<PRE><B>Question 267515</B>
a rectangle has an area that is numerically twice its perimeter.
:
L*W = 2(2L+2W)
:
If the length is twice the width, what are the dimensions?
:
Replace L with 2W
(2W)*W = 2[2(2W) + 2W)]
2W^2 = 2(4W + 2W)
2W^2 = 2(6W)
;
divide both sides by 2
W^2 = 6W
W^2 - 6W = 0
:
factor out W
W(W - 6) = 0
:
Two solutions
W = 0, meaningless
and
W = 6 units is the width
then
2(6) = 12 units is the length
:
:
Is this true?
12*6 = 2(2(12) + 2(6))
72 = 2(24 + 12), </PRE>