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If the length of a rectangle is 8 cm more than 3 times its width and the perimeter is 64 cm what are the dimensions of the rectangle?
Wiki User
∙ 15y ago
This is an algebra problem.
Let's call the length L and the width W
L = 3W + 8 is an equation describing the first part of the problem
2L + 2W = 64 is an equation describing the second part of the problem.
Now, lets substitute the (2L) in the second equation with twice the value of L in the first equation
2(3W + 8) + 2W = 64, and now we'll work out the contents of the brackets.
6W + 16 + 2W = 64, now add the values of W
8W + 16 = 64, now bring all the numeric values to the same side of the equation
8W = 64 - 16
8W = 48, now divide both sides by 8
W = 6
Now we can substitute 6 for W in the first equation
L = 3(6) + 8, now we'll work out the contents of the brackets
L = 18 + 8 now we'll add the numbers
L = 26
So now we have two values and we need to ensure that they add up to 64
2(26) + 2(6) = 52 + 12 = 64
So, the rectangle is 26 x 6
Wiki User
∙ 15y ago
Anonymous
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The length and width of a triangle are each multiplied by 4 find how the perimeter and the area of the rectangle change?
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