Math Problem Statement

Here are two arrangements of identical rectangles. Determine the dimensions, a and b, of one rectangle.

Solution

The problem you're working on involves two arrangements of identical rectangles, and the task is to find the dimensions of one rectangle. In the image:

  1. The first arrangement has a perimeter of 64 units, and it appears to be two rectangles placed side by side (the length is doubled).
  2. The second arrangement has a perimeter of 56 units, and it appears to have four rectangles stacked vertically.

To solve this, we'll use the perimeter formula for a rectangle:

P=2a+2bP = 2a + 2b

where aa is the length and bb is the width of the rectangle.

Step 1: Set up equations for each arrangement.

First arrangement (two rectangles side by side):

The perimeter is 64 units, and the length is 2a2a (since two lengths are added together), while the width is bb.

2(2a+b)=642(2a + b) = 64

Simplifying:

4a+2b=644a + 2b = 64 2a+b=32(Equation 1)2a + b = 32 \quad \text{(Equation 1)}

Second arrangement (four rectangles stacked):

The perimeter is 56 units, and the length is aa (just one length), while the width is 4b4b (since four widths are stacked).

2a+2(4b)=562a + 2(4b) = 56

Simplifying:

2a+8b=562a + 8b = 56 a+4b=28(Equation 2)a + 4b = 28 \quad \text{(Equation 2)}

Step 2: Solve the system of equations.

We now have two equations:

  1. 2a+b=322a + b = 32
  2. a+4b=28a + 4b = 28

To solve this, we can use substitution or elimination. Let's solve Equation 1 for bb:

b=322ab = 32 - 2a

Substitute this expression for bb into Equation 2:

a+4(322a)=28a + 4(32 - 2a) = 28

Simplifying:

a+1288a=28a + 128 - 8a = 28 7a+128=28-7a + 128 = 28 7a=28128-7a = 28 - 128 7a=100-7a = -100 a=14.29a = 14.29

Now substitute a=14.29a = 14.29 back into b=322ab = 32 - 2a:

b=322(14.29)=3228.58=3.42b = 32 - 2(14.29) = 32 - 28.58 = 3.42

Final dimensions:

  • a14.29a \approx 14.29 units
  • b3.42b \approx 3.42 units

Would you like more clarification on any steps or additional questions? Here are some related questions to explore further:

  1. What would happen if we had a third arrangement with a different number of rectangles?
  2. How would the perimeter change if the length aa was doubled but the width remained the same?
  3. Can we calculate the area of each rectangle using these dimensions?
  4. What would happen if we rearranged the rectangles in different shapes, like a 3x3 grid?
  5. How would the perimeter change if the width bb was tripled?

Tip: Always double-check the dimensions in a perimeter problem to ensure the formula is applied correctly for each arrangement!

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Math Problem Analysis

Mathematical Concepts

Algebra
Systems of Linear Equations
Geometry
Perimeter of Rectangles

Formulas

Perimeter of a rectangle: P = 2a + 2b

Theorems

Properties of Perimeter for Composite Shapes

Suitable Grade Level

Grades 6-8