4 feet less?

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- Jan 6th 2007, 06:31 PM #1

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## strory problem help

Sam lives on a lot that he thought was a square, 157 feet by 157 feet. When he had it surveyed , he discovered that one side was actaully 2 feet longer then he thought and the other was actually 2 feet shorter then he thought. How much less area does he have then he thought he had?

I was coming up with this anwer 457 ft less then he thought, would that be right?

- Jan 6th 2007, 06:53 PM #2

- Jan 6th 2007, 07:45 PM #3

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Hello, fancyface!

This doesn't require any algebra at all.

So exactly*where*is your difficulty?

Sam lives on a lot that he thought was a square, 157 feet by 157 feet.

When he had it surveyed, he discovered that

one side was actaully 2 feet longer then he thought

and the other was actually 2 feet shorter then he thought.

How much less area does he have then he thought he had?

Sam thought he had: ft².

Instead, his lot was actually 15**9**feet by 15**5**feet.

. . So he had: ft².

Let's see, that's a difference of . . . um . . .*(Where's my calculator?)*

- Jan 6th 2007, 08:26 PM #4

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It depends on where those two sides with errors are on the lot.

a) If two opposite sides were actually 159 ft each, and so the other two opposites were actually 155 ft each, then,

He thought the area was (157)(157) = 157^2

But actual area =

= (157 +2)(157 -2)

= 157^2 -2^2

= (Area of square lot he thought) minus 4.

Therfore, the actual lot is 4 sq. ft. less than he thought.

b) If the two sides with errors are adjacent to each other, then the actual lot area can be computed by dividing the actual lot into two triangles. One triangle is an isosceles rigth triangle; the other is a not a right triangle but whose 3 sides are 159, 157 and the hypotenuse of the isosceles right triangle. Not as easy to do. Lots of square roots. Heron's formula.

c) If the two erroneous sides are opposite each other, then there are two possible ways, depending on which erroneous side is perpendicular to one of the 157-ft side. Again, same method as in part b) above for each way.