Download 7-3-formulas-involving-polygons-ppt

When solving problems
involving polygons,
sometimes “given” information
must be determined using the
of the polygon.
Heptagon
8 sides -
Nonagon -------- 9 sides
Decagon -------- 10 sides
Undecagon----- 11 sides
Dodecagon ----- 12 sides
13-gon -----------13 sides
14-gon -----------14 sides
Pentadecagon-- 15 sides
n-gon ------------ n-sides




Step 1: Draw as many diagonals possible from one
vertex of the polygon.
Step 2: Count the number of triangles formed by
the diagonals.
Step 3: Multiply the number of triangles by 180⁰.
Step 4: Your product equals the sum of the interior
angles of the polygon.
GOAL: See if you can discover a pattern and write a
rule for finding the sum of the angles in any n-gon.
No diagonals possible!
triangle
1 diagonal possible!
quadrilateral
2 diagonals possible!
pentagon
5 diagonals possible!
octagon
X
1
0
X
X
X
X
2
5
X
X
2
1
X
X
X
X
3
6
X
Number of SIDES
3
4
5
6
7
8
n
Number of TRIANGLES
1
2
3
4
5
6
___?___
Calculation
1(180 ̊)
2(180 ̊)
3(180 ̊)
4(180 ̊)
5(180 ̊)
6(180 ̊)
(?)(180⁰)
?
Sum of Interiors
180 ̊
360 ̊
540 ̊
720 ̊
900 ̊
1080 ̊
Si =
?(180)
The sum of the angles in
every triangle is 180⁰!
Si = (n - 2) 180⁰
The number of triangles
is always 2 less than the
number of sides!
Theorem 55:
The sum (Si)of the measures of the angles of a
polygon with n sides is given by the formula -
Si = (n - 2) 180

Tri angle: Si = 180⁰
120⁰
60⁰
110⁰ 70⁰
120
130
+110
Se = 360⁰
50⁰
130⁰

110
quadrilateral: Si = 360⁰
80
50
110⁰
+120
70⁰
Se = 360⁰
100⁰
80⁰
120⁰ 60⁰
130⁰
50⁰

pentagon: Si = 540⁰
105⁰
30⁰
150⁰
75⁰
75⁰
105⁰
60⁰ 120⁰
120⁰
60⁰
105
30
105
60
+ 60
Se = 360⁰
Notice that every interior angle supplements
an exterior angle!
So one method to prove the exterior angle sum is to
multiply the number of supplements and then
subtract the interior angle sum!
(# of Sides)(Supp) = (Si + Se ) - Si
= Se
(3 sides) (180) =
540
- 180 = 360
(4 sides) (180) =
720
- 360 = 360
(5 sides) (180) =
900
- 540 = 360
(6 sides) (180) =
1080
- 720 = 360
Theorem 57:
If one exterior angle is taken at each vertex, the
sum Se of the measures of the exterior angles
of a polygon is given by the formula
X
1
0
X
X
X
X
2
5
X
0
X
X
X
X
X
X
2
Notice that when drawing diagonals, you can never use the vertex you are “ON”
(Because a diagonal is a segment – and every segment needs two endpoints!)
And you can’t connect to either adjacent vertex to draw a diagonal either!
(Because diagonals are segments found INSIDE a polygon.
If you connect the vertex you are ON to consecutive vertices the result is a SIDE!)
X
2
X
How many diagonals can be drawn from each vertex of the polygon below?
So, if all of these diagonals were UNIQUE, there would be (5)(2) = 10 diagonals
2
2
2
2
2
This
isthe
too
Label
many! to
vertices
see why
Some
of
we will
thoseto
need
segments
adjust
our
were
initial
already
total.
there!!!
How many diagonals can be drawn from each vertex of the polygon below?
So, if all of these diagonals were UNIQUE, there would be (5)(2) = 10 diagonals
The 5 unique diagonals are:
PN
PT
ET
EA
PN, PT, ET, EA, and NA
NP
NA
E2
TE
TP
AN
AE
P 2
2
A
N
2
2T
So when we
Label
Every
the
multiply the
number ofto
vertices
segment
vertices by the
see
can
why
be
number
of
diagonals
named
we willwe
in
can draw from
two
need
ways!
each
one,to
we
counted
adjust
PN is every
the
our
diagonal twice!
initial
same
How
could you
segment
total.
easily adjust
asnumber?
NP!
that
It’s getting a little
crowded in this
diagram – so let’s
analyze the pattern!
5
5
5
5
5
5
5
5
1) In this polygon, there are 8 vertices
2) We can only draw diagonals to 5 other vertices from each vertex.
3) Every diagonal we draw can be named in two ways!
4) Rule: (number of vertices) times (3 less than the number of vertices)
divided by 2
5 sides = 5(5 – 3) = 5(2)
2
2
6(6 – 3) =
2
3
6(3)__ =
2
8(8 – 3) =
2
4 8(5)__ =
2
Theorem 57:
The number d of diagonals that can be drawn in
a polygon of n sides is given by the formula
•
Find the number of diagonals
d = n(n – 3)
2
d = 11(11 – 3)
2
d = 11(8) 4
2
diagonals = 44
Find the sum of the exterior angles
180 (11) – 1620 = 1980 – 1620 =  You really don’t have to do this!
Se = 360
Always!