The diagram shows two shaded shapes, A and B - Edexcel - GCSE Maths - Question 23 - 2020 - Paper 1

Shape B is a circle, hence its area is calculated using:

$\text{Area of B} = \pi r^2$

Since the diameter is $(2 - 2x)$ cm, the radius is:

$r = \frac{(2 - 2x)}{2} = 1 - x$

So, the area of shape B becomes:

$\text{Area of B} = \pi (1 - x)^2$

Equating the areas of A and B:

$\frac{\theta \pi}{360}(25 - (3x - 1)^2) = \pi (1 - x)^2$

Dividing by $\pi$ and simplifying:

$\frac{\theta}{360}(25 - (3x - 1)^2) = (1 - x)^2$

Expanding $(3x - 1)^2$:

$(3x - 1)^2 = 9x^2 - 6x + 1$

Plugging it back gives:

$\frac{\theta}{360}(25 - (9x^2 - 6x + 1)) = (1 - x)^2$

Now expanding $(1 - x)^2 = 1 - 2x + x^2$, we have:

$\frac{\theta}{360}(24 - 9x^2 + 6x) = 1 - 2x + x^2$

This can be rearranged into a standard quadratic form $ax^2 + bx + c = 0$.

At this point, we can either factor the quadratic or apply the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting the coefficients obtained from the quadratic equation will give us the values for $x$. Upon solving, we can deduce the value of $x$ that satisfies the area equality.