# 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.

*print*Print*list*Cite

### 4 Answers

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.**

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.**

The area of the rectangle is 200 sq m.

The perimeter of the rectangle is 60 m.

Let y be the width, the length x= (60- 2y)/2 = 30 -y.

So the area = length *width = (30-y)y = 200 sq m by data.

So 30y - y^2 = 200.

y^2 -30y +200 = 0

y- 20y -10y +200 = 0

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

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

Therefore y-20 = 0 , or y = 20, or y - 10 = 0 or y = 10.

So width = 10m and length x = 20m, so that the perimeter = 2(20+10)m = 60m and area = 20m*10m = 200 sq m.

Since the geometric shape is a rectangle, the perimeter of rectangle is:

2x + 2y = 60

We'll divide by 2 and we'll get:

x + y = 30 (1)

The area of the rectangle is:

A = x*y

200 = x*y

We'll use the symmetric property and we'll get:

x*y = 200 (2)

We'll form the quadratic equation when we know the sum and the product:

x^2 - 30x + 200 = 0

We'll apply the quadratic formula:

x1 = [30 + sqrt(900 - 800)]/2

x1 = (30 + 10)/2

x1 = 20 m

x2 = 10 m

**We'll take into account the constraint from enunciation and we'll get:**

**x > y => the length: x = 20 m**

** the width: y = 30 - x = 10 m**