the dimension of a square and rectangle where the sum of the perimeter is 44?

Question

and the recatangle has and area of 120

Answer 1

It is not a square,

11 X 11 is 121

therefore it is a rectangle 10 X 12

2x + 2y = 44
x * y = 120
x+y=22
y=22-x

x*(22-x) = 120
22x - x^2 + 120=0
(x-12) * (x-10) = 0

x = 10 or x=12

Therefore the dimensions are 10X12

Answer 2

Assume a rectangle with sides of length a & b. We know the perimeter is 44 and the area 120 so

2(a + b) = 44
ab = 120

Rearrange the second equation and subsitute it for b in the first

2(a + (120/a)) = 44 or
a + 120/a = 22

Multiply thru by a, then rearrange as a quadratic

a*a + 120 = 22a
a*a - 22a + 120 = 0

Use the standard equation (-b +/- sqrt (b*b - 4ac)/2a to solve

a = (22 +/- sqrt (-22*-22 - 4 * 1 * 120)) / 2 * 1
a = (22 +/- sqrt (484 - 480)) / 2
a = (22 +/- sqrt (4)) / 2
a = (22 +/- 2) / 2
a = 11 +/- 1
a = 10 or 12

From second equation b is 120 / a so the solutions are

a = 10, b = 12 or
a = 12, b = 10

Both solutions are the same shape rectangle with different rotations. There is no square (e.g. a and b would be equal) that fits the given perimeter and area.

Answer 3

Square:
Perimeter= 4*length of sides
side=44/4 = 11
Dimension of square: 11x11

Rectangle:
Perimeter= 2*l(Length+Breadth)
l+b=22
b=22-l
lb=120
22l-l^2=120
l=10,12
b=12,10
dimensions of rectangle:12x10

Answer 4

square = 11 x11
rectangle = 10 x 12

Answer 5

You really should ask your teacher explain how to find the answer rather than ask someone else to do your homework for you. The purpose of school is not to get all the answer right but to learn how to answer these quesions yourself.


Peace

Answer 6

Square is 11x11
Rectangle is 10x12