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A box is being designed to contain 4 stacks of chocolate wafers. Each chocolate wafer is in the shape of a regular hexagon of side length 3 cm. What is the area of the rectangular base of the box? Give your answer as k√3 where k is irrational

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Oct 20, 2017

$k = \frac{315}{4}$

Explanation:

The measure of each interior angle $x$ of a regular polygon is given by : $x = \frac{\left(n - 2\right) \cdot 180}{n}$, where n is the number of sides.
So in a regular hexagon,
interior angle $x = \frac{\left(6 - 2\right) \cdot 180}{6} = {120}^{\circ}$
Fig 1 shows a regular hexagon with side $s$,
$B M = M C = s \cdot \sin 60 = \frac{\sqrt{3} s}{2}$
$A M = s \cdot \cos 60 = \frac{s}{2}$
As shown in Fig 2, $E F G H$ is the rectangular base of the box,
given that side length of the regular hexagon $s = 3$ cm,
$\implies E F = 2 s + 3 \cdot s \cdot \cos 60 = 2 \times 3 + 3 \times 3 \times \frac{1}{2} = \frac{21}{2}$ cm
$\implies E H = 5 \cdot s \cdot \sin 60 = 5 \times 3 \times \frac{\sqrt{3}}{2} = \frac{15 \sqrt{3}}{2}$ cm
Given area of the rectangular base $= k \sqrt{3}$,
$\implies k \sqrt{3} = \frac{21}{2} \cdot \frac{15 \sqrt{3}}{2} = \frac{315 \sqrt{3}}{4} {\text{ cm}}^{2}$
$\implies k = \frac{315}{4}$,
As $k$ can be expressed as $\frac{p}{q}$, where $p \mathmr{and} q$ are integers and $q \ne 0$, $k$ is rational.

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