I am currently taking the Scottish Mathematical Council's Mathematical Challenge 2016/17. I am unable to solve problem S4.

For this problem, I need to find the area of this irregular polygon.

SMC Mathematical Challenge 2016/17 S4

I have attempted to split it up into smaller shapes (triangles and rectangles) but am still unable to solve it.


closed as off-topic by Silvia Ghinassi, iadvd, Ethan Bolker, Shailesh, Leucippus Sep 28 '16 at 1:13

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Show that the right vertex and the left vertex are at the same height, and that the top vertex is $\frac12$ above them. Then we can split the shape into

  • a $1\times 1$ square
  • a right triangle with legs $1$ and $\sqrt 3$
  • a triangle with base $1+\sqrt 3$ and height $\frac12$

Using the law of sine and the law of cosine, you should be able to solve for some angles and sides.

$$S = \frac{1}{2}(1)(1) + \frac{1}{2}\sqrt{2}\sqrt{3}\sin(75) + \frac{1}{2}(1)(2)\sin(60)$$


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