Lesson A problem on equiangular but not equilateral octagon

A problem on equiangular but not equilateral octagon

Problem 1

Let's write the sequence of side lengths in the row

      EF     FG     GH     HI     IJ    JK     KL    LE

       1   sqrt(2)   1   sqrt(2)   1   sqrt(2)  1   sqrt(2)   <<<---===  (1)


You see the repeating pattern as a cycle.


All interior angles are   =  =  =  = 45*3 = 135 degrees.


You can calculate the length of the diagonal EG using the cosine law formula

     =  =  =  = 5.

so  EG = .

Obviously, all such sides EG, GI, IK and KE have the same length    due to the same reason (the same logic).



Next, this octagon has a remarkable symmetry: if you rotate it in a way that vertex E goes to vertex G

    E ---> G,

then the new octagon (the image under this rotation) will coincide with the original octagon.
(simply because the sequence of side lengths (1) will be the same and all interior angles are congruent).


It means that the quadrilateral FHJK will map into and onto itself.
It means that this quadrilateral is a square.


   If such reasoning confuses you, you can  notice that sides FH and HJ of the quadrilateral FHJK are orthogonal
   since two times angle 135 degs is 270 degs.  And, similarly, quadrilateral FHJK has all his consecutive sides 
   orthogonal with equal lengths, so this quadrilateral is a square.


Now, the area of the square FHJK is the square of its side, i.e.   = 5.


Now you can calculate the area of triangle EFG

     =  =  =  = .


We have 4 such triangles as EFG, so their total area is   = 2.


Now the total area of the octagon EFGHIJKL is the sum of the area of square FHGK PLUS four triangles

    5 + 2 = 7.


At this point, the solution is complete.

The area of the octagon EFGHIJKL is 7 square units.    ANSWER