SOLUTION: The length of a rectangle is 5 cm longer than the width. The perimeter of the rectangle is 50 cm. Find the length and width of the rectangle.

Algebra ->  Rectangles -> SOLUTION: The length of a rectangle is 5 cm longer than the width. The perimeter of the rectangle is 50 cm. Find the length and width of the rectangle.      Log On


   

Question 866589: The length of a rectangle is 5 cm longer than the width. The perimeter of the rectangle is 50 cm. Find the length and width of the rectangle.
Found 2 solutions by JulietG, josgarithmetic:
Answer by JulietG(1812) About Me  (Show Source):
You can put this solution on YOUR website!
2L + 2W = 50
L = W+5
Substitute the value of L from the second equation into the first.
2(W+5) + 2W = 50
2W+10 + 2W = 50
4W + 10 = 50
Subtract 10 from each side.
4W = 40
Divide each side by 4
W = 10
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If the width is 10, then the length is 10+5, or 15.

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
w=width
L=length
Formula for perimeter, p=2(w+L)
Given: L=w%2B5
p=50

2%28w%2BL%29=p
w%2BL=p%2F2
L=p%2F2-w
Substitute,
w%2B5=p%2F2-w
w%2Bw=p%2F2-5
2w=p%2F2-5
w=%28p%2F2-5%29%2F2
highlight_green%28w=p%2F4-5%2F2%29
Actual correct steps can vary some, but w, if in symbolic form must be equivalent to the green-outlined formula shown. Just substitute for the given p value of 50, and compute w.
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You could do all this without being in pure symbolic form - it's up to you; but it depends on how many of this general type of rectangle exercise you have. This is a very frequent type of exercise problem and generalizing its solution can be more efficient if you have many of these.