SOLUTION: The perimeter of a rectangle is 54cm. Two times the height is 3cm more than the width. Find the length of the height and the length of the width.
this is the way i've been tryin
Question 144606: The perimeter of a rectangle is 54cm. Two times the height is 3cm more than the width. Find the length of the height and the length of the width.
this is the way i've been trying to solve this problem: P=2l+2w, right?
we have the perimeter. 54=(2L+3)+2W. Is this right? I'm stuck! Found 3 solutions by ankor@dixie-net.com, shahid, Alan3354:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! The perimeter of a rectangle is 54cm. Two times the height is 3cm more than the width. Find the length of the height and the length of the width.
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Let x = the width
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It says,"Two times the height is 3cm more than the width." so we can say:
2h = (x + 3}
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Given perimeter:
2h + 2w = 54
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Substitute (x+3) for 2h and x for w and we have:
(x+3) + 2x = 54
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3x = 54 - 3
x =
x = 17 cm is the width.
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Find the length from the equation 2h = (x+3)
2h = 17 + 3
2h = 20
h = 10 cm
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Check solution by finding the perimeter using these values
You can put this solution on YOUR website! 54 = 2H+2W (Height and Width) Length would work the same, but it says height.
2H = W+3, so W = 2H - 3
Sub for W
54 = 2H + 2(2H-3)
54 = 2H + 4H - 6
60 = 6H
H = 10
W = 17
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