Algebra Rectangle Word Problem Game

Algebra Rectangle Word Problem Quiz/Game
This game focuses on solving algebra word problems that involve rectangles, covering length, width, area, and perimeter. Scroll down the page for a more detailed explanation.

 

 

How to Play the Algebra Architect: Rectangles Game

1. Look at the Problem: Read the problem carefully. Solve it and select one of the answers.
   
2. Check Your Work: If you selected the right answer, it will be highlighted in green. If you are wrong, it will be highlighted in red and the correct answer will be highlighted in green. A hint will be given to help you find the correct answer.
   
3. Get a New Problem: Click “Next Problem” for a new problem.
   Your score is tracked, showing how many you’ve gotten right.
   
4. Finish Game When you have completed 10 questions, your final score will be displayed.
    
   

Solving Rectangle Word Problems with Algebra
Rectangle word problems in algebra usually ask you to find dimensions (length and width) based on relationships between them and a given perimeter or area.
The Essential Formulas
Before starting, you must know these two formulas by heart:
Perimeter (P): P = 2L + 2W (or P = 2(L + W))
Area (A): A = L × W

The Step-by-Step Strategy
Step 1: Identify the “Base” Variable
Most problems describe one dimension in terms of the other.
Example: “The length is 3 more than the width."
Action: Let the width be x. Then the length is x + 3.

Step 2: Translate Phrases into Expressions
Look for keywords to build your algebraic expressions:
“More than” / “Greater than”: Addition (+)
“Less than”: Subtraction (-) — Be careful with order! “5 less than width” is W - 5.
“Times” / “Twice”: Multiplication

Step 3: Set Up the Equation
Plug your expressions into the relevant formula (P or A).
If the problem gives you Perimeter, use 2L + 2W = P.
If the problem gives you Area, use L × W = A.

Step 4: Solve and Answer
Solve for x, then plug that value back into your expressions to find both the length and the width.

Worked Examples
Case A: Perimeter Problem
Problem: The perimeter of a rectangle is 44 cm. The length is 2 cm less than three times the width. Find the dimensions.
Define Variables:
Width = w
Length = 3w - 2
Set up Equation (P = 2L + 2W):
2(3w - 2) + 2(w) = 44
Solve:
6w - 4 + 2w = 44
8w - 4 = 44
8w = 48
w = 6
Find Dimensions:
Width = 6 cm
Length = 3(6) - 2 = 16 cm

Case B: Area Problem
Problem: The area of a rectangle is 40 square inches. The length is 3 inches more than the width. Find the dimensions.
Define Variables:
Width = x
Length = x + 3
Set up Equation (L × W = A):
(x + 3)(x) = 40
Solve (Quadratic):
x² + 3x = 40
x² + 3x - 40 = 0
(x + 8)(x - 5) = 0
x = -8 or x = 5 (Length cannot be negative, so x = 5)
Find Dimensions:
Width = 5 in
Length = 5 + 3 = 8 in
 

Common Pitfalls to Avoid

• Forgetting the “2”: In perimeter problems, students often write L + W = P. Remember there are two of each side.
  
• Units: Ensure your final answer includes the units (cm, in, m, etc.).
  
• Negative Solutions: In area problems (quadratics), you will often get a negative x value. Discard it; distances must be positive.
   
  

This video gives a clear, step-by-step approach to explain how to solve a word problem involving the area between rectangles.

• Show Video

 

Check out our most popular games!

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.