square PQRS is a rectangle. PQ + QR = 7 and t | vedclass

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Question

(A) Given that $\square PQRS$ is a rectangle.Area of rectangle $PQRS = PQ \cdot QR = 12$.Sum of adjacent sides $PQ + QR = 7$.In $	riangle PQR$,$\angl

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Given that $\square PQRS$ is a rectangle.Area of rectangle $PQRS = PQ \cdot QR = 12$.Sum of adjacent sides $PQ + QR = 7$.In $	riangle PQR$,$\angle Q = 90^\circ$.By Pythagoras theorem,$PR^2 = PQ^2 + QR^2$.We know that $(PQ + QR)^2 = PQ^2 + QR^2 + 2(PQ \cdot QR)$.Therefore,$PQ^2 + QR^2 = (PQ + QR)^2 - 2(PQ \cdot QR)$.Substituting the given values: $PR^2 = (7)^2 - 2(12)$.$PR^2 = 49 - 24 = 25$.Taking the square root on both sides,$PR = 5$.

$\square PQRS$ is a rectangle. $PQ + QR = 7$ and the area of $\square PQRS$ is $12$. Then,$PR = \ldots$

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