To solve the problem, we'll need to set the areas of the two figures equal to each other and form a quadratic equation in terms of $x$.

1. The area of the half-circle with radius $x$: $A_1 = \frac{1}{2} \pi x^2$

2. The area of the quarter-circle with radius $x + 2$: $A_2 = \frac{1}{4} \pi (x + 2)^2$

Since the areas are equal: $\frac{1}{2} \pi x^2 = \frac{1}{4} \pi (x + 2)^2$

Dividing both sides by $\pi$: $\frac{1}{2} x^2 = \frac{1}{4} (x + 2)^2$

Multiply both sides by 4 to clear the fraction: $2x^2 = (x + 2)^2$

Expand the right-hand side: $2x^2 = x^2 + 4x + 4$

Subtract $x^2 + 4x + 4$ from both sides: $2x^2 - x^2 - 4x - 4 = 0$

This simplifies to: $x^2 - 4x - 4 = 0$

Now, we'll solve this quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = -4$, and $c = -4$:

1. Calculate the discriminant: $\Delta = b^2 - 4ac = (-4)^2 - 4(1)(-4) = 16 + 16 = 32$

2. Calculate the square root of the discriminant using the given approximation for $\sqrt{2}$: $\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2} \approx 4 \times 1.41 = 5.64$

3. Apply the quadratic formula: $x = \frac{-(-4) \pm 5.64}{2(1)} = \frac{4 \pm 5.64}{2}$

This gives us two solutions: $x = \frac{4 + 5.64}{2} = \frac{9.64}{2} = 4.82$ $x = \frac{4 - 5.64}{2} = \frac{-1.64}{2} = -0.82$

Since $x$ represents a radius, it must be positive: $x \approx 4.8$

So, the value of $x$ correct to the first decimal place is $4.8$.

Would you like a more detailed explanation or have any questions about this solution? Here are some related questions to explore:

1. How do you derive the formula for the area of a circle?
2. What is the significance of the quadratic formula in solving equations?
3. Can you explain why we only consider the positive root for $x$?
4. How do you convert a fraction of a circle's area to a whole circle?
5. What other methods can be used to solve quadratic equations?
6. How would the solution change if we had different fractions of the circle?
7. What are some practical applications of solving quadratic equations in geometry?
8. Can the quadratic formula be used for equations with coefficients that are not real numbers?

Tip: Always check the discriminant ($b^2 - 4ac$) before solving a quadratic equation to ensure it has real solutions.