a rectangular fields perimeter is 220 meters the area is 2,856 sq meters what are its dimensions dimentions?

Answer 1

2x + 2y = 220
xy = 2856
x = 2856 / y

x + y = 110
2856/y + y = 110
2856 + y² = 110y
y ² - 110y + 2856 = 0
y = [110 ±√ (110 ² - 11424) ] / 2
y = [110 ± 26 ] / 2
y = 68 , 42
Sides are 42 m and 68 m

Answer 2

let x = length
let y = width of the rectangular field

...rectangular fields perimeter is 220 meters ...
2*(x + y) = 220
x + y = 110 (Eq 1)

...the area is 2,856 sq meters...
x * y = 2856 (Eq 2)

(Eq 1) x + y = 110
x = 110 - y

plug this into Eq 2
x * y = 2856
(110 - y) * y = 2856
110 y - y^2 = 2856
y^2 - 110 y + 2856 = 0
(y-68) * (y-42) = 0

y= 68 and y = 42 are the two answers, plugging back into Eq 1:

x + y = 110
x = 110 - y

x = 110 - 68 = 42 and
x = 110 - 42 = 68

thus the rectangle should have dimensions 68 and 42 meters

Answer 3

Duel equations that both solve the given requirements.
Since it is rectangular then it has to have four sides with two of each equal to each other.
2A + 2B = 220 m
A + B = 110 m
A * B = 2856 m^2
Use one equation to get one variable equal to a number and the other variable. So,
A = 110 - B
Then replace the A in the second equation with the new equation. This gets you-
(110 - B)*B = 2856
110B - B^2 = 2856
So,
B^2-110 B + 2856 =0 which is a quadratic eqation.
The other way would give you:
A = 2856/B
then
(2856/B) + B = 110 which is also a quadratic. So you can't get around it. Use the quadratic equation and see if you get reasonable answers. You may not.

Answer 4

Length = l, Width = w
Perimeter = 2l + 2w = 2(l + w) = 220
<==> l + w = 110

Area = l*w = 2856
l + w = 110
(l + w)^2 = 110^2 = 12100
l^2 + w^2 + 2*l*w = 12100
but l*w = 2856 ==> 2*l*w = 5712 and 4*l*w = 11424

l^2 + w^2 + 2*l*w - 4*l*w = 12100 - 11424
l^2 + w^2 - 2*l*w = 676
(l - w)^2 = 676
l - w = 26

Thus,
(Eq.1) : l + w = 110
(Eq.2) : l - w = 26
Let's add them up in order to isolate "l"
l + w + l - w = 110 + 26
2l = 136
l = 136/2 = 68

Substitute in (Eq.1)
l + w = 110 and l = 68
68 + w = 110
w = 110 - 68 = 42

In conclusion : l = 68, w = 42

Answer 5

perimeter is the lenght all the way around the rectangle so we have to make an equation to fit that standard to move on

let x be length and y be width

2(x+y)=220
/2 /2
x+y=110

could be any combination of numbers that equal 110 but because there is an equation for area you have to find the right combo by plugging in numbers
****note that it does say RECTANGLE so start plugging with a big number and a small number

area
x*y=2856

after about 12 plugs.....
68*42=2856
68+42=110
*2=220 the perimeter

Answer 6

P = 2x + 2y ==> 220 = 2x + 2y ==> 110 = x + y ==> x = 110 -y
A = xy ==> 2856 = xy ==> 2856 = y*(110-y) ==>
2856 = 110y-y^2 ==> y^2-110y+2856=0
solving for y = [-(-110)+-sqrt((-110)^2-(4*1*2856))]/(2*1)

y = 68 and x = 42
or
y = 42 and x = 68

Answer 7

LET ONE SIDE=X
AS AREA=2856
OTHER SIDE=2856/X

NOW PERIMETER=2(X+2856/X)=220
OR X^2+2856=110X
OR X^2-110X+2856=0
OR X^2-68X-42X+2856=0
ORX(X-68)-42(X-68)=0
OR X=68
OTHER SIDE=42

Answer 8

68 x 42