SOLUTION: A rectangle parcel of land is 50ft wide. The length of a diagonal between opposite corners is 10ft more than the length of the parcel. What is the length of the parcel?
x= lengt
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Question 256212: A rectangle parcel of land is 50ft wide. The length of a diagonal between opposite corners is 10ft more than the length of the parcel. What is the length of the parcel?
x= length in ft of the parcel
area = length x width
x^2 + 50^2 = (x + 100)^2
x^2 + 2500 = x^2 + 100
Found 2 solutions by solver91311, scott8148:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
I'm not sure why you introduced the idea of the area of the rectangle into this. You are dealing with a right triangle which has sides of 50 and 
and a hypotenuse of 
At that, you almost had it right. Using Pythagoras:
^2)
I'm assuming that the first relationship you wrote just had a typo in it and that you meant what I just wrote above.
Now here is where you really got off track. You tried to say that ^2\ =\ x^2\ +\ 10^2)
. No! No! No! 1000 times No!
In general, ^2\ =\ \left(a\,+\,b\right)\left(a\,+\,b\right)\ =\ a^2\ +\ 2ab\ +\ b^2)
(Remember FOIL?)
So, your next step SHOULD have been:
Which simplifies to:
John
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
x= length in ft of the parcel
area = length x width ___ ???
x^2 + 50^2 = (x + 100)^2 ___ almost, but not quite ___ x^2 + 50^2 = (x + 10)^2
x^2 + 2500 = x^2 + 20x + 100
2400 = 20x ___ 120 = x
this is a 5, 12, 13 Pythagorean triple
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