In Delta PQR,angle Q = 90^{circ},QR = 21 text | vedclass

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

Vedclass · Accepted Answer

$210$

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(D) Given that $\Delta PQR$ is a right-angled triangle with $\angle Q = 90^{\circ}$.According to the Pythagorean theorem,$PQ^2 + QR^2 = PR^2$.Substituting the given values: $PQ^2 + 21^2 = 29^2$.$PQ^2 + 441 = 841$.$PQ^2 = 841 - 441 = 400$.$PQ = \sqrt{400} = 20 	ext{ cm}$.The area of a right-angled triangle is given by $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.Here,$	ext{base} = QR = 21 	ext{ cm}$ and $	ext{height} = PQ = 20 	ext{ cm}$.$	ext{Area} = \frac{1}{2} 	imes 21 	imes 20 = 21 	imes 10 = 210 	ext{ cm}^2$.

In $\Delta PQR$,$\angle Q = 90^{\circ}$,$QR = 21 \text{ cm}$ and $PR = 29 \text{ cm}$,then find the area of $\Delta PQR$ in $\text{cm}^2$.

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