SOLUTION: the perimeter of a rectangle is 160ft.one fourth the length is the same as twice the width.find the dimensions of the rectangle.

Algebra ->  Algebra  -> Systems-of-equations -> SOLUTION: the perimeter of a rectangle is 160ft.one fourth the length is the same as twice the width.find the dimensions of the rectangle.      Log On

Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

   

Question 118928: the perimeter of a rectangle is 160ft.one fourth the length is the same as twice the width.find the dimensions of the rectangle.
Answer by MathLover1(3376) About Me  (Show Source):
You can put this solution on YOUR website!


the perimeter of a rectangle is 160ft.one fourth the length is the same as twice the width.find the dimensions of the rectangle.

Given:
the perimeter P=160ft
one fourth the length L is the same as twice the width W:
%281%2F4%29L+=+2W…..=>…%281%2F4%29L%2F2+=+W…..=>…+%281%2F2%29L+=+W
So, we have:
P=2%28L%2BW%29
160ft+=+2%28L%2BW%29……………..substitute W
160ft+=+2%28L%2B%281%2F2%29L%29……………..
160ft+=+2L%2B+2%281%2F2%29L%29……………..
160ft+=+2L%2B+cross%282%29%281%2Fcross%282%29%29L%29……………..
160ft+=+2L%2B+L……………..
160ft+=+3L……………..
160ft%2F3+=+L……………..

+L+=+53.3ft……………..

+W+=+%281%2F2%29L+
+W+=+%281%2F2%29+53.3ft+
+W+=+%281%2F2%29+53.3ft+
W+=+26.65ft
Check:
P=2%28L%2BW%29
P=2%2853.3ft+%2B26.65ft%29

P=2%2879.98ft%29
P=2%2880ft%29
P=160ft