SOLUTION: This is the problem: "A garden area is 30ft long and 20ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400ft(squared), what is the width

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: This is the problem: "A garden area is 30ft long and 20ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400ft(squared), what is the width      Log On


   

Question 29806: This is the problem:
"A garden area is 30ft long and 20ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400ft(squared), what is the width of the path?"
My biggest challenge with this word problem is how to set it up using the quadratic equation. If I could figure out the values for a, b and c, than I could finish the problem.
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
OK...with questions like these...a diagram is a MUST!

Draw a rectangle (i shall call it ABCD but you do not need to label it) and then a slightly larger one around it (i shall call it KLMN but you do not need to label it).

The outer rectangle, KLMN, is the garden.
The inner rectangle, ABCD, is the lawn? whatever it is...the Q doesn't say.
The path is the "pathlike" area between the 2 rectangles.

LABELLING
OK, label lengths of KLMN as 20 and 30.
Label the width of the path as x.

So... width of long edge of inner rectangle is 30 less 2 x's --> (30-2x)
and the width of the short edge of inner rectangle is 20 less 2 x's --> (20-2x).

And we know that the area of the inner rectangle is 400. So we have

(30-2x)(20-2x) = 400
+600+-+60x+-+40x+%2B+4x%5E2+=+400+
+600+-+100x+%2B+4x%5E2+=+400+
+200+-+100x+%2B+4x%5E2+=+0+
+4x%5E2+-+100x+%2B+200+=+0+
+x%5E2+-+25x+%2B+50+=+0+

solve this to find x.

jon.