11.6 Areas of Regular Polygons
Warm Up
Lesson Presentation
Lesson Quiz
11.6 Warm-Up
1. An isosceles triangle has side lengths 20 meters,
26 meters, and 26 meters. Find the length of the
altitude to the base.
ANSWER
24 m
2. Solve 18 = 12
ANSWER
36
x.
11.6 Warm-Up
3. Evaluate (4 cos 10º)(10 sin 20º).
ANSWER
13.47
4. Evaluate
10 .
tan 88º
ANSWER
0.3492
11.6 Example 1
In the diagram, ABCDE is a regular
pentagon inscribed in F. Find
each angle measure.
a.
m
AFB
SOLUTION
a.
AFB is a central angle, so m
72°.
°
360
AFB =
, or
5
11.6 Example 1
In the diagram, ABCDE is a regular
pentagon inscribed in F. Find
each angle measure.
b.
m
AFG
SOLUTION
b.
FG is an apothem, which makes it an altitude of
isosceles ∆AFB. So, FG bisects
AFB and m
AFG = 1 m AFB = 36°.
2
11.6 Example 1
In the diagram, ABCDE is a regular
pentagon inscribed in F. Find
each angle measure.
c.
m
GAF
SOLUTION
c.
The sum of the measures of right ∆GAF is 180°.
So, 90° + 36° + m GAF = 180°, and m GAF = 54°.
11.6 Guided Practice
In the diagram, WXYZ is a square
inscribed in P.
1.
Identify the center, a radius,
an apothem, and a central
angle of the polygon.
ANSWER
2.
Find m
ANSWER
P, PY or XP, PQ, XPY
XPY, m
XPQ, and m
90°, 45°, 45°
PXQ.
11.6 Example 2
DECORATING You are decorating
the top of a table by covering it with
small ceramic tiles. The table top is a
regular octagon with 15 inch sides
and a radius of about 19.6 inches.
What is the area you are covering?
SOLUTION
STEP 1 Find the perimeter P of the table top. An
octagon has 8 sides, so P = 8(15) = 120
inches.
11.6 Example 2
STEP 2
Find the apothem a. The apothem
is height RS of ∆PQR. Because
∆PQR is isosceles, altitude RS
bisects QP .
So, QS = 1 (QP) = 1 (15) = 7.5 inches.
2
2
To find RS, use the Pythagorean
Theorem for ∆ RQS.
a = RS ≈ √19.62 – 7.52 = √327.91 ≈ 18.108
11.6 Example 2
STEP 3
Find the area A of the table top.
1
Formula for area of
A = 2 aP
regular polygon
1
≈ (18.108)(120)
2
Substitute.
≈ 1086.5
Simplify.
So, the area you are covering with tiles is about
1086.5 square inches.
11.6 Example 3
A regular nonagon is inscribed in a
circle with radius 4 units. Find the
perimeter and area of the nonagon.
SOLUTION
360°
The measure of central JLK is 9 , or 40°.
Apothem LM bisects the central angle, so
m KLM is 20°. To find the lengths of the legs,
use trigonometric ratios for right ∆ KLM.
11.6 Example 3
4
LM
LK
sin 20° = MK
LK
cos 20° =
sin 20° = MK
4
cos 20° = LM
4
sin 20° = MK
4
cos 20° = LM
The regular nonagon has side length
s = 2MK = 2(4 sin 20°) = 8  sin 20° and apothem
a = LM = 4  cos 20°.
So, the perimeter is P = 9s = 9(8 sin 20°) = 72 sin 20°
≈ 24.6 units,
1
1
and the area is A =
aP =
(4 cos 20°)(72 sin 20°)
2
2
≈ 46.3 square units.
11.6 Guided Practice
Find the perimeter and the area of the regular polygon.
3.
ANSWER
about 46.6 units, about 151.6 units2
11.6 Guided Practice
Find the perimeter and the area of the regular polygon.
4.
ANSWER
70 units, about 377.0 units2
11.6 Guided Practice
Find the perimeter and the area of the regular polygon.
5.
ANSWER
30 3  52.0 units, about 129.9 units2
11.6 Guided Practice
6.
Which of Exercises 3–5 above can be solved
using special right triangles?
ANSWER
Exercise 5
11.6 Lesson Quiz
1.
Find the measure of the central angle of a
regular polygon with 24 sides.
ANSWER 15°
11.6 Lesson Quiz
Find the area of each regular polygon.
2.
ANSWER 110 cm2
3.
ANSWER 374.1 cm2
11.6 Lesson Quiz
Find the perimeter and area of each regular polygon.
4.
ANSWER 99.4 in. ; 745.6 in.2
5.
ANSWER 22.6 m; 32 m2