The length of a rectangle is 8 in. more than twice its width. If the perimeter of the rectangle is 58 in., find the width of the rectangle.
2007-05-25 07:27:25 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Now, the perimeter is defined as sum of the sides.
So, if x is the width, then the length is 2x+8
Now, the perimeter = x + x + 2x + 8 + 2x + 8 =58
So, 6x+16 = 58 or
6x = 42
x=7
2007-05-25 07:31:24 · answer #1 · answered by Anonymous · 0⤊ 0⤋
1/2 the perimeter = (w+2w+8)
whole perimeter = 2(w+2w+8)
so
2(w+2w+8) = 58
2w+4w+16 = 58
6w+16 = 58
6w = 42
w = 7 Width = 7 Length = 7x2+8 = 22
Perimeter = 7+22+7+22 = 58
2007-05-25 14:57:45 · answer #2 · answered by const. king 2 · 0⤊ 0⤋
Hi, Formula setup and answer as follows:
w + w + (2w +8) + (2w+8) = 58
therefore 6w + 16 = 58
therfore 6w = 42
therfore w = 7
and to follow:
L= 2w = 8
therfore L = (2 x 7) + 8 = 22
to check : P = 2L + 2w
or P= (2 X 22) + (2 X 7)
or P= 44 + 14
or P = 58
2007-05-25 14:53:28 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Let the width=x.
Thus length=8+2x.
The perimiter = 2*length+2*width.
so 58=16+4x+2x=16+6x.
Thus 42=6x or x=7.
so the width is 7 inches.
2007-05-25 14:32:40 · answer #4 · answered by Jon 2 · 0⤊ 0⤋
w = width
l = length
p = perimeter
p = 2(l + w)
58 = 2(2w + 8 + w)
58 = 6w + 16
42 = 6W
w = 7
l = 22
2007-05-25 14:34:17 · answer #5 · answered by cscokid77 3 · 0⤊ 1⤋
w= width
so l =2w+8
Perimeter = 2w+2l = 2w +2(2w+8)
58 = 2w +4w + 16
42 = 6w
w = 7 inches
2007-05-25 14:33:51 · answer #6 · answered by ironduke8159 7 · 0⤊ 0⤋
L = length
w = width
L = 8 + 2w
2L + 2w = 58
Solve the two equations...
2(8+2w) + 2w = 58
solve for w
2007-05-25 14:34:17 · answer #7 · answered by mark r 4 · 0⤊ 0⤋
length=8+2x
width=x
perimeter=2l+2w
58=2(8+2x)+2(x)
58=16+4x+2x
58=16+6x
42=6x
x=7
width=7 in
2007-05-25 14:32:17 · answer #8 · answered by JO 3 · 0⤊ 0⤋