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 $\alpha :\beta \text{ then }\alpha +\beta =$

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 $$\begin{gathered} \mathrm{\Delta }OMN\text{ is }\frac{3a^2}{8} \\ \frac{Area of square}{Area of triangle}\text{=}\frac{a^2}{{\displaystyle \frac{3a^2}{8}}}\text{=}\frac{8}{3} \\ \text{Now α+β=11} \end{gathered}$$