SOLUTION: In a rectangle, the length is 12cm longer than its width. If the perimeter of the rectangle 84cm, what are the dimensions?

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Question 1057349: In a rectangle, the length is 12cm longer than its width. If the perimeter of the rectangle 84cm, what are the dimensions?

Found 2 solutions by solve_for_x, math_helper:
Answer by solve_for_x(190) (Show Source):
You can put this solution on YOUR website!
Let x represent the width of the rectangle.

Since the length is 12 cm longer, the length can be represented as x + 12.

The perimeter of a rectangle is 2 times the sum of the length and the width.

Since the perimeter of this rectangle must be 84 cm, you can write the following equation:

2(length + width) = 84

2[(x + 12) + x] = 84

Dividing both sides by 2 leaves:

(x + 12) + x = 42

Collecting and combining like terms on the left side leaves:

2x + 12 = 42

Subtracting 12 from both sides then gives:

2x = 30

Dividing both sides by 2 gives:

x = 15

The width of the rectangle is 15 cm.

From this you can determine the length.

Answer by math_helper(2461) (Show Source):
You can put this solution on YOUR website!
Let l = length
Let w = width
Need to find l and w. Two unknowns so look for two equations.
(1) l = w+12 ("length is 12cm longer than its width")
(2) 2l + 2w = 84 ("perimeter is 84cm")
...
Eq (1) allows us to subs "w+12" for "l" in (2):
2(w+12) + 2w = 84 (eq (2) w/substitution made)
2w + 24 + 2w = 84 (multiplied 2 into parens)
4w + 24 = 84 (combined like terms)
4w = 60 (subtracted 24 from both sides)
w = 60/4 (divided both sides by 4)
w = 15cm
From (1), l = 15+12 = 27cm
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Ans: The rectangle has dimensions 27cm x 15cm
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Check:
We can plug into (2) to check:
2(27) + 2(15) = 54 + 30 = 84 (ok)