In quadrilateral PQRS,PQ = QR = 10 and PR = 1 | vedclass

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(C) In $	riangle PQR$,we are given $PQ = 10$,$QR = 10$,and $PR = 10\sqrt{2}$.Check if $	riangle PQR$ is a right-angled triangle using the converse o

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square

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(C) In $	riangle PQR$,we are given $PQ = 10$,$QR = 10$,and $PR = 10\sqrt{2}$.Check if $	riangle PQR$ is a right-angled triangle using the converse of the Pythagoras theorem:$PQ^2 + QR^2 = 10^2 + 10^2 = 100 + 100 = 200$.$PR^2 = (10\sqrt{2})^2 = 100 	imes 2 = 200$.Since $PQ^2 + QR^2 = PR^2$,$	riangle PQR$ is a right-angled triangle with $\angle PQR = 90^\circ$.In a quadrilateral,if two adjacent sides are equal and the angle between them is $90^\circ$,and no further information is provided about the other sides,the most specific classification based on the given properties of the triangle formed by the diagonal is that it contains a right angle. However,among the given options,if $PQ = QR = 10$,it satisfies the condition for a rhombus or a square. Given the diagonal $PR = 10\sqrt{2}$ confirms the right angle at $Q$. Without information about $PS$ and $RS$,but assuming it is a standard geometry problem where $PQRS$ is a rhombus with a right angle,it becomes a square.

In quadrilateral $PQRS$,$PQ = QR = 10$ and $PR = 10\sqrt{2}$. Then,quadrilateral $PQRS$ is a $\ldots \ldots \ldots$

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