In Delta PQR,m angle Q = 90^{circ} and overli | vedclass

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(B) Let $PD = x$. Since $PR = 34$,then $DR = 34 - x$.In $\Delta PQR$,$\overline{QD}$ is the altitude to the hypotenuse $PR$. By the Geometric Mean The

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$5 \sqrt{34}$

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(B) Let $PD = x$. Since $PR = 34$,then $DR = 34 - x$.In $\Delta PQR$,$\overline{QD}$ is the altitude to the hypotenuse $PR$. By the Geometric Mean Theorem (or properties of similar triangles),$QD^2 = PD \cdot DR$.Substituting the given values: $15^2 = x(34 - x)$.$225 = 34x - x^2$.$x^2 - 34x + 225 = 0$.Solving the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{34 \pm \sqrt{34^2 - 4(1)(225)}}{2} = \frac{34 \pm \sqrt{1156 - 900}}{2} = \frac{34 \pm \sqrt{256}}{2} = \frac{34 \pm 16}{2}$.Two possible values for $x$ are $x = \frac{50}{2} = 25$ or $x = \frac{18}{2} = 9$.In $\Delta PQD$,by the Pythagorean theorem,$PQ^2 = QD^2 + PD^2$.If $PD = 9$,then $PQ^2 = 15^2 + 9^2 = 225 + 81 = 306$. $PQ = \sqrt{306} = \sqrt{9 \cdot 34} = 3 \sqrt{34}$.If $PD = 25$,then $PQ^2 = 15^2 + 25^2 = 225 + 625 = 850$. $PQ = \sqrt{850} = \sqrt{25 \cdot 34} = 5 \sqrt{34}$.Comparing with the given options,$5 \sqrt{34}$ is the correct answer.

In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QD}$ is an altitude to the hypotenuse $PR$. If $QD = 15$ and $PR = 34$,find $PQ$.

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