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The Peimeter of a rectangle is 56 M if the width were doubled and the length were increased by 8 M, then the perimeter would be 82m. What are the dimensions?

Respuesta :

KhaRNate

l + w + l + w = 56

2l + 2w = 56

2w = 56 - 2l

(2w) + (2w) + (l + 8) + (l + 8) = 82

2(2w) + 2(l + 8) = 82

2(56 - 2l) + 2l + 16 = 82

112 - 4l + 2l + 16 = 82

112 - 82 + 16 = 4l - 2l

46 = 2l

46/2 = l

23 = l

2w = 56 - 2l

2w = 56 - 2(23)

2w = 56 - 46

2w = 10

w = 10/2

w = 2

length(l) = 23 m and width(w) = 2 m


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