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

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(D) In $\Delta PQR$,$\angle Q = 90^{\circ}$ and $\overline{QD} \perp \overline{PR}$.By the property of geometric mean in a right-angled triangle,we ha

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$3$

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(D) In $\Delta PQR$,$\angle Q = 90^{\circ}$ and $\overline{QD} \perp \overline{PR}$.By the property of geometric mean in a right-angled triangle,we have $QD^2 = PD 	imes DR$.Also,in $\Delta PQD$ and $\Delta PQR$,by similarity criteria,we have $PQ^2 = PD 	imes PR$.Given $PD = 9 DR$,we can write $PR = PD + DR = 9 DR + DR = 10 DR$.Substituting these into the expression for $PQ^2$:$PQ^2 = (9 DR) 	imes (10 DR) = 90 DR^2$.Similarly,in $\Delta QDR$ and $\Delta PQR$,we have $QR^2 = DR 	imes PR$.$QR^2 = DR 	imes (10 DR) = 10 DR^2$.Now,find the ratio $\frac{PQ^2}{QR^2} = \frac{90 DR^2}{10 DR^2} = 9$.Taking the square root on both sides,we get $\frac{PQ}{QR} = \sqrt{9} = 3$.Therefore,$PQ = 3 	imes QR$.

In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QD}$ is an altitude. If $PD = 9 DR$,then $PQ = \ldots \ldots \ldots \times QR$.

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