# A rectangle has two parallel sides 4y+8 and 20x-16. The other parallel side are x^2 and yFind the perimeter if one side has an odd integer length.

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### 2 Answers

The parallel sides of the rectangle are 4y+8 and 20x-16.

The other parallel sides are x^2 and y.

Since parallel sides are equal in a rectangle,

4y+8 = 20x-16.......(1).

x^2= y....(2)

Substituting y = x^2 in eq (1), we get 4x^2 = 20x-16.

We divide by 4.

x^2-5x+4 = 0

( x-4)(x-1).

Therefore x = 4 or x= 1

Therefore from (2) y = x^2.

y = 4^2 = 16 Or y = 1^2 = 1

1 is odd. So one of the parallel sides which is y is 1 . Therefore another of the parallel sides is 20x- 16 = 21*1 -16 = 4.

Therefore the perimeter = 2 (sum of the parallel sides) = 2(1+4) = 10

Multiplication by an even number makes an odd number even

so, if y+8 and 20x-16 are even we need y^2 and x^2 to be odd.

We'll put the lengths of parallel sides in the relations:

4y+8 = 20x-16 (1)

and

y=x^2 (2)

We'll work over the relation (1):

We'll subtract 20x and 8 both sides:

4y - 20x = -8 - 16

4y - 20x = -24

From (2), we'll have y=x^2.

4x^2- 20x = -24

We'll add 24 both sides:

4x^2- 20x + 24 = 0

We'll divide by 4:

x^2 - 5x + 6 = 0

We'll apply quadratic formula:

x1 = [5+sqrt(25-24)]/2

x1 = (5+1)/2

x1 = 3

x2 = (5-1)/2

x2 = 2

We'll re-write the quadratic as a product of linear factors:

x^2 - 5x + 6 = (x-2)(x-3)

If x1 = 3 => y = x1^2 => y1 = 9

If x2= 2=> y = x2^2 => y2 = 4

We'll choose the odd length for y: y = 9.

The perimeter is:

P = (4y+8) +(20x-16)+y+x^2

P = (4*9 + 8) + (20*3 - 16) + 9 + 9

P = 44 + 44 + 2*9

P = 88 + 18

**P = 106 units**