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# Help with Geometry

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Thanks!

Mar 30, 2019

#1
+6043
+2

$$\text{Just eyeballing it you've got a regular hexagon with side length equal to}\\ \text{one third the side of the triangle. Let that triangle side length be }s\\ A_t = \dfrac{\sqrt{3}}{4}s^2\\ A_h = \dfrac{3\sqrt{3}}{2}\left(\dfrac{s}{3}\right)^2 = \dfrac{\sqrt{3}}{6}s^2\\ \dfrac{A_h}{A_t} = \dfrac{\frac{\sqrt{3}}{6}}{\frac{\sqrt{3}}{4}}=\dfrac 2 3$$

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Mar 30, 2019
#2
+4323
+2

An easier way is to split the regular hexagon into six equilateral triangles. There are three more in the triangle $$ABC$$, so $$\frac{6}{9}=\boxed{\frac{2}{3}}.$$

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Mar 30, 2019
#3
+10541
+1

Here's the picture.

Mar 31, 2019