dimensions 13

Question

A security fence encloses a rectangular area on one side of a park in a city. Three sides of fencing are? used, since the fourth side of the area is formed by a building. The enclosed area measures 800
square feet. Exactly 80
feet of fencing is used to fence in three sides of this rectangle. What are the possible dimensions that could have been used to construct this? area

Answer 1

Hi Shawn.

 

Because the security fence is only on 3 sides, 2 sides will be of length x, the 3rd side will be of length y.

 

So the amount of fencing you need is 2x + y = 80 feet of fencing used.

 

To find x and y, they give you a hint by way of the area.  

 

The enclosed area = 800 square feet, and the area's formula is one side (x) * the other side (y).

 

So we have that (x)(y)=800   and that  2x + y = 80.

 

There are many ways to solve this, the easiest would be to solve for y (in the 2x + y = 80 equation) and then substitute it into the other (xy=800) equation.

 

Doing this, y = 80 - 2x

 

So xy = 800 becomes x(80-2x) = 800

 

Distribute and get that 80x - 2x² = 800.

 

To solve this, put all of the information on the left and let the right side =0.  In other words, subtract 800 from both sides to get

 

-2x² + 80x - 800 = 0.

 

I don't like working with negative x²'s, so I am going to divide EVERY term by (-2).

 

x² - 40x +400 = 0

 

This factors into (x-20)(x-20) = 0

 

So x-20 = 0, or x = 20.  This is the dimension of your two matching sides.

 

To find y, just go back to where you rearranged to get y = 80 - 2x, and put 20 in for the x.

 

y = 80 - 2(20) = 80 - 40 = 40

 

So the dimension of the side opposite the building, is 40 feet.  The dimension of the other two parallel sides is 20 feet.

 

Notice, 20 * 40 = Area = 800

and 2(20) + 40 = 80 = amount of fencing.

 

 

Final answer:  Dimensions are 20 ft x 40 ft