# GeometryA 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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### 2 Answers

The length of the rectangle is x and the width is y. The area of a rectangle is x*y and the perimeter is 2(x + y)

Here we have area = 200

=> xy = 200

Perimeter = 60

=> 2(x + y) = 60

=> x + y = 30

=> x = 30 - y

Substitute in xy = 200

(30 - y)y = 200

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

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

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

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

y = 20 or y = 10

**As y is the width y = 10 and the length = 20.**

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