If the perimeter of the rectangle is 52in find the width of the rectangle.
2007-04-02 10:33:32 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
width = x
length = 12 + 2x
P = 2(length + width)
52 = 2(12+2x + x)
52 = 2(12 + 3x)
26 = 12 + 3x
14 = 3x
14/3 = x
So the width is 14/3 = 4.667 in
2007-04-02 10:43:07 · answer #1 · answered by Mathematica 7 · 0⤊ 0⤋
Perimeter = 2*(l+w)
l=2*w+12
Substitute for l in first equation
==> Perimeter=52=2*((2*w+12)+w)
==> 52 = 6*w+24
==> w = (52-24)/6
==> w = 28/6 inches ==> 4.67 inches
2007-04-02 17:44:16 · answer #2 · answered by naveen 1 · 0⤊ 0⤋
first you start off with the formula for the perimeter which is
two times the length plus two times the width
2L + 2W = P
using the info above
Length = 2W+12
so just substitute that in for L and you get
2(2W+12) + 2W = 52
4W+24+2W=52
6W+24=52
6W=28
W=4.7in
then go back to the other 1st equation to get L
2L+2W=52
2L+2(4.7)=52
2L+9.4=52
2L=42.6
L=21.3
That is the length and width of your rectangle
2007-04-02 17:41:45 · answer #3 · answered by Ally 2 · 0⤊ 1⤋
w = width
2w+12=length
2w + 2(2w + 12)= 52 in.
2w + 4w + 24 = 52
6w = 52 - 24
6w = 28
w = 4 2/3 inches
2007-04-02 17:39:44 · answer #4 · answered by ecolink 7 · 0⤊ 1⤋