I think the answer is 119.29cm^2?
I was thinking of how to do this for ages and eventually had to look up apothems and the formula for the area of any regular polygon to work it out…hopefully I’ve learnt something here?! Thanks for the puzzles and the site-will be great for my high attaining students (and for me!!)

Your answer is correct ðŸ™‚ I think you made it too complicated though. You can solve it with simultaneous equations. Start by drawing a square around the shape.

Great website, I solved it using simultaneous equations.
Let the length of each side of the octagon be called a and put a square around it and let the remaining lengths be called b then the length of the square is a+2b=12 and the hypotenuse of each triangle using pythag is sqrt(2)*b which equals a.
Then solve them simultaneously to get b=12/(2+swrt(2)). Then finally subtract the area of all 4 triangles from the area of the square to get 119.29…. cm^2.

I did it first using trapezia etc. but simpler to just take the four corner triangles from the square as above. Really nothing simultaneous needed, if you want you can keep an unknown such as the side of the polygon until evaluation stage. Exact answer much more elegant.

I think the answer is 119.29cm^2?

I was thinking of how to do this for ages and eventually had to look up apothems and the formula for the area of any regular polygon to work it out…hopefully I’ve learnt something here?! Thanks for the puzzles and the site-will be great for my high attaining students (and for me!!)

I think the answer is 119.29cm^2?! Fingers crossed…fantastic site by the way-very challenging!

Your answer is correct ðŸ™‚ I think you made it too complicated though. You can solve it with simultaneous equations. Start by drawing a square around the shape.

Great website, I solved it using simultaneous equations.

Let the length of each side of the octagon be called a and put a square around it and let the remaining lengths be called b then the length of the square is a+2b=12 and the hypotenuse of each triangle using pythag is sqrt(2)*b which equals a.

Then solve them simultaneously to get b=12/(2+swrt(2)). Then finally subtract the area of all 4 triangles from the area of the square to get 119.29…. cm^2.

That’s the method is would use too. Correct answer!

I did it first using trapezia etc. but simpler to just take the four corner triangles from the square as above. Really nothing simultaneous needed, if you want you can keep an unknown such as the side of the polygon until evaluation stage. Exact answer much more elegant.