EXAMPLE 2 Find the area of a regular

EXAMPLE 2 Find the area of a regular polygon 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.

EXAMPLE 2 Find the area of a regular polygon 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

EXAMPLE 2 Find the area of a regular polygon STEP 3 Find the area A of the table top. 1 Formula for area of regular polygon A = 2 a. P 1 ≈ (18. 108)(120) 2 Substitute. ≈ 1086. 5 Simplify. ANSWER So, the area you are covering with tiles is about 1086. 5 square inches.

EXAMPLE 3 Find the perimeter and area of a regular polygon 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.

EXAMPLE 3 Find the perimeter and area of a regular polygon 4 sin 20° = MK LK LM cos 20° = LK sin 20° = MK 4 cos 20° = LM 4 sin 20° = MK 4 cos 20° = LM The regular nonagon has side length s = 2 MK = 2(4 sin 20°) = 8(sin 20°) and apothem a = LM = 4(cos 20°).

EXAMPLE 3 Find the perimeter and area of a regular polygon ANSWER So, the perimeter is P = 9 s = 9(8 sin 20°) = 72 sin 20° ≈ 24. 6 units, 1 1 and the area is A = a. P = (4 cos 20°)(72 sin 20°) 2 2 ≈ 46. 3 square units.

GUIDED PRACTICE for Examples 2 and 3 Find the perimeter and the area of the regular polygon. 3. SOLUTION 360 The measure of the central angle is = or 72°. 5 Apothem a bisects the central angle, so angle is 36°. To find the lengths of the legs, use trigonometric ratios for right angle.

GUIDED PRACTICE for Examples 2 and 3 sin 36° = b hyp b sin 36° = 8 8 sin 36° = b The regular pentagon has side length = 2 b = 2 (8 sin 36°) = 16 sin 36° 20° So, the perimeter is P = 5 s = 5(16 sin 36°) = 80 sin 36° ≈ 46. 6 units, and the area is A = 1 a. P = 1 6. 5 46. 6 2 2 ≈ 151. 5 units 2.

GUIDED PRACTICE for Examples 2 and 3 Find the perimeter and the area of the regular polygon. 4. SOLUTION The regular nonagon has side length = 7. So, the perimeter is P = 10 · s = 10 · 7 = 70 units

GUIDED PRACTICE for Examples 2 and 3 360 The measure of central is = or 36°. Apothem 10 a bisects the central angle, so angle is 18°. To find the lengths of the legs, use trigonometric ratios for right angle. opp tan 18° = adj tan 18° = a= 3. 5 a 3. 5 tan 18° ≈10. 8 and the area is A = 1 a. P = 1 10. 8 2 2 ≈ 377 units 2. 70

GUIDED PRACTICE for Examples 2 and 3 5. SOLUTION 360° The measure of central angle is 3 = 120°. Apothem a bisects the central angle, so is 60°. To find the lengths of the legs, use the trigonometric ratios.

GUIDED PRACTICE a cos 60° = x x cos 60° = 5 for Examples 2 and 3 b sin 60° = 10 10 sin 60° = b x 0. 5 = 5 x = 10 The regular polygon has side length s = 2 (10 sin 60°) = 20 sin 60° and apothem a = 5.

GUIDED PRACTICE for Examples 2 and 3 So, the perimeter is P = 3 s = 3(20 sin 60°) = 60 sin 60° = 30 1 and the area is A = a. P 2 1 = 2 × 5 30 = 129. 9 units 2 3 3 units

GUIDED PRACTICE for Examples 2 and 3 Which of Exercises 3– 5 above can be solved using special right triangles? 6. ANSWER Exercise 5 can be solved using special right triangles. The triangle is a 30 -60 -90 Right Triangle