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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)

 Jun 13, 2021
 #1
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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

 

\({area(ABCDEF)\over area(△PQR)} = {3\over 2} = 3:2\)

 

 

smiley

 Jun 14, 2021

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