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I see you've uploaded a geometry problem involving a triangle with labeled side lengths and angles. Let me guide you through solving it.

### Question 4a:
You are asked to calculate the value of \( x \), given that:
- The area of triangle \( ABC \) is \( 12\sqrt{2} \).
- \( \angle A = 45^\circ \).
- The sides of the triangle are \( AC = x - 4 \) and \( AB = x - 2 \).

We can use the formula for the area of a triangle involving two sides and the included angle:
\[
\text{Area} = \frac{1}{2}ab\sin(C)
\]
Where:
- \( a = AC = x - 4 \),
- \( b = AB = x - 2 \),
- \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \).

Thus, the area equation becomes:
\[
12\sqrt{2} = \frac{1}{2} \times (x - 4) \times (x - 2) \times \frac{\sqrt{2}}{2}
\]

### Solving for \( x \):
Let's simplify the equation step by step. First, we multiply both sides of the equation by 2 to eliminate the fraction:
\[
24\sqrt{2} = (x - 4)(x - 2) \times \frac{\sqrt{2}}{2}
\]
Next, multiply through by 2 to simplify the equation:
\[
48 = (x - 4)(x - 2)
\]

Now expand the right-hand side:
\[
48 = x^2 - 2x - 4x + 8
\]
\[
48 = x^2 - 6x + 8
\]

Subtract 48 from both sides:
\[
0 = x^2 - 6x - 40
\]

Now, solve the quadratic equation:
\[
x^2 - 6x - 40 = 0
\]

We can solve this using the quadratic formula:
\[
x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-40)}}{2(1)}
\]
\[
x = \frac{6 \pm \sqrt{36 + 160}}{2}
\]
\[
x = \frac{6 \pm \sqrt{196}}{2}
\]
\[
x = \frac{6 \pm 14}{2}
\]

Thus, we have two possible values for \( x \):
\[
x = \frac{6 + 14}{2} = 10 \quad \text{or} \quad x = \frac{6 - 14}{2} = -4
\]

Since a side length cannot be negative, we discard \( x = -4 \), leaving:
\[
x = 10
\]

### Question 4b:
We are asked to find the side \( BC \). Now that we know \( x = 10 \), we can substitute \( x \) into the expressions for \( AB \) and \( AC \):

- \( AB = x - 2 = 10 - 2 = 8 \),
- \( AC = x - 4 = 10 - 4 = 6 \).

To find \( BC \), we can use the **Law of Cosines**:
\[
BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(\angle A)
\]
Substituting the known values:
\[
BC^2 = 8^2 + 6^2 - 2(8)(6)\cos(45^\circ)
\]
\[
BC^2 = 64 + 36 - 96 \times \frac{\sqrt{2}}{2}
\]
\[
BC^2 = 100 - 48\sqrt{2}
\]

This is the expression for \( BC^2 \). You can now approximate or leave the result in terms of \( \sqrt{2} \).

---

Do you want further clarification on this, or do you have any questions?

### 5 Related Questions:
1. What is the Law of Cosines, and how is it used in triangle calculations?
2. Can we solve for \( BC \) using any other methods?
3. How does the sine rule differ from the cosine rule?
4. How would the problem change if the angle \( A \) were different?
5. What are some real-life applications of triangle area calculations?

### Tip:
When solving problems involving angles and side lengths, the Law of Sines or Cosines can often simplify your calculations significantly.

The area of triangle ABC is 12√2. Calculate the value of x. Also, find BC.

Learn how to calculate the value of x and the length of side BC in triangle ABC using trigonometry and algebra. This problem involves key concepts like the Law of Cosines and the quadratic equation, making it suitable for students in Grades 9-11.

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