A right-angled triangle is formed by the diameters of three semicircular regions, A, B and C as shown in the diagram - Edexcel - GCSE Maths - Question 14 - 2022 - Paper 1

To solve this problem, we start by using the Pythagorean Theorem, which states that for a right-angled triangle with sides of lengths corresponding to the diameters of the semicircles, the relationship can be expressed as:

$d_A^2 = d_B^2 + d_C^2$

Where:

• $d_A$ is the diameter of region A
• $d_B$ is the diameter of region B
• $d_C$ is the diameter of region C

Next, we can express the area of each semicircular region in terms of their diameters:

• Area of region A: A_A = rac{1}{2} \pi \left(\frac{d_A}{2}\right)^2 = \frac{\pi d_A^2}{8}

• Area of region B: $A_B = \frac{1}{2} \pi \left(\frac{d_B}{2}\right)^2 = \frac{\pi d_B^2}{8}$

• Area of region C: $A_C = \frac{1}{2} \pi \left(\frac{d_C}{2}\right)^2 = \frac{\pi d_C^2}{8}$

Now, substituting the expressions for the areas into the equation from the Pythagorean theorem:

• We need to show that: $A_A = A_B + A_C$

Substituting the areas:

$\frac{\pi d_A^2}{8} = \frac{\pi d_B^2}{8} + \frac{\pi d_C^2}{8}$

This simplifies to:

$d_A^2 = d_B^2 + d_C^2$

This confirms the area relationship by verifying that the area of region A is indeed equal to the sum of the areas of regions B and C, as required.