SOLUTION: The width of the rectangle is 4/5 of its dimension. Find the dimension of the rectangle if the perimeter become 62cm when each dimension is increased by 2cm.

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Question 1105957: The width of the rectangle is 4/5 of its dimension. Find the dimension of the rectangle if the perimeter become 62cm when each dimension is increased by 2cm.
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
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The width of the rectangle is 4/5 of its cross%28dimension%29 LENGTH. Find the dimension of the rectangle if the perimeter become 62cm when each dimension is increased by 2cm.
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system%28w=%284%2F5%29L%2C2%28w%2B2%29%2B2%28L%2B2%29=62%29

2w%2B4%2B2L%2B4=62
2w%2B2L=54
4w%2B4L=108
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5w=4L
108-4w=5w
108=9w
highlight%28w=12%29---------width, original rectangle
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L=%285%2F4%29%2A12
highlight%28L=15%29----------length, original rectangle

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


If each dimension was increased by 2cm, then the perimeter was increased by 8cm. So the original perimeter was 62-8=54cm.

The width is 4/5 the length, so call the width 4x and the length 5x.

Width plus length is half the perimeter:
4x%2B5x=54%2F2+=+27
9x=27
x=3

The length is 4x=12cm; the length is 5x=15cm.