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# polygons question

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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.

Nov 21, 2018

#1
+702
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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

Nov 21, 2018
#2
+701
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Oh yea...the percent in LaTeX does not work.  $$72$$%. Must write with the regular "%".

.
Nov 22, 2018
edited by PartialMathematician  Nov 22, 2018
edited by PartialMathematician  Nov 22, 2018
#3
+22358
+10

(the LaTeX percent doesn't work) = ?

Write \% in Latex to get % : $$\text{Percent in Latex: }~ \%$$

Nov 22, 2018