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

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(D) In $\Delta PQR$,since $m \angle Q = 90^{\circ}$ and $\overline{QS} \perp \overline{PR}$,by the properties of right-angled triangles,we have $PQ^2

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(D) In $\Delta PQR$,since $m \angle Q = 90^{\circ}$ and $\overline{QS} \perp \overline{PR}$,by the properties of right-angled triangles,we have $PQ^2 = PS \cdot PR$ and $QR^2 = SR \cdot PR$.Subtracting these two equations: $PQ^2 - QR^2 = (PS - SR) \cdot PR$.Given $PQ^2 - QR^2 = 260$ and $PS - SR = 10$.Substituting these values: $260 = 10 \cdot PR$.Therefore,$PR = \frac{260}{10} = 26$.

In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QS}$ is an altitude. If $PS - SR = 10$ and $PQ^2 - QR^2 = 260$,then $PR = \dots$

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