# Polygon area method

I saw this problem in a puzzle book. Just wondering if anyone can explain the principle behind this method.

A rectilinear figure of any number of sides can be reduced to a triangle of equal area, and as $$\angle AGF$$ happens to be a right-angle the thing is quite easy in this way:

• Continue the line $$GA$$.

• Now lay a parallel ruler from $$A$$ to $$C$$, run it up to $$B$$ and mark the point $$1$$.

• Then lay the ruler from $$1$$ to $$D$$ and run it down to $$C$$, marking point $$2$$.

• Then lay it from $$2$$ to $$E$$, run it up to $$D$$ and mark point $$3$$.

• Then lay it from $$3$$ to $$F$$, run it up to $$E$$ and mark point $$4$$.

If you now draw the line $$4$$ to $$F$$ then $$\triangle G4F$$ is equal in area to the irregular field.

As our scale map shows $$GF$$ to be $$7$$ inches (rods), and we find the length $$G4$$ in this case to be exactly $$6$$ inches (rods), we know that the area of the field is $$\frac 12 (7\times 6)$$ or $$21$$ square rods.

The simple and valuable rule I have shown should be known by everybody-but is not