Question 928071
The sum of the measures of the interior angles of a convex polygon with {{{n}}} sides is
{{{(n-2)*180^o}}}
In this case,
{{{(n-2)*180^o=540^o}}}-->{{{n-2=540^o/180^o}}}-->{{{n-2=3}}}-->{{{n=3+2}}}-->{{{highlight(n=5)}}} .
So this regular polygon has {{{highlight(5)}}} sides. It is a pentagon.
Five sides means {{{5}}} vertices.
Form each vertex you can draw diagonals to other vertices.
There are {{{6-1=4}}} other vertices, but the {{{2}}} lines connecting to the next vertices to either side are not diagonals, but sides of the polygon.
So, from each vertex, you can draw {{{5-1-2=2}}} diagonals connecting to other vertices.
That would add up to {{{5*2=10}}} diagonals, but we would be counting each diagonal twice, for example once as a line from vertex A to vertex C, and once more as a line from vertex C to vertex A.
So there are just {{{highlight(5)}}} possible diagonals .
The {{{2}}} diagonals from each vertex split the pentagon into {{{3}} isosceles triangles. one of them has one side of the pentagon as its base and two diagonals as its legs. The other two isosceles triangles have one diagonal of the pentagon as its base and two sides of the pentagon as its legs.
If we need to count all possible triangles whose sides are entire sides and entire diagonals of the pentagon, there are {{{5}}} isosceles triangles with a side of the pentagon as a base, and {{{5}}} with a diagonal of the pentagon as a base, for a total of {{{5+5=10}}} large triangles.
We can also figure out that there are {{{10}}} different large triangles, because there are {{{highlight(10)}}} different sets of 3 vertices that can be made from the 6 vertices of the pentagon.
The diagonal also form {{{10}}} more, smaller triangles with one or two vertices at the intersections of diagonals inside the pentagon.
{{{drawing(350,350,-1.75,1.75,-.2,3.3,
line(-1,0,1,0),line(1,0,1.618,1.902),line(-1,0,-1.618,1.902),
line(0,3.078,1.618,1.902),line(0,3.078,-1.618,1.902),
green(line(1,0,0,3.078)),green(line(1,0,-1.618,1.902)),
green(line(-1,0,0,3.078)),green(line(-1,0,1.618,1.902)),
green(line(1.618,1.902,-1.618,1.902))
)}}}
Connecting the vertices of the regular pentagon to its center we form {{{5}}} congruent triangles with {{{5}}} angles at the center of the pentagon called "central angles".
Each central angle measures {{{360^o/5=highlight(72^o)}}}
The measure of each of the {{{5}}} interior angles of the pentagon is
{{{540^o/5=highlight(108^o)}}} .
The perimeter of the pentagon is {{{5*(5cm)=highlight(25cm)}}} .
The area of the regular pentagon can be calculated as the sum of the areas of the {{{5}}} isosceles triangles formed by the lines connecting the vertices to the center.
For each of those isosceles triangles, we can take the side of the pentagon as the base, and the line connecting the midpoint of that base to the center of the pentagon as the altitude. The measure of that altitude is the height of the triangle, and the apothem of the regular pentagon.
In general, the area of the regular polygon with {{{n}}} sides is {{{n*(side*apothem/2)=(n*side)*apohem/2=perimeter*apothem/2}}}
{{{drawing(350,350,-1.75,1.75,-.2,3.3,
green(triangle(0,0,1,0,0,1.376)),line(-1,0,1,0),
line(1,0,1.618,1.902),line(-1,0,-1.618,1.902),
line(0,3.078,1.618,1.902),line(0,3.078,-1.618,1.902),
green(line(1,0,0,1.376)),green(line(0,1.376,-1.618,1.902)),
green(line(0,1.376,0,3.078)),green(line(-1,0,0,1.376)),
green(line(1.618,1.902,0,1.376)),red(arc(-1,0,1,1,-108,0)),
green(rectangle(0,0,0.1,0.1)),red(arc(0,1.376,1,1,126,198)),
red(arc(0,1.376,1,1,54,90)),locate(-.1,0,5cm),locate(0.2,.15,2.5cm),
arrow(-0.2,-0.1,-1,-0.1),arrow(0.2,-0.1,1,-0.1),arrow(0.4,0.4,0.01,0.4),
locate(0.4,0.5,apothem),locate(-1.1,0.5,red(108^o)),
locate(-0.45,1.37,red(72^o)),locate(0.1,0.95,red(36^o))
)}}} {{{apothem=2.5cm/tan(36^o)=highlight(3.44cm)}}} (rounded), and {{{area=(25cm)(3.44cm)/2=highlight(34)}}}{{{cm^2}}} .