Examkin

Question

Difficulty: HardQuadratic Equations and Factoring

A rectangular plot of land has a perimeter of 56 meters and an area of 180 square meters. A border of uniform width xx meters is constructed around the inside edge of the plot, reducing the remaining inner area to 84 square meters. What is the value of xx?

Answer: 2 meters

Answer

The border width x is 2 meters.
Solving the system L+W=28L + W = 28 and LW=180LW = 180 gives outer dimensions of 18 and 10 meters. The inner area equation (18−2x)(10−2x)=84(18 - 2x)(10 - 2x) = 84 simplifies to x2−14x+24=0x^2 - 14x + 24 = 0, which factors into (x−2)(x−12)=0(x - 2)(x - 12) = 0. Since 2x2x must be less than the total width of 10 meters (x<5x < 5), the extraneous root x=12x = 12 is discarded, leaving x=2x = 2.

Step-by-Step Solution

1
Determine the outer dimensions of the plot.
Length = 18 meters, Width = 10 meters
The semi-perimeter is L + W = 56 / 2 = 28, and the area is L * W = 180. Forming the quadratic equation t^2 - 28t + 180 = 0 yields roots t = 18 and t = 10.
2
Write the quadratic expression for the inner area in terms of x.
(18 - 2x)(10 - 2x) = 84
A uniform border of width x along all four sides reduces both the length and width by 2x.
3
Expand and simplify the quadratic equation into standard form.
x^2 - 14x + 24 = 0
Expanding yields 180 - 56x + 4x^2 = 84. Subtracting 84 gives 4x^2 - 56x + 96 = 0, which divides by 4 to give x^2 - 14x + 24 = 0.
4
Factor the quadratic equation to solve for possible values of x.
x = 2 or x = 12
Factoring x^2 - 14x + 24 = 0 gives (x - 2)(x - 12) = 0.
5
Eliminate the extraneous root based on physical geometry constraints.
x = 2
The total width is 10 meters, so 2x must be less than 10 (x < 5). Thus, x = 12 is physically impossible and discarded.

Key Concept

Quadratic Equations, Factoring, and Extraneous Solutions
Rate this question

Topics

Quadratic Equations and Factoring

More questions from this topic