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GEOMETRY
INTERIOR ANGLES OF REGULAR POLYGONS
In the previous assignment, many students estimated the coordinate positions of the vertices.
This is unnecessary as the program can perform any calculations required to position these
vertices exactly. All we need to do is give the program the mathematical formulas. To do this,
however, we need to figure out the interior and exterior angles for the pentagon (imidazole) and
hexagon (pyrimidine) shapes.
You know that the sum of the three interior angles of a ∆ = 180°. If we draw segments from the
center of a regular polygon with n sides (a regular polygon is one whose sides all have equal
length) to each of its vertices, we will create n congruent isosceles ∆s (the ∆s are congruent
because their 3 corresponding sides are congruent: SSS). So, a square will yield 4 ∆s, a pentagon
5∆s and a hexagon 6∆s.
If you sum the interior angles in all of these triangles, you will get n x 180° (Figure 3A).
Figure 3A
Now, if you sum just the vertex angles of these n ∆s (the angles around the center point) you will
always get 360°. Refer to the blue circles around the center points in Figure 3B.
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Figure 3B
Subtracting the sum of these vertex angles (360°) from n x 180° gives you the sum of all of the
polygon's interior angles.
Dividing the sum of the polygon's interior angles by n will give the measure of a single interior
angle (Figure 3B).
There is an alternate way to calculate the measure of a single interior angle of a regular polygon.
Dividing the sum of the vertex angles of the n ∆s (again: the angles around the center point,
which always sum to 360°) by n gives the measure of a single ∆ vertex angle.
Because each ∆ is isosceles, its base angles are congruent.
Summing the 3 angles of the ∆ in Figure 3C gives:
360/n + b + b = 180
360/n + 2b = 180
2b = 180 - 360/n
Notice that 1 interior polygon angle is comprised of 2 base ∆
angles. Therefore, a single interior polygon angle = 180 - 360/n.
Figure 3C
GEOMETRY
COORDINATES OF HEXAGON VERTICES
We can divide a hexagon into 6 congruent equilateral
∆s, as in Figure 4A. These 6 triangles are
equilateral for the following reason: Each base angle
in each ∆ measures ½(120°)= 60°; so the vertex angle
is also 60°. Therefore in a hexagon, the 6 isosceles
triangles are also equiangular.
Because all equiangular ∆s are also EQUILATERAL,
these 6 ∆s are also equilateral.
Let's call the length of each side of the hexagon s.
Remember: the 3 sides of the equilateral ∆s are
congruent, and each angle measures 60°.
Figure 4A
Let's focus on just one of the ∆s.
If we draw a segment (the blue segment labeled x) so that it bisects the vertex angle of the ∆,
it will create two congruent right ∆s (per the ASA, SAS, or Hypotenuse-Leg theorems).
Because the shortest legs of these two right ∆s are congruent (CPCTC), each of the legs is
therefore half the length (s/2) of the hypotenuse (s). Because we now know the lengths of two
sides of a right ∆, we can find the length of the 3rd side (x) using the Pythagorean Theorem,
which states: The sum of the squares of the lengths of the two legs = The square of the length of
the hypotenuse
x2 + (s/2)2 = s2
Solving for x:
x2 = s2 - (s/2)2
x2 = s2 - s2/4
x2 = 3/4(s2)
x = s(√3/2)
(Figure 4B)
Figure 4B
If you look at the Guanine Drawing, you can determine the length of the side of the hexagon in
the molecule by examining either of the horizontal sides of the hexagon. From the discussions
above, you now know how to express the lengths of each of the two legs of the 30-60-90 ∆s in
terms of the length of the side (the hypotenuse of the 30-60-90 ∆s = side).
You can put all of this information together to find the x-y coordinates of each of the 6 vertices
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of the hexagon. Instead of using hard-coded numbers, express the coordinates using expressions
in terms of the hexagon's side.
(Circumnavigating Regular Polygons): EXTERIOR ANGLES
In Figure 6, imagine that the pen first draws the top edge of each polygon (segment AB), so it
starts out facing to the right (due east, 90°). At point B, the pen turns clockwise a certain angle so
that it is facing the same direction as segment BC before drawing it. This angle (labeled 2 in
Figure 6) is an EXTERIOR angle. How do you calculate the size of an exterior angle?
Figure 6
Remember, that for a polygon, the exterior and interior angles are supplementary, i.e., they sum
to 180°. In a hexagon, since each interior angle measures 120°, each exterior angle will therefore
measure 60°.
When drawing the hexagon, the pen would turn 6 times in order to reach its original position and
direction. 60° x 6 = 360°, the number of degrees in a full circle. To emphasize this point, in
Figures 6, all of the exterior angles for both the hexagon and pentagon have been positioned
next to one another (to the bottom right of each polygon) to show that, in both cases, they sum to
360°. Therefore, another way to calculate an exterior angle is to divide 360° by the polygon's
number of sides, or 360/n.