Oxford PAT 2014, Question 5



5 thoughts on “Oxford PAT 2014, Question 5

    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.

    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.

Add your comments

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s