SOLUTION: if the perimeter of a rectangle is 94yd, and the area of the rectangle is 90 yd^2. What are the dimensions of the rectangle.

Algebra ->  Rectangles -> SOLUTION: if the perimeter of a rectangle is 94yd, and the area of the rectangle is 90 yd^2. What are the dimensions of the rectangle.      Log On


   

Question 648540: if the perimeter of a rectangle is 94yd, and the area of the rectangle is 90 yd^2. What are the dimensions of the rectangle.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
let's the length e L and width W
the perimeter of a rectangle is P=2%28L%2BW%29 is 94yd, and
the area of the rectangle is A=L%2AW 90 yd^2.
given:
P=+94yd........1
A=90yd%5E2.......2
_____________________________
2%28L%2BW%29=+94yd........1
L%2AW=90yd%5E2.......2
_____________________

2%28L%2BW%29=+94........1...solve for L
L%2BW=+94%2F2
L=+47-W.......substitute in 2
%2847-W%29%2AW=90.......2...solve for W
47W-W%2AW=90
47W-W%5E2=90
0=W%5E2-47W%2B90 or
W%5E2-47W%2B90=0......use quadratic formula to find W
W+=+%28-%28-47%29+%2B-+sqrt%28%28-47%29%5E2-4%2A1%2A90+%29%29%2F%282%2A1%29+
W+=+%2847+%2B-+sqrt%282209-360+%29%29%2F2+
W+=+%2847+%2B-+sqrt%281849%29%29%2F2+
W+=+%2847+%2B-+43%29%2F2+...the roots are
W+=+%2847+%2B+43%29%2F2+
W+=+90%2F2+
highlight%28W+=+45%29+
or
W+=+%2847+-+43%29%2F2+
W+=+4%2F2+
highlight%28W+=+2%29+
now find L

L=+47-W..
L=+47-45
highlight%28L=+2%29
or
L=+47-W..
L=+47-2
highlight%28L=+45%29

one pair solutions is: highlight%28L=+2%29 and highlight%28W+=+45%29+
another pair is highlight%28L=45%29 and highlight%28W+=+2%29+