There is a rectangular sheet of dimension (2m | vedclass

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(D) rectangle is formed by choosing two horizontal lines and two vertical lines. To have odd side lengths,the distance between the chosen horizontal l

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$m^2n^2$

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(D) rectangle is formed by choosing two horizontal lines and two vertical lines. To have odd side lengths,the distance between the chosen horizontal lines must be odd,and the distance between the chosen vertical lines must be odd.Let the horizontal lines be $x_0, x_1, \dots, x_{2m-1}$ and vertical lines be $y_0, y_1, \dots, y_{2n-1}$.For the horizontal side length to be odd,we must choose two lines $x_i$ and $x_j$ such that $|i-j|$ is odd. This means one line must have an even index and the other must have an odd index.There are $m$ even indices $(0, 2, \dots, 2m-2)$ and $m$ odd indices $(1, 3, \dots, 2m-1)$.The number of ways to choose one even index and one odd index is $m 	imes m = m^2$.Similarly,for the vertical side length to be odd,we must choose one even index and one odd index from the $n$ even and $n$ odd indices available,which is $n 	imes n = n^2$.Therefore,the total number of such rectangles is $m^2 	imes n^2 = m^2n^2$.

There is a rectangular sheet of dimension $(2m-1) \times (2n-1)$,where $m > 0, n > 0$. It has been divided into squares of unit area by drawing lines perpendicular to the sides. Find the number of rectangles having sides of odd unit length.

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