if the area of a rectangle is 120sq. units and the perimeter is 44sq. units. find its length.?

Answer 1

let x = width
y = height

area of the rectangle = xy = 120
perimeter = 2x + 2y = 44

from the second equation, 2x + 2y = 44
therefore, y = (44 - 2x)/2 or 22 - x

then plug that y value into the first equation xy = 120

x(22 - x) = 120

solve for x,

22x - x^2 = 120
-x^2 + 22x - 120 = 0

now factor it

-(x^2 - 22x + 120) = 0
x^2 - 22x + 120 = 0 (0 divided by -1 equals 0)
(x - 10)(x - 12) = 0
x = 10, 12

or use quadratic formula,

x = (-b (+/-) sqrt(b^2 - 4ac))/(2a)
x = (-22 (+/-) sqrt(484 - 4(-1)(-120)))/(2(-1))
x = (22 (+/-) sqrt(4))/2
x = 11 (+/-) 1
x = 10, 12

then plug x values back into either the area equation or the perimeter equation in order to find the y values,

when x = 10, y should equal 12
when x = 12, y should equal 10

therefore, one side is 10 and another side is 12.

Answer 2

12 units

Answer 3

A=wL
P= 2w+2L

44=2w+2L
120=wl

w= 120/L

44= 2(120/L)+2L
44= 2(120/L+L)
22=120/L+L
22-L= 120/L
22L-L^2= 120
L^2-22L+120
(L-10)(L-12)

L= 10 or 12

Answer 4

l*b=120
l=120/b
2(l+b)=44
120/b+b=22
120+b^2=22b
b^2-22b+120=0
b^2-10b-12b+120=0
b=10,then l=12
since b can not be 12bcoz then it will be greater than l which is not possible
therefore l=12 units

Answer 5

l*b = 120
b=120/l
2( l+b) = 44
l+b = 22
l + 120/l=22
l^2 + 120 = 22l
l^2 - 22l +120 =0
solve it