Horrible Hexagons 1

Geometry Level 3

The diagram above shows a regular hexagon ${ H }_{ 1 }$ with two equilateral triangles inscribed in it.

If the shaded region has area $x$ $\text{ unit}^{ 2 }$ , what is the area of ${ H }_{ 1 }$ in terms of $x$ ?

2x\ \text{unit}^2

\frac { 6 }{ \sqrt { 3 } } x \ \text{unit}^2

\frac { 5 }{ 2 } x\ \text{unit}^2

2\sqrt { 3 } x\ \text{unit}^2

3x \ \text{unit}^2

\frac { 3\sqrt { 3 } }{ 2 } x\ \text{unit}^2

1 solution

Brian Charlesworth
Apr 4, 2015

Method 1:

Each of the $12$ "distinct" triangles encircling the shaded region have the same area, and the shaded region can be formed by "folding" the $6$ of these triangles that are equilateral about the sides they share with the shaded region. In this way we see that $H_{1}$ is composed of $18$ non-overlapping triangles of the same area, and thus $H_{1} = \frac{18}{6}x = \boxed{3x}$ units $^{2}.$

Method 2:

By symmetry we see that the shaded area is a regular hexagon with side lengths that are $\frac{1}{2}\sec(30^{\circ}) = \frac{1}{\sqrt{3}}$ that of side lengths of $H_{1}.$ Since the area of a regular hexagon is proportional to the square of the length of its sides, we see that the area of the shaded region is $(\frac{1}{\sqrt{3}})^{2} = \frac{1}{3}$ that of $H_{1},$ and hence the area of $H_{1}$ is $\boxed{3x}$ units $^{2}.$

Horrible Hexagons 1

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