Foo - Geometry Question

Ok it's been 30+ years since I took geometry and I was just asked to figure out something so I thought I would turn to all you math people.

My buddy has a pentagon shaped canopy. The sides are equal length and the only measurement he knows is the it is 15' from pole to pole at the widest point. how long are the sides?

Why not just measure a side? ;)

Anyway, here we go... for the sake of being able to follow, let's orient the pentagon so that one edge is parallel to the horizontal (on a piece of paper), and the vertex opposite that edge is pointing up. We'll number the corners starting from the vertex at the top, which will be corner number 1.

Draw a line segment between corners 2 and 5. This should be parallel to the horizontal. We also know that this distance is 15'. Draw a second line segment from corner 1 straight down to intersect the newly drawn line segment at a right angle. We now have two right triangles.

Since the interior angle of a regular pentagon is 108 degrees (540 degrees total divided by 5 corners), the corner of each right triangle at corner 1 is 54 degrees. The length of the leg of the triangle opposite this corner is 7.5'. Using simple trig, sin (54 deg) = 7.5' / x where x is the hypotenuse of the right triangle, which happens to also be the edge of the pentagon.

Solving for x, x = 7.5' / (sin (54 deg)) = about 9.27'.

Amazing!

According to:
http://mathworld.wolfram.com/Pentagon.html the angle beteen line drawn from widest point (which divides pentagon into a triangle and square) is PI/5 (or 180/5=36 degrees).

Since the square is length 15, the triangle can be divided into two right trangles with bottom side of 7.5 feet.

Knowing simple trig, we can use Cos(PI/5) = 7.5/Side
Side = 9.27 feet

Damn it....Jschen beat me to it!

hehe :p

Damn you guys are good Thanks :D