11.6

Warm UpWarm Up

Lesson QuizLesson Quiz

Lesson PresentationLesson Presentation

Areas of Regular Polygons

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.

122. Solve 18 = x.

ANSWER 24 m

ANSWER 36

11.6 Warm-Up

3. Evaluate (4 cos 10º)(10 sin 20º).

4. Evaluate .10tan 88º

ANSWER 13.47

0.3492ANSWER

11.6 Example 1

a. m AFB

In the diagram, ABCDE is a regular pentagon inscribed in F. Find each angle measure.

SOLUTION

a. AFB is a central angle, so m AFB = , or 72°.

360°

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11.6 Example 1

In the diagram, ABCDE is a regular pentagon inscribed in F. Find each angle measure.

SOLUTION

b. m AFG

b. FG is an apothem, which makes it an altitude of isosceles ∆AFB. So, FG bisects AFB and m

AFG = m AFB = 36°.12

11.6 Example 1

In the diagram, ABCDE is a regular pentagon inscribed in F. Find each angle measure.

SOLUTION

c. m GAF

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.

P, PY or XP, PQ, XPYANSWER

2. Find m XPY, m XPQ, and m PXQ.

90°, 45°, 45°ANSWER

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

So, QS = (QP) = (15) = 7.5 inches.12

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To find RS, use the Pythagorean Theorem for ∆ RQS.

a = RS ≈ √19.62 – 7.52 = 327.91 ≈ 18.108 √

Find the apothem a. The apothem is height RS of ∆PQR. Because ∆PQR is isosceles, altitude RS bisects QP .

11.6 Example 2

STEP 3 Find the area A of the table top.12A = aP Formula for area of

regular polygon

≈ (18.108)(120)12 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.

SOLUTION360°

The measure of central JLK is , 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.

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11.6 Example 3

sin 20° = MKLK

sin 20° = MK4

4 sin 20° = MK

cos 20° = LMLK

cos 20° = LM4

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,and the area is A = aP = (4 cos 20°)(72 sin 20°) ≈ 46.3 square units.

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11.6 Guided Practice

3.

Find the perimeter and the area of the regular polygon.

about 46.6 units, about 151.6 units2ANSWER

11.6 Guided Practice

Find the perimeter and the area of the regular polygon.

4.

70 units, about 377.0 units2ANSWER

11.6 Guided Practice

Find the perimeter and the area of the regular polygon.

5.

30 3 52.0 units, about 129.9 units2ANSWER

11.6 Guided Practice

6. Which of Exercises 3–5 above can be solved using special right triangles?

Exercise 5ANSWER

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