Find the ratio of the area of a regular hexagon with sides of 1 unit to the area of an equilateral triangle with sides of 2 units. (Nots: "regular" means that all of the sides and angles are equal)
+526 
ABCDEF is a regular hexagon and PQR is an equilateral triangle.
In ABCDEF,
area(△AOB) = area(△BOC) = area(△COD) = area(△DOE) = area(△EOF) = area(△FOA)
⇒area(ABCDEF) = area(△AOB + △BOC + △COD + △DOE + △EOF + △FOA)
\(= {\sqrt3 \over 4}+ {\sqrt3 \over 4}+ {\sqrt3 \over 4}+ {\sqrt3 \over 4}+ {\sqrt3 \over 4}+ {\sqrt3 \over 4}\)
\(=6× {\sqrt3 \over 4}\)
\(={3\sqrt3 \over 2}\) sq. units
⇒area(△PQR) \(={\sqrt3 \over 4}×4\)
\(=\sqrt3\) sq. units
\({area(ABCDEF)\over area(△PQR)} = {3\over 2} = 3:2\)
