SOLUTION: A RECTANGLE HAS DIMENSIONS OF 18 FEET BY 13 FEET. A GRAVEL PATH OF UNIFORM WIDTH IS TO BE BUILT AROUND THE GARDEN. HOW WIDE CAN THE PATH BE IF THERE IS ENOUGH GRAVEL FOR 516 SQUARE

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: A RECTANGLE HAS DIMENSIONS OF 18 FEET BY 13 FEET. A GRAVEL PATH OF UNIFORM WIDTH IS TO BE BUILT AROUND THE GARDEN. HOW WIDE CAN THE PATH BE IF THERE IS ENOUGH GRAVEL FOR 516 SQUARE      Log On


   

Question 164449: A RECTANGLE HAS DIMENSIONS OF 18 FEET BY 13 FEET. A GRAVEL PATH OF UNIFORM WIDTH IS TO BE BUILT AROUND THE GARDEN. HOW WIDE CAN THE PATH BE IF THERE IS ENOUGH GRAVEL FOR 516 SQUARE FEET?
(I DON'T KNOW HOW TO SET THIS UP AS A QUADRATIC EQUATION.)
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
To help you set it up:
Imagine the 18 x 13 rectangle is already on the ground. Now imagine a border around that rectangle. What you have is a 'rectangle inside a rectangle'. The area you care about is the difference in the areas of the two rectangles. Assume the path is the same width all the way around. Let that width = X
Area of rectangle is A+=+Length%2AWidth
The inside rectangle is given as 18 x 13 = 234
The area of the outer rectangle is given as
Width+=+13+%2B+2W
Length+=+18+%2B+2W
AreaOfBorder+=+AreaOfOuterRecntangle+-+AreaOfInnerRectangle+
516+=+%2813%2B2W%29%2818%2B2W%29+-+234+
516+=+234+%2B+62W+%2B+4W%5E2+-+234
Now collect like terms, move everything to one side and solve
If you need more help, let me know