OPQR is a square and point Q is (alpha, alpha | vedclass

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(C) The vertices of the square $OPQR$ are $O(0, 0)$,$P(\alpha, 0)$,$Q(\alpha, \alpha)$,and $R(0, \alpha)$.Area of square $OPQR = \alpha^2$.$M$ is the

Vedclass · Accepted Answer

$8 : 3$

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(C) The vertices of the square $OPQR$ are $O(0, 0)$,$P(\alpha, 0)$,$Q(\alpha, \alpha)$,and $R(0, \alpha)$.Area of square $OPQR = \alpha^2$.$M$ is the midpoint of $PQ$,so $M = (\frac{\alpha + \alpha}{2}, \frac{0 + \alpha}{2}) = (\alpha, \frac{\alpha}{2})$.$N$ is the midpoint of $QR$,so $N = (\frac{0 + \alpha}{2}, \frac{\alpha + \alpha}{2}) = (\frac{\alpha}{2}, \alpha)$.The coordinates of the vertices of $	riangle OMN$ are $O(0, 0)$,$M(\alpha, \frac{\alpha}{2})$,and $N(\frac{\alpha}{2}, \alpha)$.Area of $	riangle OMN = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$= \frac{1}{2} |0(\frac{\alpha}{2} - \alpha) + \alpha(\alpha - 0) + \frac{\alpha}{2}(0 - \frac{\alpha}{2})|$$= \frac{1}{2} |\alpha^2 - \frac{\alpha^2}{4}| = \frac{1}{2} |\frac{3\alpha^2}{4}| = \frac{3\alpha^2}{8}$.Ratio = $\frac{	ext{Area of square}}{	ext{Area of } 	riangle OMN} = \frac{\alpha^2}{\frac{3\alpha^2}{8}} = \frac{8}{3} = 8 : 3$.

$OPQR$ is a square and point $Q$ is $(\alpha, \alpha)$. If $M$ and $N$ are the midpoints of sides $PQ$ and $QR$ respectively,then what is the ratio of the area of the square to the area of triangle $OMN$?

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