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The figure $ABCDEF$ is a regular hexagon. Evaluate the quotient $\dfrac { \text{Area of hexagon } ABCDEF } { \text{Area of } \triangle BCE }$.

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Answer:

$3$

Step by Step Explanation:

1. Let $O$ be the center of the regular hexagon $ABCDEF.$
   
   Therefore, the area of hexagon $ABCDEF = 6 \times \text{ the area of } \triangle EOC$
2. In $\triangle BCE,$
   $O$ is the midpoint of $BE.$ Therefore $OC$ is a median of the triangle $\triangle BCE$.
   We know that the median of a triangle divides the triangle into two triangles with equal areas.
   Therefore, the area of $\triangle EOC = \text{ area of } \triangle BOC$
   or the area of $\triangle BCE = 2 \times$ the area of $\triangle EOC$
3. $\dfrac { \text{Area of hexagon } ABCDEF } { \text{Area of } \triangle BCE } = \dfrac { 6 \times \text{Area of } \triangle EOC } { 2 \times \text{Area of } \triangle EOC } = 3$

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