Word problems on Quadratic Equations (4)

Area of a rectangle is `110cm^2`and 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`

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