   SEARCH HOME Math Central Quandaries & Queries  Question from Jason, a student: Two rectangular plots have the same area. The first has a perimeter of 40m while the second has a length 2m less than the first and a width 1m greater. Find the length and breadth of the first rectangle Hi Jason.

Start by assigning variables:
Let L be the length and B be the breadth of the first rectangle.

Then the area of the first rectangle is BL, right? Length x Breadth = Area.

The perimeter of the first rectangle is the total of its sides: B+B+L+L = 40m. So 2B+2L = 40.

The second rectangle has a length 2m less than the first, so its length is (L-2). The Width (breadth) is 1m greater, so the breadth is (B+1). But we know the area is this length times this breadth, and that is (L-2)x(B+1).

Now the question said that the area of the first rectangle [BL] and the area of the second rectangle [(L-2)(B+1)] are the same. That means they are equal.

BL = (L-2)(B+1)

But we also determined that

2B + 2L = 40

This gives us "two equations with two unknowns". So now we can solve it using either the substitution method or the elimination method. You get to decide which one you want to use (they will both give you the same answer, of course).

If you need examples of how to do this, then use the Quick Search in our Quandaries and Queries to look for "elimination method" or look for "substitution method".

Hope this helps,
Stephen La Rocque.     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.