SOLUTION: The perimeter of a rectangle is 16 feet. If twice its length is 4 times its width, find the width and the length of a rectangle.
My Work:
Equation 1: 2L=4W
Equation 2: L+W=16
Algebra.Com
Question 1177440: The perimeter of a rectangle is 16 feet. If twice its length is 4 times its width, find the width and the length of a rectangle.
My Work:
Equation 1: 2L=4W
Equation 2: L+W=16
W=16/3
L=10/6
Found 3 solutions by Boreal, mananth, MathLover1:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
perimeter is 2L +2W=16; L+W=8, the half-perimeter, not 16
2L=4W, so L=2W
6W=16
W=8/3 feet
4W=32/3 feet, and 2L=32/3 feet so L=16/3 feet
width=8/3 feet
Half-perimeter is 24/3 feet (8)
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
My Work:
Equation 1: 2L=4W
Equation 2: 2L+2W=16
Therefore 4w+2w=16
6w=16
w=16/6 =8/3
l=2*8/3=16/3
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
The perimeter of a rectangle:
if the perimeter of a rectangle is feet, then
............eq.1
if twice its length is times its width, then
.......solve for
.......eq.2, substitute in eq.1
............eq.1
go to
.......eq.2, substitute
the width is ft and the length ft
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