throwback maths 8

throwback 8
Solution

At each vertex, the Internal angles of the polygons must add to $360^\circ$. The internal angle of the square is $90^\circ$, the hexagon has an internal angle of $120^\circ$ and so the third shape must have an internal angle of $360^\circ - (90^\circ+120^\circ) = 150^\circ$.

The external angle at each corner of the unknown polygon is $180^\circ-150^\circ = 30^\circ$. It follows that the polygon must have $$\frac{360}{30} = 12\;\mbox{sides}$$