rectangular wall

Question

The area of a rectangular wall of a barn is
30
square feet. Its length is
4
feet longer than twice its width. Find the length and width of the wall of the barn.

Answer 1

 

Let w = width of barn

Let 2w + 4 = length of barn .... "4 feet longer than twice the width"

 

The area is 30 ft²

 

   Area = Length • Width

   30  = (2w+4)(w)

   30 = 2w² + 4w

   2w² + 4w - 30 = 0

   w²  + 2w - 15 = 0 

   (w - 3)(w + 5) = 0

 

    w = 3 or w = -5 ... discard the -5;  cant have a negative side measure

 

The length is 2w + 4

    = 2(3) + 4

    = 6 + 4

    = 10

    

CHECK:

 

    Length • Width = Area

    10(3) = 30

 

Answer 2

In this problem, you don't know the length and width, but you know their relation to each other.

 

You know that the length (L) is 4 feet longer than 2x the width (W).

 

So L = 4+2W.

 

And you also know that length x width is the formula for area, so you know that W x L = 30 (the area of the barn).

 

The two equations look like this:

L = 4+2W

W x L = 30.

 

Since L and (4+2W) are equal to each other, you can substitute one for the other in any equation. You substitute the L in the second equation for (4+2W) and then you can solve for W.

 

L = 4+2W

W x (4+2W) = 30.

 

So now you use the distributive property to get rid of the parentheses:

 

W(4+2W) = 30.

4W + 2W² = 30.

 

Now, the only way to solve this is to use the quadratic formula. First, you have to rearrange the equation so it fits the standard a² + b + c = 0. You do this by subtracting 30 from both sides.

 

2W² + 4W - 30 = 0.

 

Now, you use the quadratic formula. The two solutions you get are 3 and -5. A negative answer doesn't make sense, because we're talking about the length of a wall, so the answer for W (width) is 3. Now, plug 3 back in to the equation for W and solve for L:

 

3 x L = 30.

L = 10.