The smallest possible area of the new rectangle is 52 square units.
In the question, we are given that a rectangle has side lengths of integral value.
We assume the lengths to be x and y units respectively, such that x and y are integers.
In the new arrangement, one pair is increased by 30%, and the other pair is decreased by 20%, such that the new sides are also integers.
We assume a 30% increase in x.
New side = x + 30% of x = x + 30/100 of x = x + 3x/10 = 13x/10.
For this to be an integer, x needs to be a multiple of 10.
Since we are considering the smallest area, we will consider the smallest multiple of 10, that is, 10 itself.
Hence, side 1 = 13 units.
We assume a 20% decrease in y.
New side = y - 20% of y = y - 20/100 of y = y - y/5 = 4x/5.
For this to be an integer, y needs to be a multiple of 5.
Since we are considering the smallest area, we will consider the smallest multiple of 5, that is, 5 itself.
Hence, side 2 = 4 units.
Now, the area of this rectangle is the product of its sides = 13*4 sq. units = 52 sq. units.
Therefore, the smallest possible area of the new rectangle is 52 square units.
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