Bob wants to build a doghouse that is 2 feet longer than it is wide. He is building it on a concrete slab that will leave 2 feet of concrete slab visible ob each side. If the area of the house is 48 square feet, what are the dimensions of the concrete slab ?
Let's start by using algebra to solve for the dimensions of the doghouse.
First, we know that the area of the house is 48 square feet. We can use the formula for the area of a rectangle (length x width = area) to set up an equation:
(length)(width) = 48
Next, we know that the length of the doghouse is 2 feet longer than the width. We can use algebra to represent this relationship:
length = width + 2
We can substitute this expression for length in the area equation:
(width + 2)(width) = 48
Now we just need to solve for the width.
Expanding the expression on the left:
width^2 + 2width = 48
Moving all the terms to one side:
width^2 + 2width - 48 = 0
We can use the quadratic formula to solve for the width:
width = (-b ± sqrt(b^2 - 4ac)) / 2a
In this case, a = 1, b = 2, and c = -48. Plugging those values in:
width = (-2 ± sqrt(2^2 - 4(1)(-48))) / 2(1)
width = (-2 ± sqrt(4 + 192)) / 2
width = (-2 ± sqrt(196)) / 2
Taking the positive value for width:
width = (-2 + 14) / 2 = 6
So the width of the doghouse is 6 feet.
We can use the expression we derived earlier to find the length:
length = width + 2 = 8
So the dimensions of the doghouse are 6 feet by 8 feet.
To find the dimensions of the concrete slab, we need to add 2 feet on each side:
width of slab = 6 + 4 = 10 feet
length of slab = 8 + 4 = 12 feet
So the dimensions of the concrete slab are 10 feet by 12 feet.