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Corners are cut from an equilateral triangle to produce a regular convex hexagon as shown in the figure above.

The ratio of the area of the regular convex hexagon to the area of the original equilateral triangle is

  1. $2:3$
  2. $3:4$
  3. $4:5$
  4. $5:6$
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Given that, corners are cut from an equilateral triangle to produce a regular convex hexagon. The smaller triangles cut are equilateral triangles (If we cut off corners to create a regular hexagon, then each angle of the hexagon is $120^{\circ}$, and so each angle of every removed triangle is $60^{\circ}$ making these triangles equilateral)

Let the original triangle side be $3a.$

The area of regular convex hexagon $H_{1} = \frac{3\sqrt{3}}{2}s^{2},$ the area of equilateral triangle $ T_{1}= \frac{\sqrt{3}}{4}s^{2},$ where $s$ is the side length. 

Now, $H_{1} = \frac{3\sqrt{3}}{2}a^{2},A_{1} = \frac{\sqrt{3}}{4}(3a)^{2} = \frac{9\sqrt{3}}{4}a^{2}$

The required ratio $ = \dfrac{H_{1}}{A_{1}} = \dfrac{\frac{3\sqrt{3}}{2}a^{2}}{\frac{9\sqrt{3}}{4}a^{2}} = \frac{2}{3}.$

$\therefore$ The ratio of the area of the regular convex hexagon to the area of the original equilateral triangle is $2:3.$

So, the correct answer is $(A).$

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How's it proved that the smaller triangles are equilateral?
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If we cut off corners to create a regular hexagon, then each angle of the hexagon is $120^{\circ}$, meaning that each angle of each removed triangle is $60^{\circ}$, so these triangles are equilateral.

From here:https://math.stackexchange.com/questions/3726183/corners-are-cut-off-from-an-equilateral-triangle-to-produce-a-regular-hexagon-a

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