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OPQR is a square and M, N are the middle points of the sides PQ and QR, respectively, Then the ratio of the area of the square to that of triangle OMN is α:β then α+β=

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detailed solution

11

Let the coordinates of vertices O, P, Q, R be (0, 0), (a, 0), (a, a), (0, a), respectively. Then, we get the coordinates of M as (a, a/2) and those of N as (a/2, a). Therefore, the area of ΔOMN is 3a28 Area of squareArea of triangle=a23a28=83 Now α+β=11


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