In Delta PQR,m angle Q = 90^circ. If PQ = 5 a | vedclass

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(C) Given that $\Delta PQR$ is a right-angled triangle with $m \angle Q = 90^\circ$.$PQ$ and $QR$ are the legs,and $PR$ is the hypotenuse.Using the Py

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$30$

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(C) Given that $\Delta PQR$ is a right-angled triangle with $m \angle Q = 90^\circ$.$PQ$ and $QR$ are the legs,and $PR$ is the hypotenuse.Using the Pythagorean theorem: $PQ^2 + QR^2 = PR^2$.Substitute the given values: $5^2 + QR^2 = 13^2$.$25 + QR^2 = 169$.$QR^2 = 169 - 25 = 144$.$QR = \sqrt{144} = 12$.The area of a right-angled triangle is given by $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.$	ext{Area} = \frac{1}{2} 	imes PQ 	imes QR = \frac{1}{2} 	imes 5 	imes 12$.$	ext{Area} = 5 	imes 6 = 30$.

In $\Delta PQR$,$m \angle Q = 90^\circ$. If $PQ = 5$ and $PR = 13$,find the area of $\Delta PQR$.

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