In Delta PQR,mangle Q = 90^{circ} and T is th | vedclass

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(D) Given that $\Delta PQR$ is a right-angled triangle with $m\angle Q = 90^{\circ}$.Using the Pythagorean theorem,$PR^2 = PQ^2 + QR^2$.Substituting t

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$5$

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(D) Given that $\Delta PQR$ is a right-angled triangle with $m\angle Q = 90^{\circ}$.Using the Pythagorean theorem,$PR^2 = PQ^2 + QR^2$.Substituting the given values,$PR^2 = 6^2 + 8^2 = 36 + 64 = 100$.Thus,$PR = \sqrt{100} = 10$.In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.Since $T$ is the midpoint of $\overline{PR}$,$\overline{QT}$ is the median to the hypotenuse $\overline{PR}$.Therefore,$QT = \frac{1}{2} 	imes PR = \frac{1}{2} 	imes 10 = 5$.

In $\Delta PQR$,$m\angle Q = 90^{\circ}$ and $T$ is the midpoint of $\overline{PR}$. If $PQ = 6$ and $QR = 8$,then $QT = \ldots$

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