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.
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
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.
giorgiana1976 | College Teacher | (Level 3) Valedictorian
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