q = 90^circ. If PQ = 8 an | vedclass

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(D) In $\Delta PQR$,since $m\angle Q = 90^\circ$,the triangle is a right-angled triangle.According to the Pythagorean theorem,the square of the hypote

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$17$

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(D) In $\Delta PQR$,since $m\angle Q = 90^\circ$,the triangle is a right-angled triangle.According to the Pythagorean theorem,the square of the hypotenuse is equal to the sum of the squares of the other two sides.Here,$PR$ is the hypotenuse.So,$PR^2 = PQ^2 + QR^2$.Given $PQ = 8$ and $QR = 15$.$PR^2 = 8^2 + 15^2$.$PR^2 = 64 + 225$.$PR^2 = 289$.Taking the square root on both sides,$PR = \sqrt{289} = 17$.

In $\Delta PQR$,$m\angle Q = 90^\circ$. If $PQ = 8$ and $QR = 15$,then $PR = \ldots$

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