proof on regular hexagon???

Question

http://img339.imageshack.us/my.php?image=hexagonpc9.png

The diagram shows a regular hexagon ABCDEF, with diagonal FD and FC draw. Supposed AB = 6

a. find the area of triangle FED
b. find the area of triangle FCD
c. find the area of trapazoid ABCF


help please???

Answer 1

The key to finding these areas is to note that a regular hexgon is made up of six equalateral triangles with side length equal to the side length of the hexagon (they are formed by connecting the center of the hexagon to the vertices). Since the area of an equalateral triangle is (sqrt(3)/4)*side^2, the area of each of these triangles is 9*sqrt(3).

Now notice that when you connect D and F, the triangle DEF consists of two halves of an equalateral triangles, so its area is 9*sqrt(3). The triangle CDF consists of a whole equalateral triangle plus two halves, so its area is 18*sqrt(3). Finally, the trapazoid, ABCF consists of three equalateral trianlges, so its area is 27*sqrt(3).

Answer 2

The first thing to do is draw out your hexagon, with the diagonals FD and FC drawn.

triangle FED is a isosceles triangle with the two sides = 6 and the angles = 120, 30 and 30

the area of a triangle is (ab)/2 where a is the altitude and b is the base of the triangle

you need to figure out both a and b. a is the altitude from E to the line FD, you can figure out a using sin(30) = a/FE => a = sin(30)*6
and b = FE = 2cos(30)*6

or can realize that half of FED is a 30-60-90 triangle so the sides are in the ratio 1-2-SQRT(3) with 2 being the longest or the hypotenuse (FE=6), so a and b would be (3 and 3*SQRT(3))

and the area of FED would be (3*(3*SQRT(3)*2)/2 = 9*SQRT(3)

I would then deteremine the area of the whole hexagon, and your trapezoid would be 1/2 of the area of the hexagon, and the area of FCD would be equal to (1/2 are of hexagon - area of FED)

area of the hexagon = ((S^2)*6)/(4*tan(pi/6)) where S = side size =6
area = (36*6)(4*tan(pi/6)) (note this is when tangent in Randians
in degrees
area = (36*6)(4*tan(30)) tan(30) = 1/SQRT(3)

area = 93.53

the rest just follows.

good luck

Answer 3

Very nice drawing. To start, the vertex angles of the regular hexagon are 120 deg. So extend DE to a point G, and drop an altitude from the vertex F to DG. The area of FED is 1/2 (base)(altitude), where we know the base is 6. Triangle FEG is 30/60/90, so if the hypotenuse FE=6, the altitude is 3 sqrt(3). So the area of FED is 9 sqrt(3).
For triangle FCD, we already have the altitude from D to base FC, we just need to find FC. If you accept some arm-waving, FC = 12, and the area is 18 sqrt(3). For the trapezoid, you have the two bases and the altitude, so just substiute into the trapezoid area formula.

Answer 4

If you triangulate a regular polygon into triangles that have a side of the polygon as one side and the center of the polygon as a vertex, the triangles will be isoceles triangles, because at least two sides will always be equal. The central angles in any polygon add up to 360 degrees. A hexagon has 6 central angles. In a regular hexagon, they have the same measure, so each is 360/6 = 60 degrees. An isoceles triangle with one angle of 60 degrees will have all angles 60 degrees and therefore be equilateral. Note: Because there are other ways to triangulate a polygon, it is more appropriate to say that a regular hexagon CAN be divided into 6 equilateral triangles, rather than saying it IS divided into 6 equilateral triangles.

Answer 5

Answers:

a. Area FED = 9*sqrt(3)

b. Area FCD = 18*sqrt(3)

c. Area ABCF = 27*sqrt(3)