Another math question this time i'll show you what i did so some people won't accuse me of cheating. here's th

Question

Here's the problem: A rectangular pool is 8m longer than it is wide.A walkway 1m wide surrounds the pool .find the dimensions of the pool if the area of the walkway is 100m. This is how I tried to solve it:100=(w+8)(w+1)-(w)(w) =100=8wsq. +8x+1- 8xsq.=100=8x+1=100-1=8x+1-1=99=8x.Then I divide 99 by 8 , but I always get the wrong answer.Am I doing something wrong? Please help.

Answer 1

The area is 100 _m_? I don't think so - area is measured in square meters.

Assuming that that's just a typo, the proper setup is 100=(w+2)(w+10)-w(w+8) -- the area of the pool+walkway minus the area of the pool. Simplifying:

w²+12w+20-w²-8w=100
4w=80
w=20
L=28

edit: corrected typo

Answer 2

l = w + 8

(w + 8 + 1 + 1)(w + 1 + 1) - (w + 8)w = 100
(w + 10)(w + 2) - (w^2 + 8w) = 100
w^2 + 2w + 10w + 20 - w^2 - 8w = 100
(2 + 10 - 8)w + 20 = 100
(12 - 8)w = 80
4w = 80
w = 20

l = w + 20
l = 20 + 20
l = 40

ANS : Width of pool is 20m and Length of pool is 40m

Answer 3

The walkway will be 100m long by 1m wide. The inside length of the walkway will be 96m since the squares on the end overlap.

Answer 4

if you make the width equal to x, so the lenght must be (x+8) ->8 metres longer

the lenght of the area is x+10, its width x+2
so you have the whole area (x+2)(x+10)
and the area of the pool x(x+8)

and (x+2)(x+10)-x(x+8)=100
x^2+12x+20-x^2-8x=100
4x=80
so x=20m

pool's width =20m
pool's lenght = 28m

Answer 5

Assume the short one = X
then:
100={(X)*2}+{(X+10)*2}
100=2X+2X+20
80=4X
That is:
X=20

Answer 6

(w+2)(w+10)-w(w+8)=100
w^2+10w+2w+20-w^2+8w=100
10w+2w+24+8w=100
20w+20=100
-20 -20
20w=80
/20 /20
w=4
4+8=12=l