OPQR is a square and M and N are the midpoint | vedclass

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(C) Let the side length of the square be $a$. Let the coordinates of vertices $O, P, Q, R$ be $(0,0), (a, 0), (a, a), (0, a)$ respectively.The area of

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$8 : 3$

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(C) Let the side length of the square be $a$. Let the coordinates of vertices $O, P, Q, R$ be $(0,0), (a, 0), (a, a), (0, a)$ respectively.The area of the square $OPQR$ is $a^2$.$M$ is the midpoint of $PQ$,so its coordinates are $\left(a, \frac{a}{2}\right)$.$N$ is the midpoint of $QR$,so its coordinates are $\left(\frac{a}{2}, a\right)$.The area of $\Delta OMN$ with vertices $(0,0), (a, \frac{a}{2}), (\frac{a}{2}, a)$ is given by:Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$Area $= \frac{1}{2} |0(\frac{a}{2} - a) + a(a - 0) + \frac{a}{2}(0 - \frac{a}{2})|$Area $= \frac{1}{2} |a^2 - \frac{a^2}{4}| = \frac{1}{2} |\frac{3a^2}{4}| = \frac{3a^2}{8}$.The ratio of the area of the square to the area of $\Delta OMN$ is $a^2 : \frac{3a^2}{8} = 1 : \frac{3}{8} = 8 : 3$.

$OPQR$ is a square and $M$ and $N$ are the midpoints of sides $PQ$ and $QR$ respectively. Then,the ratio of the area of the square to the area of the triangle $OMN$ is:

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