We are given that the area of rectangle A is three times the area of triangle B, and we are asked to find the area of rectangle A. Here's how we can solve this:

Step 1: Find the area of triangle B

The formula for the area of a triangle is:

$\text{Area of Triangle} = \frac{1}{2} \times \text{base} \times \text{height}$

From the diagram, the base of triangle B is 3 cm and the height is $2x$ cm. Therefore, the area of triangle B is:

$\text{Area of Triangle B} = \frac{1}{2} \times 3 \times 2x = 3x \, \text{cm}^2$

Step 2: Express the area of rectangle A

The area of rectangle A is the product of its length and width. From the diagram, the length is $2x - 2$ cm and the width is 6 cm. So, the area of rectangle A is:

$\text{Area of Rectangle A} = 6 \times (2x - 2) = 6(2x - 2) = 12x - 12 \, \text{cm}^2$

Step 3: Set up the equation

We are told that the area of rectangle A is three times the area of triangle B. So we can set up the equation:

$\text{Area of Rectangle A} = 3 \times \text{Area of Triangle B}$

Substitute the expressions for the areas:

$12x - 12 = 3 \times 3x$

Step 4: Solve for $x$

Simplify the equation:

$12x - 12 = 9x$

Subtract $9x$ from both sides:

$3x - 12 = 0$

Add 12 to both sides:

$3x = 12$

Divide both sides by 3:

$x = 4$

Step 5: Find the area of rectangle A

Now that we know $x = 4$, substitute this value into the expression for the area of rectangle A:

$\text{Area of Rectangle A} = 12x - 12 = 12(4) - 12 = 48 - 12 = 36 \, \text{cm}^2$

Thus, the area of rectangle A is 36 cm².

Would you like any further details or explanations?

Here are 5 related questions for further exploration:

1. How would the problem change if the rectangle's width was variable instead of constant?
2. What would happen if the area of rectangle A was only twice the area of triangle B?
3. How could we determine the perimeter of triangle B using the value of $x$?
4. What other dimensions could satisfy the condition that rectangle A's area is three times triangle B's?
5. How would this problem change if we had a different shape, like a parallelogram, instead of a triangle?

Tip: When solving geometry problems involving relationships between areas, always remember to express all areas in terms of a single variable before setting up your equations.