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) About Me  (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:



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 . No! No! No! 1000 times No!

In general, (Remember FOIL?)

So, your next step SHOULD have been:



Which simplifies to:





John


Answer by scott8148(6628) About Me  (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