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Question 1190984: the area of a rectangle is 54 yd^2,and the length of the rectangle is 3 yd more than twice the width. Find the dimensions of the rectangle.
Answer by ikleyn(52756)   (Show Source):
You can put this solution on YOUR website! .
the area of a rectangle is 54 yd^2,and the length of the rectangle is 3 yd more than twice the width.
Find the dimensions of the rectangle.
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If the width is w yards (now unknown), then the length is (2w+3) yards,
and the area equation is
w*(2w+3) = 54. (1)
I will solve it mentally. For it, I multiply both sides by 2
2w*(2w+3) = 108.
What are two positive integer numbers that differ by 3 and produce 108, when multiplied ?
You can easy guess them mentally: 9 and 12.
So, 2w = 9; hence, w = 4.5.
ANSWER. The dimensions of the rectangle are 4.5 yards (the width) and 12 yards (the length).
CHECK. 4.5*12 = 54 square yards. ! Correct !
Solved // (mentally).
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Alternatively, you can solve equation (1) as a quadratic equation, using routine methods,
such as a quadratic formula or factoring.
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