A rectangle has a perimeter of 60 m and an area of 200 m2. Find the length x and width y, x > y, of the rectangle.

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We are given that the perimeter of the rectangle is 60m and the area is 200m^2.

The length x and width y, have to be found.

Now the perimeter = 2x+2y = 60

=> x+y = 30

=> x = 30 -y

The area is given by xy = (30-y)*y = 200

=> 30y - y^2 =200

=> y^2 - 30y + 200 = 0

=> y^2 - 20y - 10y + 200=0

=> y(y-20) - 10(y-20) =0

=> (y-10)(y-20) =0

So y can be 10 or 20, but as it is the width we take y as 10.

Therefore the length of the rectangle is 20 and the width is equal to 10.

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Perimeter = 60

Area = 200 m^2

Find x an y such that: x > y

We know that the perimeter of the rectangle is:

 P = 2x + 2y

==> 60 = 2x + 2y

Divide by 2:

==> x + y = 30

==> x = 30 - y ...............(1)

Also we know that the area of the rectangle is:

 A = x*y

==> 200 = x*y

But from (1) , we know that x = 30 - y:

==> 200 = (30-y)*y

==> 200 = 30y - y^2

==> y^2 - 30y + 200 = 0

Let us factor:

==> (y- 20) ( y- 10) = 0

==> y1= 20  ==> x1= 10

==> y2= 10 ==> x2= 20

But given that: x > y

Then x = 20   and y = 10 is the sides of the rectangle.

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