Oxford PAT 2011, Question 3

2011_page2

9 thoughts on “Oxford PAT 2011, Question 3

    1. 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

    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. 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).

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