The length of the rectangle is 15 dm and the width is 13 dm.
Let's assume that the length of the rectangle is "L" and the width is "W".
From the problem statement, we have two pieces of information:
The area of the rectangle is 195 dm²:
Area = Length x Width
195 dm² = L x W
The width is two less than the length:
W = L - 2
Now, we can substitute the second equation into the first equation to eliminate W and get an equation with only one variable:
195 dm² = L x (L - 2)
Simplifying the equation:
195 dm² = L² - 2L
L² - 2L - 195 dm² = 0
To solve for L, we can use the quadratic formula:
L = (-b ± √(b² - 4ac)) / 2a
Where a = 1, b = -2, and c = -195.
L = (2 ± √(2² + 4 x 1 x 195)) / 2 x 1
L = (2 ± √4 + 780) / 2
L = (2 ± √784) / 2
L = (2 ± 28) / 2
L = 15 or L = -13
Since the length can't be negative, the length of the rectangle is L = 15 dm.
Now we can use the equation W = L - 2 to find the width:
W = 15 dm - 2 dm
W = 13 dm
Therefore, the length of the rectangle is 15 dm and the width is 13 dm.
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