In Delta PQR,angle P is a right angle and ove | vedclass

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(A) In $\Delta PQR$,$\angle P = 90^{\circ}$ and $\overline{PX} \perp \overline{QR}$.According to the geometric mean theorem (or property of altitude t

Vedclass · Accepted Answer

$6$

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(A) In $\Delta PQR$,$\angle P = 90^{\circ}$ and $\overline{PX} \perp \overline{QR}$.According to the geometric mean theorem (or property of altitude to the hypotenuse),in a right-angled triangle,the altitude to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other.Specifically,$\Delta QXP \sim \Delta QPR$ and $\Delta PXR \sim \Delta QPR$.From the similarity $\Delta QXP \sim \Delta QPR$,we have $\frac{QX}{PQ} = \frac{PQ}{QR}$.Thus,$PQ^2 = QX \cdot QR$.Given $QX = 4$ and $RX = 5$,we have $QR = QX + RX = 4 + 5 = 9$.Substituting these values,$PQ^2 = 4 \cdot 9 = 36$.Therefore,$PQ = \sqrt{36} = 6$.

In $\Delta PQR$,$\angle P$ is a right angle and $\overline{PX}$ is an altitude. If $QX = 4$ and $RX = 5$,then $PQ = \ldots \ldots$

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