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15x3

Area = 3x15=45

Perimeter = 3+3+15+15 = 36

This is solved algebraically by substituting the value of the length in terms of the width

2L + 2W = 36

L = 18-W

L x W = 45 and (18-W) x W = 45

gives the equation

W2+18W-45 = 0

(W-15)(W-3) = 0 and the width and length are 15 and 3

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