The Algebra Word Problem Solver solves Area of Rectangle Word Problems.

Find AREA
Let A=AREA
Let L=LENGTH
Let W=WIDTH
Given A=LW
Given L=4
Given W=6
A=4W                 //  Substitute 4 for L in A=LW
A=24
A=24                 //  Substitute 6 for W in A=4W
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Solving Rectangle Word Problems

• Look for the words rectangle or rectangular. These words most often identify the problem type as pertaining to rectangles.
• Find what is given. Look for phrases similar to the following:
  • Length is 12 cm. This translates to L = 12 cm
  • Width of 42 cm...Area is ... or Perimeter is...
    
• See what the problem is asking to be found. Look for phrases similar to the following:
  • Find the Length, or Find the Width or Find the Area or Find the Perimeter.
  • Find the dimensions. This means find the length and width.
• Use the Rectangle Area and Rectangle Perimeter Formulas:
  • Rectangle Area: Area = Length times Width or A=LW
  • Rectangle Area: Perimeter = 2 times (Length plus Width) or P = 2(L+W).

Examples of Rectangle Problems solved by the Algebra Word Problem Solver:
* Find the width of a rectangle if the length is 12 cm and the perimeter is 42 cm.

* The length of a rectangle is 3 inches less than twice the width. If each dimension is increased by 4 inches, the area is increased by 88 square inches.
Find the dimensions of the original rectangle.

>>> Please see below for a sample word problem * solved by the Algebra Word Problem Solver * with solution.