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The length of the rectangle is one more than the width.
If the dimensions are both decreased by 2 units, the area of the new rectangle
is 30 sq. units less than the area of the original rectangle.
Find the area of the original rectangle.
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Let w be the width of the rectangle; then its length is (w+1) unit, according to the condition.
The hypothetical rectangle has the width of (w-2) units and the length of ((w+1)-2) = (w-1) units;
so its area is this product (w-2)*(w-1).
And now, we have THIS equation for the areas
(w-2)*(w-1) = w*(w+1) - 30.
Simplify; then solve for w
w^2 - 2w - w + 2 = w^2 + w - 30
-3w + 2 = w - 30
2 + 30 = w + 3w
32 = 4w
w = 32/4 = 8
Thus we found that the dimensions of the original rectangle are w= 8 (the width) and 8+1 = 9 (the length).
Hence, the area of the original triangle is 8*9 = 72 square units.
Solved.
To CHECK : the dimensions of the smaller rectangle are 6 and 7 units; its area is 6*7 = 42 square units;
it is 30 units less than 72 square units, the area of the original rectangle. ! Checked !