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# A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length

- m
^{2 }– n^{2}
- m (n + 1) n (n + 1)
- 4
^{m + n – 2}
- m
^{2}n^{2}

^{2 }– n^{2}^{m + n – 2}^{2}n^{2}Hint:

### The internal angles of a rectangle, which has four sides, are all exactly 90 degrees. At each corner or vertex, the two sides come together at a straight angle. The rectangle differs from a square because its two opposite sides are of equal length. We have to find the number of rectangles possible with odd side length.

## The correct answer is: m^{2}n^{2}

### Now we have given that a rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having an edge length of one unit.

There are 2n horizontal lines and 1 2 m vertical lines (numbered 1,2.......2n).

Two horizontal lines, one with an even number and one with an odd number, as well as two vertical lines must be chosen in order to create the necessary rectangle.

Then the number of rectangles will be:

(1+3+5+......+(2m−1))(1+3+5+......+(2n−1))=m^{2}n^{2}

So it is m^{2}n^{2}.

Here we used the concept of number system and the rectangle, we can also solve it by permutation and combination. herefore, we get the number of rectangles possible with odd side length = m^{2}n^{2}.

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