Question 1207466
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The regular hexagon ABCDEF has sides of length 2. The point P is the midpoint of AB. 
Q is the midpoint of BC and so on. Find the area of the hexagon PQRSTU
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<pre>
The area of the regular hexagon ABCDEF is 6 times the area of an equilateral triangle 
with the side length 2.


So, the area of ABCDEF is  {{{6*2^2*(sqrt(3)/4)}}} = {{{6*sqrt(3)}}} square units.


To get the area of the regular hexagon PQRSTU, we should subtract from the area ABCDEF 
the area of triangle PBQ 6 times.


PBQ is an isosceles triangle with the lateral side length of 1 and the concluded angle of 120 degrees.
Therefore, its area is

    {{{(1/2)*1*1*sin(120^o)}}} = {{{sqrt(3)/4}}}.


Thus the area of PQRSTU is  

    {{{6*sqrt(3)}}} - {{{6*(sqrt(3)/4)}}} = {{{(6-6/4)*sqrt(3)}}} = {{{4.5*sqrt(3)}}} = 7.794228634 square units.    <U>ANSWER</U>
</pre>

Solved.