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'Unusual Polygon' printed from https://nrich.maths.org/
The area of square
BDFG is $6\times 6 = 36$ square
units.
So the total area of the three triangles
ABG,
BCD and
DEF is also $36$ square
units.
These three triangles are congruent and so each has an area of $12$
square units.
The area of each triangle is $\frac{1}{2}\times base \times \
height$ and the base is $6$ units and hence we
have $\frac{1}{2}\times 6 \times \ height = 12$,
so the height is $4$ units.
Let $X$ be the midpoint of
BD. Then
CX is perpendicular to the base
BD (since
BCD is an isosceles
triangle).
By Pythagoras' Theorem, $BC = \sqrt{3^2+4^2} = 5$ units.
Therefore the perimeter of
ABCDEFG is $6\times 5+ 6 = 36$
units.