A regular hexagon is truncated to form a regular dodecagon (12-gon) by removing identical isosceles triangles from its six corners. What percent of the area of the original hexagon was removed? Express your answer to the nearest tenth.
+1252 Let the side of the hexagon be 1. Let \(s\) be the length of each of the equal sides in the removed isosceles triangles. Define points A, B, C, D, E, and F. \(\Delta CDB\) is a 30-60-90, so \(CD=s/2\) and \(DB=s\sqrt3/2 \) . \(AB=1-2s\) because \(CF=1\) . For the dodecagon, we must solve \( AB=2\times BD\)
\(1-2s=s\sqrt3, 1=2s+s\sqrt3, 1=s(2+\sqrt3), s=\frac{1}{2+\sqrt3}\)
Rationalize the denominator to get \(s=2-\sqrt3\)
The area of the hexagon is \(\frac{3\sqrt3}{2}\) and dodecagon is \(\frac{3s^2\sqrt3}{2}\).
The deleted area is \(s^2=0.072=7.2%\) percent
7.2% (the LaTeX percent doesn't work)
You are very welcome!
:P
+775 Oh yea...the percent in LaTeX does not work. \(72\)%. Must write with the regular "%".
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