# Oxford PAT 2011, Question 3

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## 9 thoughts on “Oxford PAT 2011, Question 3”

1. Edward2016 says:

Could you explain the second part please

1. Hi Edward — see below, where I’ve just right now answered that exact same question.

2. Jaime says:

Hi, could you explain your answer to the second part of the question? Thanks!

1. Jaime says:

And also, are these the minimum areas that you found? I.e if x was at a/2 as opposed to where it is on the diagram then that would be the max possible area for the 0≤x≤a/2 ?
Thanks

1. Not sure I understand what you’re saying here? Can you explain?

2. Sure.

First, we are expressing y in terms of x. When x goes between a/2 and a, we can see that y goes from (a√3)/2 to zero. So we get the equation y = (a-x)√3.

Second, we are saying that the area covered can be calculated by taking the area of the whole equilateral triangle (which is (a^2√3)/4 and subtracting the area of the triangle that we haven’t yet reached (which is (a-x)y/2).

We can use the first equation to substitute for y in the second equation, and then we just need to simplify the expression for A.

1. Edward2016 says:

Thank you! This is really useful!

3. Frances says:

Hi, it states at the top that y= x * sqrt(3), I don’t really understand where this comes from, can you explain?

1. Hi Frances, thanks for your question. The triangle is equilateral, so when x <= a/2 the line is sloping up at π/3 radians (60 degrees). So the gradient of the line is tan π/3, which is equal to sqrt(3), so when x <= a/2, y = x * sqrt(3).