Question 924425
{{{drawing(300,350,-3.3,3.3,-2.2,5.5,
triangle(-3,0,3,0,0,5.2),
triangle(-3,3.464,3,3.464,0,-1.733)
)}}} The superimposed part is a hexagon,
and it is surrounded by six small equilateral triangles that form the points of the star.
Because it is all so symmetrical, the sides of all the little equilateral triangles,
and the sides of the hexagon
are all congruent.
Since three such sides add up to a side of the large triangles,
the sides of the small triangles measure {{{3cm/3=1cm}}} .
The area of the star is the area of one of the large triangles plus the area of 3 small triangles/star points.
The area of a triangle is calculated as
{{{(1/2)side[1]*side[2]*sin(angle_in_between)}}}
Since all these triangles are equilateral,
{{{side[1]=side[2]}}} , and
{{{sin(angle_in_between)=sin(60^o)=sqrt(3)/2}}} .
So the are of each triangle is
{{{(1/2)side^2*(sqrt(3)/2)}}} .
For the large triangle, the area is
{{{(1/2)*3^2*(sqrt(3)/2)=(1/2)*9*(sqrt(3)/2)=9sqrt(3)/4}}} .
For each small triangle, the area is
{{{(1/2)*1^2*(sqrt(3)/2)=(1/2)(sqrt(3)/2)=sqrt(3)/4}}}
So the total area of the star is
{{{9sqrt(3)/4+3*(sqrt(3)/4)}}}={{{9sqrt(3)/4+3sqrt(3)/4}}}={{{12sqrt(3)/4}}}={{{highlight(3sqrt(3))}}}