In Delta PQR,m angle Q : m angle R : m angle | vedclass

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(C) Given that in $\Delta PQR$,$m \angle Q : m \angle R : m \angle P = 1 : 2 : 1$.Let the angles be $x, 2x, x$.Since the sum of angles in a triangle i

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$2 \sqrt{3}$

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(C) Given that in $\Delta PQR$,$m \angle Q : m \angle R : m \angle P = 1 : 2 : 1$.Let the angles be $x, 2x, x$.Since the sum of angles in a triangle is $180^{\circ}$,we have $x + 2x + x = 180^{\circ}$,which gives $4x = 180^{\circ}$,so $x = 45^{\circ}$.Thus,$m \angle Q = 45^{\circ}$,$m \angle R = 90^{\circ}$,and $m \angle P = 45^{\circ}$.Since $m \angle P = m \angle Q$,the sides opposite to these angles are equal,so $QR = PR$.In the right-angled $\Delta PQR$ (where $m \angle R = 90^{\circ}$),by the Pythagorean theorem:$PQ^2 = PR^2 + QR^2$Since $QR = PR$,we have $PQ^2 = PR^2 + PR^2 = 2PR^2$.Given $PQ = 2 \sqrt{6}$,so $PQ^2 = (2 \sqrt{6})^2 = 4 	imes 6 = 24$.Therefore,$24 = 2PR^2$,which implies $PR^2 = 12$.Taking the square root,$PR = \sqrt{12} = 2 \sqrt{3}$.

In $\Delta PQR$,$m \angle Q : m \angle R : m \angle P = 1 : 2 : 1$. If $PQ = 2 \sqrt{6}$,then $PR = \ldots$

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