The length of a rectangle is ten inches longer then twice the width. If the perimeter of the rectangle is 344 inches, how do I find the dimensions?
2007-09-14 04:26:36 · 9 answers · asked by k_artiaco 1 in Science & Mathematics ➔ Mathematics
P = 2L + 2W
L = 2W + 10
344 = 2(2W + 10) + 2W
==> distribute
344 = 4W + 20 + 2W
==> subtract 20 from both sides and combine like terms
324 = 6W
==> divide both sides by 6
54 = W
==> plug this into the equation for length:
L = 2(54) + 10 = 108 + 10 = 118
The dimensions are 118 inches x 54 inches.
2007-09-14 04:30:40 · answer #1 · answered by C-Wryte 4 · 0⤊ 1⤋
If you don't learn how to do any of this stuff for your self, rather than asking us to do all your homework for you, you're going to flunk all of your tests.
The whole point is for you to learn how to figure this stuff out.
You need to take the information that you do have, and the thing you're looking for, and express them algebraically.
Try drawing a rectangle, and labeling it with the information you know, and he unknowns.
Then write out the equation.
Since the lenght is expressed in terms of the width, the width is the unknown thing you need to figure out first.
Once you have it, you can figure out the length.
You know the whole perimeter is 344. That means that w1 + l1 + w2 + l2 (w = width; l = length) = 344.
Since it's a rectangle, you also know that w1 = w2, and l1 = l2.
So why don't you work with -- instead of both sides, and the top and bottom, one of each. That would be half the perimeter.
So, divide the perimeter in half.
That means that one of the lengths plus one of the widths equals that number (half the perimeter).
l + w = (344/2)
l= 2w - 10
so (2w - 10) + w = (344/2)
3w - 10 = (344/2)
3w = (344/2) + 10
Solve for width.
Plug that number into the equation for length.
Then plug the width and length into the equation for perimeter, and make sure what you get is 344 -- that is, check your work.
2007-09-14 15:31:45 · answer #2 · answered by tehabwa 7 · 0⤊ 0⤋
Perimeter Formula
P = 2L + 2W
Let
w = Width
2w + 10 = Length
344 = the perimeter
- - - - -
The equation
344 = 2(2w + 10) + 2w
The distributive property
344 = 4w + 20 + 2w
Collect like terms
344 = 6w + 20
Transpose 20
324 - 20 = 6w + 20 - 20
324 = 6w
Divide both sides of the equation by 6
324 / 6 = 6 / 6
324 / 6 = w
54 = w
- - - - - - -
The width is 54 inches
The Length is 2w + 10
2(54) + 10 =
108 + 10 =
118 Inches
The length is 118 inches
- - - - - - -s-
2007-09-14 11:42:42 · answer #3 · answered by SAMUEL D 7 · 0⤊ 0⤋
Let the width of the rectangle be 'w' inches
Therefore,the length is 2w+10 inches
We know that perimeter of a retangle
=2(legtn+width)
Therefore,according tothe problem,
2(2w+10+w)=344
or,2(3w+10)=344
or 3w+10=344/2=172
or,3w=172-10=162
or,w=162/3=54
Therefore,the width of the rectangle=54 inches
The length=2*54+10=118 inches
2007-09-14 11:41:55 · answer #4 · answered by alpha 7 · 0⤊ 0⤋
L = length
w = width
L = 10 + 2w
Perimeter = 2L + 2w
344 = 2(10+2w) + 2w
344 = 20 + 4w + 2w
324 = 6w
6w = 324
w = 54
L = 10 + 2w
L = 10 + 2(54)
L = 10 + 108
L = 118
width = 54 inches and length = 118 inches
2007-09-14 11:35:04 · answer #5 · answered by coolesteugene 2 · 1⤊ 0⤋
you're told that l=10 +2w and P=344 inches
since P=2(l+w) then
344=2(10 +2w +w)
172=10 +3w
162=3w
54=w then l=10 +2(54)
l=118", w=54"
2007-09-14 11:38:08 · answer #6 · answered by marcus101 2 · 0⤊ 0⤋
if the length L is 10 then the perimeter is 20+ 2width W= 344
20+2W=344
2W=324
W=162
L=10
2007-09-14 11:41:04 · answer #7 · answered by mich 1 · 0⤊ 0⤋
Perimeter = 2(length) + 2(width)
From your problem, length = 2(width) + 10
Plug in that substitution for length, and solve.
2007-09-14 11:34:17 · answer #8 · answered by Woodpecker 2 · 0⤊ 0⤋
beats me
2007-09-14 11:30:42 · answer #9 · answered by Jazz 3 · 0⤊ 1⤋