SOLUTION: The length of a rectangle is 6 cm longer than its width. If the perimeter of the rectangle is 48 cm, find its area.

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Question 622247: The length of a rectangle is 6 cm longer than its width. If the perimeter of the rectangle is 48 cm, find its area.
Answer by math-vortex(648) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, there--
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Here is one way to solve this problem.
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Let W be the width of the rectangle.
Let L be the length of the rectangle.
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We need to write two equations that use the information in the problem to model the situation.
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The perimeter is 48 cm. Perimeter is the distance around the the rectangle: two widths and two lengths.
[two widths of the rectangle] + [two lengths of the rectangle] = [the perimeter]
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In algebra, we write
2W%2B2L=48
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We also know that the length of a rectangle is 6 cm longer than its width.
[the length of the rectangle] = [the width of the rectangle] + [6 more centimeters]
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In algebra, we write,
L=W%2B6
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Now we have a system of two equations. We will solve it using the substitution method. Substitute W+6 for L in the perimeter equation.
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2W%2B2L=48
2W%2B2%28W%2B6%29=48
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Simplify and solve for W. Clear the parentheses.
2W%2B2W%2B12=48
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Combine like terms.
4W%2B12=48
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Subtract 12 form both sides of the equation.
4W=48-12
4W=36
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Divide both sides of the equation by 4.
W=9
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The width of the rectangle is 9 cm. Since the length of the rectangle is 6 more the the width, the length is 15 cm.
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The formula for the area of a rectangle is A=L*W. The area is 135 square cm since 9*15=135.
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The last step to to check our answer against the original problem. The perimeter should be 48 cm. We see that 2 widths is 18 cm, and 2 lengths is 30 cm; 18+30=48. Check!
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Please email me if I need to explain any part of this better. By the way, there is a way to solve this problem without using two variables and a system of equations. Let me know if you'd like to see that method.
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Ms.Figgy
math.in.the.vortex@gmail.com