a 174 ft long fence is placed around the perimeter of a rectangular pool that has a 3 ft sidewalk around it. the actual pool w/o the sidewalk is twice as long as it is wide. what is the dimensions of the pool w/o the sidewalk?
thanks for the help.
2006-09-15 00:40:15 · 6 answers · asked by shih rips 6 in Science & Mathematics ➔ Mathematics
the sides of the pool are x and 2x. If you take into consideration the sidewalk, the perimeter will be (x+6) and (2x+6). Set up the equation for the perimeter to be 2(x+6)+2(2x+6)=174. x=25 so the dimensions of the pool without the sidewalk is 25ft x 50ft
2006-09-15 01:04:20 · answer #1 · answered by samantha m 1 · 0⤊ 1⤋
27 ft and 54 ft
let
if x is the width and y be the length of the pool with sidewalk
then as you said pool is twice as long as it is wide so
x = y/2 let this be equation (a)
now the sidewalk is 3ft so the length and width of the pool with sidewalk becomes (y+3) and (x+3).
and according to the definition of perimeter
2* (x+3+y+3) = 174 equation (b)
solving the two equations we have the answer.
length=54 ft
width= 27 ft
2006-09-15 07:53:54 · answer #2 · answered by Arif 2 · 1⤊ 0⤋
pool is 54' x 27'
2006-09-15 07:52:04 · answer #3 · answered by B-rad 2 · 0⤊ 0⤋
The length is 54 ft.
The width is 27 ft.
If you workk it backwards, this answer is correct.
2006-09-15 10:46:07 · answer #4 · answered by b0b0link 2 · 0⤊ 0⤋
You only want the answer, not the solution ?
Ok, it is 25 ft x 50ft.
2006-09-15 08:06:14 · answer #5 · answered by Jahar 1 · 0⤊ 1⤋
50' x 25'
BUT, is the teacher going to want you to show your work?
2006-09-15 09:11:04 · answer #6 · answered by boardintooblivian2 2 · 0⤊ 0⤋