33.
The length of a rectangle is 8 in. more than twice its width. If the perimeter of the rectangle is 64 in., find the width of the rectangle.
A) 10 in. B) 7 in. C) 8 in. D) 9 in.
2006-09-24 07:19:00 · 6 answers · asked by pjaxdriver13 1 in Science & Mathematics ➔ Mathematics
The answer is C, 8 in.
The way to solve this problem is to set up a rectangle based on the problem.
You are trying to solve for the width, so let x=the width.
If the length is 8 inches more than twice its width, then the length must be 2x+8.
So to find the perimeter, add together all of the sides. Remember that there are two of each side. This gives you 2x+8+2x+8+x+x or 6x+16.
You already know that the perimeter is 64, so set 6x+16 equal to 64.
If 6x+16=64, then 6x=64-16, or 48. x=48/6 or 8.
2006-09-24 07:36:13 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Let the length be 'l' and the width 'w'.
The given conditions are :
l = 2w + 8 ........(1)
2(l + w) = 64 ......(2)
Re-writing (1) & (2) :
l - 2w = 8 ........(3)
2l + 2w = 64 ........(4)
Adding (3) & (4) :
3l = 72 or l = 24
Substituting this value of l in (1) :
24 = 2w + 8 or 2w = 16 and w = 8
2006-09-24 14:29:47 · answer #2 · answered by Problem Child 2 · 0⤊ 0⤋
the answer is C) 8 inches - - - L is length, & W is width. SO - - -W X 2 + 8 = L
2006-09-24 14:32:46 · answer #3 · answered by Dottie B 1 · 0⤊ 0⤋
C.) 8
2006-09-24 14:30:35 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Hi. Set it up like this:
2(2x+x+8)=64 where x=width
2x+x+8=32
2x+x=24
3x=24
x=???
Oh oh, looks like I messed up. Problem child is better at math I guess!
2006-09-24 14:25:56 · answer #5 · answered by Cirric 7 · 0⤊ 0⤋
24in
2006-09-24 15:18:48 · answer #6 · answered by chinni s 1 · 0⤊ 0⤋