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.