The area of a rectangular field is 2275 m2. the field is enclosed by 200 m of fence. what are the dimensions of the field?

So it is enclosed by 200 m of  fence
perimiter=200 m
it is a rectangle to
p=legnth+legnth+width+width or
p=l+l+w+w or
p=2l+2w or
p=2(l+w)
so
p=200
200=2(l+w)
divide both sides by 2
100=l+w
area=l times w
area=2275
lw=2275
l+w=100
combine and solve
l+w=100
subtract w
l=100-w
subisutute 100-w for l in other eqution
(100-w)(w)=2275
distribute
-w^2+100w=2275
add (w^2-100w) to both sides
0=w^2-100w+2275
factor
find what 2 numbers add up to -100 and multiply to get 2275
guess (or factor 2275 and find factors that add up to -100)
figure out that they are -65 and -35
0=(w-65)(w-35)
set each to zero
0=w-65
0=w-35

solve for w
w=65 or 35
65>35 so
65 m=legnth
35 m=width
 

c) 6

step-by-step explanation:

let pq = qr = rs = x, and xz = y and zu = z.

if the horizontal lines are parallel, then corresponding segments are in proportion:

substitute:

therefore 3 = y = z.

xu = y + z → xu = 3 + 3 = 6