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The length of a rectangle is 8 cm more than two times the width. If the perimeter of the rectangle is 40 cm, what are the dimensions?

User Rwx
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1 Answer

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Let's call x to the length of the rectangle and y to its width, then, from the given information we can formulate the following equations:

Since the length "x" equals 8 times more than two times the width "y", we just have to add 8 to the product of y and 2 and then make it equal to x, then we get:

x = 8 + 2×y

We can also formulate another equation, that is the perimeter of the rectangle, the perimeter of a rectangle is given by the following expression:

P = 2y + 2x

In this case, P equals 40, by replacing 40 for P and 8 + 2y into the above equation, we get:

40 = 2y + 2(8 + 2y)

40 = 2y + 2×8 + 2×2y

40 = 2y + 16 + 4y

Combining like terms:

40 = 6y + 16

Solving for y:

40 -16 = 6y + 16 - 16

24 = 6y

24/6 = 6y/6

4 = y

y = 4

Then y equals 4. In order to find the value of x, we just have to replace 4 for y into the equation x = 8 + 2×y, like this:

x = 8 + 2×4

x = 8 + 8

x = 16

Then, x equals 16 and y equals 4. Since x represents the length and y represents the width, then the length of the rectangle equals 16 cm and its width equals 4.

User Hetzroni
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