SOLUTION: A rectangle has base length x+3, altitude length x+1, and diagonals of length 2x each. What are the lengths of its base, altitude, and diagonals? I don't really know how I'm su

Algebra ->  Pythagorean-theorem -> SOLUTION: A rectangle has base length x+3, altitude length x+1, and diagonals of length 2x each. What are the lengths of its base, altitude, and diagonals? I don't really know how I'm su      Log On


   



Question 808496: A rectangle has base length x+3, altitude length x+1, and diagonals of length 2x each. What are the lengths of its base, altitude, and diagonals?
I don't really know how I'm supposed to draw the rectangle, but what I've done is draw it short and long and put my x+3 across the bottom, my two diagonals going across, and then my altitude going up to where the diagonals intersect. Then I've tried using the usual a^2+b^2=c^2 I don't know if the diagonal I'm supposed to be using would just be x instead of 2x, since they're being cut in half by each other, but using 2x, I get x^2+8x+10=x^2 and using x I get 2x^2+8x+10=x^2, but neither of those give me the correct answer.

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A rectangle has base length x+3, altitude length x+1, and diagonals of length 2x each.
What are the lengths of its base, altitude, and diagonals?
:
I think you have the right idea, one little mistake, it should be (2x)^2
(x+3)^2 + (x+1)^2 = (2x)^2
FOIL
x^2 + 6x + 9 + x^2 + 2x + 1 = 4x^2
Collect like terms
x^2 + x^2 - 4x^2 + 6x + 2x + 9 + 1 = 0
-2x^2 + 8x + 10 = 0
easier to factor of we multiply by -1
2x^2 - 8x - 10 = 0
Factors to
(2x+2)(x-5) = 0
The positive solution is all we want here
x = 5
:
:
See if that checks out
8^2 + 6^2 = 10^2