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April 5th, 2007, 04:08 PM   #1
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Percent of Square's Area

If the length of each side of a square is increased by 10%, by what percent does the area of the square increase?

MY WORK:

A = side^2

A = (0.10)^2

A = 0.20

Area is increased by 20%.

However, the textbook tells me that the correct answer is 21%.

How can this be?
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April 5th, 2007, 05:08 PM   #2
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A = side^2

new side = old side + 0.1*old side = 1.1*old side

A = (1.1(s))^2

A = (1.21)s^2

Since s^2 was the old area, the new area is 21% greater than the old area.

Area is increased by 21%.
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April 5th, 2007, 05:18 PM   #3
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ok

Where did you get 1.1?
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April 5th, 2007, 05:23 PM   #4
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Because the new side is 10% longer than the old side. So, the new side is equal to the (old side + 0.1*old side), since (0.1*old side) = 10% of the old side.
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April 5th, 2007, 05:30 PM   #5
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ok

This is a tricky question.

All this new side versus the old side.
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April 6th, 2007, 09:32 AM   #6
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If the length of each side of a square is increased by 10%, what makes you think the result is a square? Nothing in the question tells you that.
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April 6th, 2007, 09:50 AM   #7
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Why wouldn't the result be a square?
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April 6th, 2007, 02:19 PM   #8
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What is there to "hold the square together"?
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April 6th, 2007, 02:23 PM   #9
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True. However, if we specified that one corner of the square held intact, and that the sides that grew were at right angles to each other, and the square was re-drawn from there, that would better represent this situation.
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April 6th, 2007, 02:34 PM   #10
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Okay - so your viewpoint is now one corner and angles are maintained. What holds the opposite corner together?
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