Edexcel Pure Paper 1 2024 Question 11

Start with what you know.

The semicircle has diameter 10, so radius is 5.

Also OB is from a circle centred at A, radius 5.

So you’ve got:

OA = OB = AB = 5

That’s the moment things simplify.

If all three sides are 5, then triangle AOB isn’t just any triangle — it’s equilateral.

So the angle is:

\frac{\pi}{3}

A lot of people didn’t spot that, and that’s where things got messy.

Now, the shaded region isn’t one shape.

It’s easier to think:

“what big piece do I have, and what do I need to take away?”

First part

Take the sector centred at O.

The angle there is:

\frac{2\pi}{3}

Area:

\frac{1}{2} \cdot 25 \cdot \frac{2\pi}{3}

= \frac{25\pi}{3}

That’s most of the region already.

Now the bit to remove

This is where people sometimes rush.

It’s not just a triangle. It’s a segment.

So you need:

sector − triangle

Sector first (angle \frac{\pi}{3}):

\frac{1}{2} \cdot 25 \cdot \frac{\pi}{3} = \frac{25\pi}{6}

Triangle:

\frac{1}{2} \cdot 5 \cdot 5 \cdot \sin\left(\frac{\pi}{3}\right)

= \frac{25}{2} \cdot \frac{\sqrt{3}}{2}

= \frac{25\sqrt{3}}{4}

So that curved section is:

\frac{25\pi}{6} – \frac{25\sqrt{3}}{4}

Put it together

Now subtract it:

\frac{25\pi}{3} – \left(\frac{25\pi}{6} – \frac{25\sqrt{3}}{4}\right)

Be careful here — easy place to drop a sign.

= \frac{25\pi}{3} – \frac{25\pi}{6} + \frac{25\sqrt{3}}{4}

Convert:

\frac{25\pi}{3} = \frac{50\pi}{6}

So:

= \frac{25\pi}{6} + \frac{25\sqrt{3}}{4}

That’s it.

✅ Final Answer

\frac{25\sqrt{3}}{4} + \frac{25\pi}{6}