SOLUTION: I think this is right for this problem. Can someone pls check?
50/4=12.5x2+7=32
32/2=16-7=9
W=32
L=9
The length of a rectangle is 7" greater than its width. The perimete
Algebra ->
Customizable Word Problem Solvers
-> Geometry
-> SOLUTION: I think this is right for this problem. Can someone pls check?
50/4=12.5x2+7=32
32/2=16-7=9
W=32
L=9
The length of a rectangle is 7" greater than its width. The perimete
Log On
Question 408536: I think this is right for this problem. Can someone pls check?
50/4=12.5x2+7=32
32/2=16-7=9
W=32
L=9
The length of a rectangle is 7" greater than its width. The perimeter of the rectangle is 50". Find the length and width of the rectangle.
You can put this solution on YOUR website! The length of a rectangle is 7" greater than its width. The perimeter of the rectangle is 50". Find the length and width of the rectangle.
------
Perimeter = 2(length + width)
---
Let width be "x".
The length is "x+7".
---
50 = 2(x+7 + x)
25 = 2x+7
2x = 18
x = 9 (width)
x+7 = 16 (length)
======================
Cheers,
Stan H.
You can put this solution on YOUR website!
The length of a rectangle is greater than its width . The perimeter of the rectangle is .
Find the length and width of the rectangle.
given: of a rectangle is greater than its width ..=>
(