Areas of Regular Polygons
Lesson 11.5
Equilateral Triangle
Remember: drop an
altitude and you create two
30-60-90 triangles.
What is the measure of the sides and altitude
in terms of one side equaling s?
Altitude = s√3
2
C
T
Given: ∆ CAT is
equilateral, and TA = s
A
Find the area of ∆CAT
S
A∆
CAT
=
=



=
1
bh
2
1
s
s(
2
2
S2
3
4
3)
Theorem 106: Area of an
equilateral triangle = the product
of 1/4 the square of a side and
the square root of 3. Where s is
the length of a side
2
S
Aeq∆ = 3
4
An equilateral triangle has
a side of 10 cm long. Find
the area of the triangle.
A=
2
10 (√3)
4
A = 25√3 cm2
Area of a regular polygon:
Remember all interior angles
are congruent and all sides
are equal.
N
E
Regular pentagon:
T
O
O is the center
OA the radius
P
M
A
OM is an apothem
You can make 5 isosceles
triangles in a pentagon.
Any regular polygon:
Radius: is a segment joining the
center to any vertex
Apothem: is a segment joining the
center to the midpoint of any side.
Apothems:
1. All apothems of a regular polygon are
congruent.
2. Only regular polygons have apothems.
3. An apothem is a radius of a circle inscribed in
the polygon.
4. An apothem is the perpendicular bisector of a
side.
5. A radius of a regular polygon is a radius of a
circle circumscribed about the polygon.
6. A radius of a regular polygon bisects an angle
of the polygon.
Theorem 107: Areg poly = ½ ap
Area of a regular polygon equals
one-half the product of the
apothem and the perimeter.
Where a = apothem
p = perimeter
A regular polygon has a perimeter
of 40 cm and an apothem of 5
cm. Find the polygon’s area.
A = ½ap
= ½(5)(40)
= 100 cm2
Find the area of a regular
hexagon whose sides are
18 cm long.
1. Draw the picture
2. Write the formula
3. Plug in the numbers
4. Solve and label units
Find the perimeter
18cm
Find each angle
Find the apothem
P = 18(6) = 108 cm
Angles = 720º/6 angles = 120º per angle
Radius breaks it into 60º angles.
30-60-90 triangle, apothem = 9√3 cm
Write the formula, and solve.
A = ½ ap
A = ½ (9√3)108
A = 486√3 cm 2
Team Challenge:
A square is
inscribed in an
equilateral
triangle as shown.
Find the area of
the shaded region.
2x + x√3 = 12
x=
12
2 + √3
x = 12(2 – √3)
x√3
x
A
(shaded)
x√3
x
= ½ (12)(6√3) – [12(2 – √3)√3]2
= 1764√3 - 3024