In Delta PQR,mangle Q = 90^{circ} and overlin | vedclass

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(C) $1$. In right-angled $\Delta PQR$,by Pythagoras theorem,$PR^2 = PQ^2 + QR^2$.$2$. $PR^2 = 8^2 + 6^2 = 64 + 36 = 100$,so $PR = 10$.$3$. The area of

Vedclass · Accepted Answer

$8.64$

Vedclass · Answer

(C) $1$. In right-angled $\Delta PQR$,by Pythagoras theorem,$PR^2 = PQ^2 + QR^2$.$2$. $PR^2 = 8^2 + 6^2 = 64 + 36 = 100$,so $PR = 10$.$3$. The area of $\Delta PQR = \frac{1}{2} 	imes PQ 	imes QR = \frac{1}{2} 	imes 8 	imes 6 = 24$.$4$. Also,area of $\Delta PQR = \frac{1}{2} 	imes PR 	imes QM = \frac{1}{2} 	imes 10 	imes QM = 5 	imes QM$.$5$. Equating the areas: $5 	imes QM = 24$,so $QM = 4.8$.$6$. In $\Delta QMR$,by Pythagoras theorem,$MR^2 = QR^2 - QM^2 = 6^2 - (4.8)^2 = 36 - 23.04 = 12.96$.$7$. $MR = \sqrt{12.96} = 3.6$.$8$. Area of $\Delta QMR = \frac{1}{2} 	imes QM 	imes MR = \frac{1}{2} 	imes 4.8 	imes 3.6 = 2.4 	imes 3.6 = 8.64$.

In $\Delta PQR$,$m\angle Q = 90^{\circ}$ and $\overline{QM}$ is an altitude to the hypotenuse $PR$. If $PQ = 8$ and $QR = 6$,find the area of $\Delta QMR$.

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