SOLUTION: Please help me solve this problem, One side of rectangle is 1 meter less than the length of the diagonal.If the area is 60 meter squared, what are its dimensions?

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Question 492750: Please help me solve this problem, One side of rectangle is 1 meter less than the length of the diagonal.If the area is 60 meter squared, what are its dimensions?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
One side of rectangle is 1 meter less than the length of the diagonal. If the area is 60 meter squared, what are its dimensions?
:
Let L = the length of the rectangle
Let W = the width of the rectangle
:
I says,"the area is 60 meter squared,", therefore:
L * W = 60
then
L = 60%2FW
:
"One side of rectangle is 1 meter less than the length of the diagonal"
Diagonal = W + 1
Using pythag:
W + 1 = sqrt%28L%5E2+%2B+W%5E2%29
Replace L with 60/W
W + 1 = sqrt%28%2860%2FW%29%5E2%2B+W%5E2%29
Square both sides
(W+1)^2 = %2860%2FW%29%5E2 + W^2
FOIL
W^2 + 2W + 2 = 3600%2FW%5E2 + W^2
W^2 - W^2 + 2W + 2 = 3600%2FW%5E2
2W + 2 = 3600%2FW%5E2
Multiply both sides by W^2
2W^3 + 2W^2 = 3600
2W^3 + 2W^2 - 3600 = 0
simplify, divide by 2
W^3 + W^2 - 1800 = 0
:
Plot this equation on your graphing calc, should look like this:
+graph%28+300%2C+200%2C+-6%2C+15%2C+-1000%2C+2000%2C+x%5E3%2Bx%5E2-1800+%29+
you can see W ~ 11.8 meters is the width
and
L = 60%2F11.8
L ~ 5.1 meters is the length
;
:
See if that checks out by finding the diagonal
d = sqrt%2811.8%5E2+%2B+5.1%5E2%29
d = 12.85 which is about 1 inch longer than the width