# Oxford PAT 2014, Question 5

## 5 thoughts on “Oxford PAT 2014, Question 5”

1. John says:

Hi,

How do you get the internal angle of pi/3?

Thanks

John

1. Hi John,

The question itself says that the triangle is an equilateral triangle (and so has internal angles of pi/3). If you need to prove that the triangle is equilateral, then consider what happens if you rotate the whole diagram repeatedly by 2pi/3: the resulting diagrams are identical to the original, and therefore the sides of the triangle must all be the same.

2. George says:

Why is the area of three segments not 2piR^2?

1. Hi George, thanks for your question.

First, some terminology — I’ve mistakenly used the word ‘segment’ for the area of the circle swept out between two radii — the correct term is ‘sector’ I think. But I think it’s obvious what we are talking about — each sector (segment as was) is the bit of area that is shared by the triangle and one of the circles.

The area of each of the circles is πr^2. Each sector has an internal angle of π/3, so the area of each sector is ((π/3)/2 π)πr^2 = (πr^2)/6. Therefore the total area of three of the sectors is 3(πr^2)/6, which equals (πr^2)/2.

1. George says:

Hi,

Thank you for that! I had a bit of a Maths block.