# GCSE Edexcel Maths Paper 1 – The Final Question

The Question

The diagram shows three circles, each of radius 4cm.

The centres of the circles are A, B and C such that ABC is a straight line and AB = BC = 4cm.

Work out the total area of the two shaded regions.

Like the question about Hannah and her sweets from 2015, this year, this question has had everyone scratching their heads!

Geometry questions like this can be particularly difficult because, on first assessing them, there are often lots of different ways to break up the regions. Students (and teachers!) can feel overwhelmed by choice. They’re often the kind of questions that are easy when you’re shown how, though.

Critical thinking was billed to be the big thing about the new maths GCSE introduced for examination in 2017. Each paper has a number of ‘you could never have predicted that’ questions, alongside the routine, easy-to-prepare-for questions. The advantages the former bring are deeper understanding, practice of higher-level reasoning and the challenge and satisfaction of solving the problem. Here’s a great set of grade 9 problem-solving questions, to ensure that your students are familiar with that kind of thinking.

For students who’ve been inspired by this and other more complex questions, we also have a full set of preparation for A Level resources.

For this question, the only value given was the radius of each circle, which was 4cm. The task was to find the total area of the shaded region in terms of π. There’s more than one way to approach it, but here’s one: \begin{aligned} \text{Height of triangle } ABX &= \sqrt{4^2 - 2^2} \\ &= \sqrt{12} \\ &= 2\sqrt{3} \end{aligned} \begin{aligned} \text{Area of triangle } ABX &= \frac{1}{2} \times 2\sqrt{3} \times 4 \\ &= 4\sqrt{3} \end{aligned} \begin{aligned} \text{Area of the circle} &= \pi \times 4^2 \\ &= 16\pi \end{aligned} \begin{aligned} \text{Area of the sector} &= \frac{1}{6} \times 16\pi \\ &=\frac{8}{3} \pi \end{aligned} \begin{aligned} \text{Area of the segment} &= \frac{8}{3} \pi - 4\sqrt{3} \end{aligned}

At this point, there are several different ways of finding the area of the shaded region. You could subtract the area of 4 ABX triangles and 8 segments from the area of the circle or you could add the area of 2 triangle ABXs and subtract the area of 2 segments. Whichever method you chose would give the same answer: \begin{aligned} \text{Area of the shaded area} &= 16\sqrt{3} - \frac{16}{3} \pi \end{aligned}