## Pages

### Word problems on Quadratic Equations (4)

Area of a rectangle is 110cm^2and perimeter is 54 cm. Determine the dimensions of rectangle by forming an equation in x.

#### Solution:

Given:
• Area of rectangle = 110 cm^2
• Perimeter of rectangle = 54 cm
• Length of rectangle = x
Since Area = length * width,
110 = x * "width"
110/x = "width"
Perimeter = 2(Length + Width)
54 = 2(x + 110/x)
54 = 2x + 220/x
54 = (2x^2 + 220)/x
54x = 2x^2 + 220
-2x^2 + 54x - 220 = 0
-(2x^2 - 54x + 220) = 0
2x^2 - 54x + 220 = 0
2x^2 - 44x - 10x + 220 = 0
2x(x - 22) - 10(x - 22) = 0
(x - 22) (2x - 10) = 0
x = 22 and 2x = 10
x = 22 and x = 5
Therefore length is 22 and width is 110/22 = 5