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The problem involves a right-angled triangle with sides labeled in terms of \( x \). The lengths of the sides are as follows:
- One leg: \( x + 5 \),
- Another leg: \( x - 2 \),
- The hypotenuse: \( 2x - 1 \).

We are tasked with finding the value of \( x \) and the area of the triangle.

### Step 1: Apply the Pythagorean theorem
For a right-angled triangle, the Pythagorean theorem states:
\[
(\text{leg}_1)^2 + (\text{leg}_2)^2 = (\text{hypotenuse})^2
\]
Substituting the values:
\[
(x + 5)^2 + (x - 2)^2 = (2x - 1)^2
\]

### Step 2: Expand both sides
Expand each square:
- \( (x + 5)^2 = x^2 + 10x + 25 \),
- \( (x - 2)^2 = x^2 - 4x + 4 \),
- \( (2x - 1)^2 = 4x^2 - 4x + 1 \).

Now substitute back into the Pythagorean equation:
\[
(x^2 + 10x + 25) + (x^2 - 4x + 4) = 4x^2 - 4x + 1
\]

### Step 3: Simplify the equation
Combine like terms on the left side:
\[
2x^2 + 6x + 29 = 4x^2 - 4x + 1
\]

Move all terms to one side:
\[
0 = 4x^2 - 4x + 1 - 2x^2 - 6x - 29
\]
\[
0 = 2x^2 - 10x - 28
\]

### Step 4: Solve the quadratic equation
Divide the equation by 2 to simplify:
\[
x^2 - 5x - 14 = 0
\]

Use the quadratic formula to solve for \( x \):
\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-14)}}{2(1)}
\]
\[
x = \frac{5 \pm \sqrt{25 + 56}}{2}
\]
\[
x = \frac{5 \pm \sqrt{81}}{2}
\]
\[
x = \frac{5 \pm 9}{2}
\]
Thus, the two possible values for \( x \) are:
\[
x = \frac{5 + 9}{2} = 7 \quad \text{or} \quad x = \frac{5 - 9}{2} = -2
\]

Since \( x \) must be positive for the side lengths to make sense, we have:
\[
x = 7
\]

### Step 5: Find the area of the triangle
The area of a right-angled triangle is given by:
\[
\text{Area} = \frac{1}{2} \times \text{leg}_1 \times \text{leg}_2
\]
Substitute the values:
- \( \text{leg}_1 = x + 5 = 7 + 5 = 12 \),
- \( \text{leg}_2 = x - 2 = 7 - 2 = 5 \).

Thus, the area is:
\[
\text{Area} = \frac{1}{2} \times 12 \times 5 = 30
\]

### Final Answer:
- \( x = 7 \),
- The area of the triangle is \( 30 \, \text{square units} \).

Would you like more details on any specific step? Here are some related questions:

1. How is the quadratic formula derived?
2. What is the significance of the Pythagorean theorem in geometry?
3. Can you explain more examples of solving quadratic equations?
4. How do you find the area of different types of triangles?
5. What are the practical applications of right-angled triangles?

**Tip:** Always check that the values for the variable make sense in the context of the problem, especially when dealing with geometric shapes.

The diagram shows a right-angled triangle. Find the value of x and hence the area of the triangle.

Solve for x in a right-angled triangle using the Pythagorean theorem and quadratic equation. Learn how to find the area of the triangle step by step. Suitable for students in Grades 9-10.

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