Home > #MTBoS, A Level, GCSE, Maths, Starters, Teaching > UKMT Maths Challenge Q16

UKMT Maths Challenge Q16

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Above is a photo of question 16 from yesterday’s UKMT Senior Maths Challenge. And what I fantastic question it is!

It shows (incase you can’t make it out on the picture) a rectangle and a circle which have the same centre. The two shorter sides of the rectangle are tangents to the circle, and the rectangle is 6×12. The task is to find the area that is inside both.

After the challenge had been completed one if my Y13s came to ask about it, and it took me a little while to work it out. I played around with some similar triangles, thinking that I would need to calculate the bit of the rectangle outside the circle and subtract them from 72. Then I had the realisation that I could easily split it up into triangles and sectors that I could calculate the area of with great ease.

First I sketched:

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Showing the rectangle split into areas and sectors. I then considered the triangles, if I split then into right angled triangles I had a hypotenuse of 6 and a height of 3, so the cosine of the top angle must be 0.5, which means the angle must be pi/3.

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I then used the sine of this angle ((rt3)/2) to calculate the opposite side as 3rt3. Then thought “why didn’t I use Pythagoras’s Theorem?!” A quick check using Pythagoras’s Theorem shows us the side is rt27, which simplifies to 3rt3. The area of the triangle then must be 9rt3.

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Knowing that the angles at the centre in the two triangles are 2pi/3, that angles round a point add up to 2pi and that the sectors are congruent means we can deduce the angle if each sector is pi/3. We also know the area of a sector is (r^2(theta))/2, which gives the area of each sector as 6pi.

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Finally we sum the areas and get 18rt3 + 12pi.

The student had looked at the answers and estimated them using the knowledge that rt3 is around 1.7 and pick is a but more than three. This meant he could eliminate c,d and e, but he didn’t get to the final answer.

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  1. November 13, 2014 at 10:52 pm

    Reblogged this on ryanjhay79.

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