SOLUTION: The perimeter of a ractangle is 48 feet the width is 6 feet less than the length what are tje dimesions of the rectangle ?

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: The perimeter of a ractangle is 48 feet the width is 6 feet less than the length what are tje dimesions of the rectangle ?      Log On


   



Question 797105: The perimeter of a ractangle is 48 feet the width is 6 feet less than the length what are tje dimesions of the rectangle ?
Answer by mohanraj04(12) About Me  (Show Source):
You can put this solution on YOUR website!
ans:
the perimeter of the rectangle is 48ft.
2(L+W)=48-------equ:1
where,
L be the length of the rectangle.
W be the width of the rectangle.
the width is 6ft. less than the length
W=L-6-----------EQU:2
by substituting equ:2 in equ:1, we get
2[L+(L-6)]=48
2[2L-6]=48
[2L-6]=48/2
[2L-6]=24
2L=24+6
2L=30
L=30/2
L=15
by substituting L=15 in equ:2, we get
W=15-6
W=9
therefore the dimensions of the rectangle are 15ft. and 9 ft.