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EXAMPLE 2

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?. Find the area of a regular polygon. EXAMPLE 2. DECORATING. SOLUTION. STEP 1.

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EXAMPLE 2

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  1. You are decorating the top of a table by covering it with small ceramic tiles. The table top is a regular octagon with 15inch sides and a radius of about 19.6inches. What is the area you are covering? Find the area of a regular polygon EXAMPLE 2 DECORATING SOLUTION STEP 1 Find the perimeter Pof the table top. An octagon has 8 sides, so P = 8(15) = 120inches.

  2. Find the apothem a. The apothem is height RSof ∆PQR. Because ∆PQRis isosceles, altitude RSbisects QP. 1 1 So,QS = (QP) = (15) = 7.5 inches. 2 2 √ a = RS ≈ √19.62 – 7.52 = 327.91 ≈ 18.108 Find the area of a regular polygon EXAMPLE 2 STEP 2 To find RS, use the Pythagorean Theorem for ∆ RQS.

  3. 1 A = aP 2 1 ≈ (18.108)(120) 2 ANSWER So, the area you are covering with tiles is about 1086.5square inches. Find the area of a regular polygon EXAMPLE 2 STEP 3 Find the area Aof the table top. Formula for area of regular polygon Substitute. ≈ 1086.5 Simplify.

  4. A regular nonagon is inscribed in a circle with radius 4 units. Find the perimeter and area of the nonagon. 360° The measure of central JLKis , or 40°. Apothem LMbisects the central angle, so m KLMis 20°. To find the lengths of the legs, use trigonometric ratios for right ∆KLM. 9 Find the perimeter and area of a regular polygon EXAMPLE 3 SOLUTION

  5. LM MK sin 20° = cos 20° = LK LK MK LM cos 20° = sin 20° = 4 4 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(cos20°). Find the perimeter and area of a regular polygon EXAMPLE 3

  6. ANSWER 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 sin20°)≈46.3 square units. 1 1 2 2 Find the perimeter and area of a regular polygon EXAMPLE 3

  7. 3. 360 The measure of the central angle is = or 72°. Apothem abisects the central angle, so angleis 36°. To find the lengths of the legs, use trigonometric ratios for right angle. 5 for Examples 2 and 3 GUIDED PRACTICE Find the perimeter and the area of the regular polygon. SOLUTION

  8. b sin 36° = 1 1 and the area is A = aP = 6.5 46.6 hyp 2 2 b sin 36° = 8 8 sin 36° = b The regular pentagon has side length = 2b= 2 (8 sin 36°) = 16 sin 36°20° So, the perimeter is P = 5s = 5(16 sin 36°) = 80 sin 36° for Examples 2 and 3 GUIDED PRACTICE ≈ 46.6 units, ≈151.5 units2.

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

  10. 3.5 tan 18° = a opp tan 18° = adj 3.5 a = 360 tan 18° The measure of centralis = or 36°. Apothem abisects the central angle, so angleis 18°. To find the lengths of the legs, use trigonometric ratios for right angle. 10 1 1 and the area is A = aP = 10.8 70 2 2 for Examples 2 and 3 GUIDED PRACTICE ≈10.8 ≈377 units2.

  11. 5. 360° The measure of central angle is =120°. Apothem abisects the central angle, so is 60°. To find the lengths of the legs, use the trigonometric ratios. 3 for Examples 2 and 3 GUIDED PRACTICE SOLUTION

  12. a b cos 60° = sin 60° = x 10 10 sin 60° = x cos 60° = 5 b x 0.5 = 5 x = 10 The regular polygon has side length s = 2 = 2 (10 sin 60°) = 20 sin 60° and apothem a = 5. for Examples 2 and 3 GUIDED PRACTICE

  13. So, the perimeter isP = 3 s = 3(20 sin 60°) = 60 sin 60° = 30 3 units 1 and the area is A = aP 2 1 = ×5 30 3 2 for Examples 2 and 3 GUIDED PRACTICE = 129.9 units2

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

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