Area and Perimeter:
Areas of Regular Polygons
Review: Inscribed Polygons
& Circumscribed Circles
Inscribed means written inside
Circumscribed means written around (the outside)
inscribed polygon
circumscribed circle
Def: A polygon is inscribed in a circle & the circle is circumscribed about
the polygon when each vertex of the polygon lies on the circle.
Def: A regular polygon is a polygon that is equiangular & equilateral.
Inscribed Regular Polygons & Triangles
Inscribed Regular Pentagon
Total of Interior Angles = 540
Each Interior Angle = 108
5 congruent isosceles triangles
Total of Central Angles = 360
Each central angle = 72
Parts of a Regular Polygon

A stands for Area




A(nonagon) is the area of a regular 9-sided figure.
n is the number of sides of a regular polygon
p is perimeter, r is radius, s is side
a is apothem
 Apothem – The line segment from the center of a
regular polygon to the midpoint of a side or the
length of this segment.
 Sometimes known as the inradius, or the radius
of a regular polygon’s inscribed circle.
Regular Polygon Area Theorem
Given: an inscribed regular n-gon (shown as an octagon)
A(n-gon) = nA( XOY )
1
 n sa
2
1
 a(ns)
2
O
a
X
s
Y
1
 ap where, p = the perimeter
2
Regular Polygon Area Theorem: The area of a regular
polygon is one half the product of the apothem & the
perimeter.
Regular Polygon Terminology
O
X
(Regular Octagon)
M Y
Center of a regular polygon - the center of the circumscribed circle (O).
Radius of a regular polygon - the distance from the center to a vertex (OX).
Central angle of a regular polygon - an angle formed by 2 radii drawn
to consecutive vertices. ( XOY )
Apothem of a regular polygon - the (perpendicular) distance from the
center of the polygon to a side. (OM)
Example: Square
r = 8 2 . Find a, p, A.
r
45
x
a
hyp  leg 2
p  ns
8 2 a 2
p  4(2x)
a8
xa8
p  4[(2)(8)]
s
p  64
1
ap
2
1
 8(64)
2
 256
A
Example: Equilateral Triangle
a = 4. Find r, p, A .
hyp  2short
r
30
r  2a
a
s
p  3(2x)
r  2(4)
r8
x
p  ns
long  3 short
x  3(4)
p  3[(2 3(4)]
p  24 3
1
ap
2
1
 4(24 3)
2
A
 48 3
Example: Regular Hexagon
a = 5 3 . Find r, p, A.
long  3 short
a 3x
r
a
5 3  3x
x5
60
x
s
hyp  2 short
r  2(5)
r  10
p  ns
 6(2x)
 6(2)(5)
 60
1
ap
2
1
 (5 3)(60)
2
A
 150 3
Regular Nonagon
r = 10; Find a, p, A.
a
r
a = 10(.9397)
a  9.397
X x
s
70
opp
sin X 
hyp
a
sin 70 
10
p  ns
 9(2x)
cos X 
adj
hyp
x
cos 70 =
10
x = 10(.3420)
x = 3.420
p = 9(2)(3.420)
p = 61.56
1
A  ap
2
1
 (9.397)(61.56)
2
= 289.24
Examples
r
r
r
a
a
A
1. 8 2
2.
8
5.
5
6.
49
3.
4.
r
6
a
a
p
6
4
7.
12
8.
9 3
A
More Examples
s
r
a
x
1. r = 4 2 , find A.
2. a = 6, find A.
r
a
x
3. a = 8, find p.
4. r = 12, find s.
r
a
x
5. s = 8, find r.