Question 126583: The length of a rectangle is 50 ft. The perimeter of the rectangle is 275 ft. more than the width of the rectangle. Find the width Found 2 solutions by checkley71, marcsam823:Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website! 2*50+2W=275-W
100+2W=275-W
2W+W=275-100
3W=175
W=175/3
W=58.33 ANSWER FOR THE WIDTH.
PROOF:
2*50+2*58.33=275-58.33
100+116.66=216.67
216.66=216.67
You can put this solution on YOUR website! 1. This problem involves writing two expressions for the perimeter of the rectangle in terms of the width (w) and setting them equal to each other.
a. Let the length of the rectangle = 50
b. Let the width of the rectangle = w
c. Let the perimeter = (given)
d. Let the perimeter = (two times the length 2(50) + two times the width)
2. Set c. and d. equal to each other:
3. Check:
Also:
Our answer is correct.
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