The length of the rectangle is 12 dm longer than the width. If the length is increased by 3 dm and the width by 2 dm

| December 30, 2020 | education

The length of the rectangle is 12 dm longer than the width. If the length is increased by 3 dm and the width by 2 dm, then the area of the rectangle will increase by 80 dm2. Find the length and width of the rectangle.

Let x dm be the width of the rectangle, then the length of the rectangle is (x + 12) dm. The area of a rectangle is equal to the product of length and width. So S = x * (x + 12)
If the length is increased by 3 dm, then we get (x + 12 + 3) dm, that is, (x + 15) dm. If the width is increased by 2 dm, then we get (x + 2) dm. The area has increased by 80 dm ^ 2, which means the area is x (x + 12) + 80. So,
(x + 15) (x + 2) = x (x + 12) + 80
x ^ 2 + 15x + 2x + 30 = x ^ 2 + 12x + 80
x ^ 2 – x ^ 2 + 17x – 12x = 80 – 30
5x = 50
x = 10
The width is 10 dm, which means the length is 10 + 12 = 22 dm.
Answer: 10 dm and 22 dm



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