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Congratulations to Nisha Doshi, Year 9, The Mount School, York for this beautifully explained solution.

If these triangles are made into a tessellation, they will form regular hexagons with circles overlaid. Each triangle consists of 3 x 1/6 = 1/2 of a circle, plus the shaded area, so to find the area of the shaded section, you can do : (area of triangle - area of 1/2 of a circle)

Using Pythagoras' Theorem the height of the triangle is $\sqrt((2r)^2 - r^2 ) = r\sqrt3$. So the area of the triangle is $r^2\sqrt3$. The area of ${1\over 2}$ a circle is ${1\over 2}\pi r^2$. So the area shaded is $r^2\sqrt3 - {1\over 2}\pi r^2$ and the proportion of the tessellation that is shaded is $$ \frac{r^2\sqrt3 - {1\over 2}\pi r^2}{r^2\sqrt3} = 1 - \frac{\pi}{2\sqrt3} $$

Well done Arwa Jamil, Year 8, the International School Brunei who calculated correctly, to 3 significant figures, that 9.31% of the floor is shaded.


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You are only given the three midpoints of the sides of a triangle. How can you construct the original triangle?

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The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice.

NRICH is part of the family of activities in the Millennium Mathematics Project.

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