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We are given that the area of the rectangle is numerically equal to the perimeter of the triangle.

### Step 1: Write the expression for the area of the rectangle
- The dimensions of the rectangle are \( 2x - 1 \) and \( x + 8 \).
- The area of the rectangle is calculated as:
\[
\text{Area of rectangle} = \text{length} \times \text{width} = (2x - 1)(x + 8)
\]
Expanding this expression:
\[
\text{Area} = 2x(x + 8) - 1(x + 8) = 2x^2 + 16x - x - 8 = 2x^2 + 15x - 8
\]

### Step 2: Write the expression for the perimeter of the triangle
- The sides of the triangle are \( 5x - 1 \), \( 6x + 8 \), and \( 8x + 15 \).
- The perimeter of the triangle is the sum of the three sides:
\[
\text{Perimeter of triangle} = (5x - 1) + (6x + 8) + (8x + 15)
\]
Simplifying:
\[
\text{Perimeter} = 5x + 6x + 8x - 1 + 8 + 15 = 19x + 22
\]

### Step 3: Set up the equation
We are told that the area of the rectangle is numerically equal to the perimeter of the triangle:
\[
2x^2 + 15x - 8 = 19x + 22
\]
Move all terms to one side:
\[
2x^2 + 15x - 8 - 19x - 22 = 0
\]
Simplify:
\[
2x^2 - 4x - 30 = 0
\]
Divide through by 2 to simplify:
\[
x^2 - 2x - 15 = 0
\]

### Step 4: Solve the quadratic equation
We now solve \( x^2 - 2x - 15 = 0 \) using the quadratic formula:
\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-15)}}{2(1)}
\]
\[
x = \frac{2 \pm \sqrt{4 + 60}}{2} = \frac{2 \pm \sqrt{64}}{2}
\]
\[
x = \frac{2 \pm 8}{2}
\]
So, \( x = \frac{2 + 8}{2} = 5 \) or \( x = \frac{2 - 8}{2} = -3 \).

Since the dimensions of the shapes must be positive, we discard \( x = -3 \) and keep \( x = 5 \).

### Final Answer:
The value of \( x \) is \( 5 \).

Would you like further details on any part of this solution?

### 5 Related Questions:
1. How do we check if the calculated value of \( x \) satisfies both the area and perimeter conditions?
2. What are other ways to solve the quadratic equation besides using the quadratic formula?
3. How do we handle such equations when multiple solutions arise, and one needs to be discarded?
4. How would the problem change if the perimeter of the triangle were instead equal to the area of a different shape?
5. How do you find the dimensions of the shapes given the value of \( x \)?

### Tip:
When solving quadratic equations, always check for both possible solutions to see which one makes sense in the context of the problem (e.g., dimensions must be positive).

The area of the rectangle, measured in cm², is numerically equal to the perimeter of the triangle, measured in cm. Work out the value of x. If your answer is a decimal, give it to 1 d.p.

This math problem involves solving for x when the area of a rectangle equals the perimeter of a triangle. Using quadratic equations and geometry concepts, the solution is explained step-by-step. Suitable for students in Grades 8-10.

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