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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.

Guest Nov 21, 2018
 #1
avatar+422 
+2

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

CoolStuffYT  Nov 21, 2018
 #2
avatar+656 
+2

Oh yea...the percent in LaTeX does not work.  \(72\)%. Must write with the regular "%".

PartialMathematician  Nov 22, 2018
edited by PartialMathematician  Nov 22, 2018
edited by PartialMathematician  Nov 22, 2018
 #3
avatar+20680 
+8

(the LaTeX percent doesn't work) = ?

 

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

 

laugh

heureka  Nov 22, 2018

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