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August 24th, 2016, 12:21 AM   #1
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What are the dimensions of the given rectangle?

The area of a rectangle is 420 cm². The length of a rectangle is one less than thrice its width. What are the dimensions of the given rectangle?
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August 24th, 2016, 02:46 AM   #2
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Let A be the length and B be the width of the rectangle
1) A x B = 420 [The area of a rectangle is 420 cm²]
2) A = (3-1)B [The length of a rectangle is one less than thrice its width]
Substituting A from (2) into (1) we have:
(3-1)B x B =420
2 x B x B =420
B x B =210
B = 14.49
A = (3-1)B = 28.98
Therefore the dimensions are 14.49 x 28.98 to 2 decimal places
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August 24th, 2016, 02:54 AM   #3
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$ab=420,\quad a=3b-1$

$a=\dfrac{3\cdot420}{a}-1$

$a^2+a-1260=0\implies a=35,b=12$
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August 24th, 2016, 06:16 AM   #4
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I prefer:

w = width, 3w-1 = length

w(3w-1) = 420
3w^2 - w - 420 = 0

w = 12
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August 24th, 2016, 06:50 AM   #5
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I submit that the question was worded badly
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August 24th, 2016, 09:06 AM   #6
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Quote:
Originally Posted by weirddave View Post
I submit that the question was worded badly
...so should be easier for you, being "weird"dave
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August 24th, 2016, 10:06 AM   #7
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On a timed test, use short cuts if possible; like:

when is 3w^2 >= 420 ?

try 10: 3w^2 = 300
try 11: 3w^2 = 363
try 12: 3w^2 = 432 ....ahhhh, so:

Apply the "-1":
12 * 3 - 1 = 35

12 * 35 = 420
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August 24th, 2016, 11:38 AM   #8
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I submit that I shouldn't do math before morning coffee!
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August 24th, 2016, 04:27 PM   #9
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With me, not before at least 2 coffees:
the one for my left eye, followed by the one for my right eye!
(extra large, Tim Horton's dark roast brand)
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