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. ^@
    A B C D E F O
    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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