1. How to find the area of a regular polygon. Chapter 10.3 & 10.5GeometryStandard/Goal 2.2

2. 1. Check and discuss the assignment from yesterday. 2. Work on Quiz 10.1-10.2 3. Work and discuss Equilateral Activity 4. Read, write, and discuss how to find the area of a regular polygon. 5. Work on assignment.

3. 1. Draw a equilateral triangle 2. Draw the altitude of the triangle. 3. Label the angle measures of all the triangles. 4. Label the lengths of the sides of the triangle, using “ x ” as the length of a side of the equilateral triangle. 5. Now find the height of the triangle. You may need to use Pythagorean theorem. 6. Use the formula of the area of a triangle to find the area of a equilateral triangle. 7. Switch the value of “x” to “s” 8. You now have the formula for the area of an equilateral triangle in terms of “ s ”, the length of a side.

4. The area of an equilateral triangle is one fourth the square of the length of the side times S S S

5. Center of the Polygon The center of its circumscribed circle. Radius of the Polygon is the distance from the center to a vertex. Apothem of the Polygon the perpendicular distance from the center to a side.

6. B E A C D G H a Center G Radius GB or GC Apothem GH F

7. The area of a regular polygon is half the product of the apothem a and the perimeter P, so: Note: P = ns

8. Central angle of a regular polygon is an angle whose vertex is the center and whose sides contain two consecutive vertices of the polygon.

9. m 1 = = 60Divide 360 by the number of sides. 360 6 m 3 = 180 – (90 + 30) = 60The sum of the measures of the angles of a triangle is 180. m 1 = 60, m 2 = 30, and m 3 = 60. A portion of a regular hexagon has an apothem and radii drawn. Find the measure of each numbered angle. Lesson 10-3 m 2 = m 1The apothem bisects the vertex angle of the isosceles triangle formed by the radii. 1212 m 2 = (60) = 30Substitute 60 for m 1. 1212

10. Find the area of a regular polygon with twenty 12-in. sides and a 37.9-in. apothem. p = ns Find the perimeter. p = (20)(12) = 240Substitute 20 for n and 12 for s. A = 4548Simplify. The area of the polygon is 4548 in. 2 A = (37.9)(240)Substitute 37.9 for a and 240 for p. 1212 A = ap Area of a regular polygon 1212 Lesson 10-3

11. Consecutive radii form an isosceles triangle, as shown below, so an apothem bisects the side of the octagon. A library is in the shape of a regular octagon. Each side is 18.0 ft. The radius of the octagon is 23.5 ft. Find the area of the library to the nearest 10 ft 2. To apply the area formula A = ap, you need to find a and p. 1212 Lesson 10-3

12. Step 2: Find the perimeter p. p = ns Find the perimeter. p = (8)(18.0) = 144Substitute 8 for n and 18.0 for s, and simplify. (continued) Step 1: Find the apothem a. a 2 + (9.0) 2 = (23.5) 2 Pythagorean Theorem a 2 + 81 = 552.25 Solve for a. a 2 = 471.25 a 21.7 Lesson 10-3

13. To the nearest 10 ft 2, the area is 1560 ft 2. (continued) Lesson 10-3 Step 3: Find the area A. A = ap Area of a regular polygon A (21.7)(144)Substitute 21.7 for a and 144 for p. A 1562.4Simplify. 1212 1212

14. 1. Draw the figure in the circle 2.Draw an isosceles triangle to the center. 3.Draw both the radii and apothem 4.Find the apothem. -Need to find the exterior angle which will be the same as the central angle. - ½ the central angle is the angle of ½ the triangle. - Use cosine to find the apothem. 5. Find the side using a ² + b ² = c ². - After finding the side multiply it by 2 to get the full length 6. Use the formula A = ½ a∙ns

15. Find the area of a regular polygon with 10 sides and side length 12 cm. Find the perimeter p and apothem a, and then find the area using the formula A = ap. 1212 Because the polygon has 10 sides and each side is 12 cm long, p = 10 12 = 120 cm. Use trigonometry to find a. Lesson 10-5 360 10 Because the polygon has 10 sides, m ACB = = 36. and are radii, so CA = CB. Therefore, ACM BCM by the HL Theorem, so CACB 1212 1212 m  ACM = m  ACB = 18 and AM = AB = 6.

16. (continued) Now substitute into the area formula. A = ap 1212 A = 120 1212 6. tan 18° Substitute for a and p. A = 360 tan 18° Simplify. 360 18 1107.966073 Use a calculator. The area is about 1108 cm 2. Lesson 10-5 tan 18° = 6a6a Use the tangent ratio. a = 6 tan 18° Solve for a.

17. The radius of a garden in the shape of a regular pentagon is 18 feet. Find the area of the garden. Find the perimeter p and apothem a, and then find the area using the formula A = ap. 1212 Lesson 10-5 Because the pentagon has 5 sides, m  ACB = = 72. CA and CB are radii, so CA = CB. Therefore, ACM BCM by the HL Theorem, so 360 5 1212 m  ACM = m  ACB = 36

18. (continued) So p = 5 (2 AM ) = 10 AM = 10 18(sin 36°) = 180(sin 36°). Lesson 10-5 Use the cosine ratio to find a. Use the sine ratio to find AM. a = 18(cos 36°) AM = 18(sin 36°) Use the ratio. Solve. cos 36° = a 18 sin 36° = AM 18 Use AM to find p. Because ACM BCM, AB = 2 AM. Because the pentagon is regular, p = 5 AB.

19. (continued) Finally, substitute into the area formula A = ap. 1212 A = 18(cos 36°) 180(sin 36°) 1212 Substitute for a and p. A = 1620(cos 36°) (sin 36°)Simplify. A 770.355778 Use a calculator. The area of the garden is about 770 ft 2. Lesson 10-5

20. A triangular park has two sides that measure 200 ft and 300 ft and form a 65° angle. Find the area of the park to the nearest hundred square feet. Use Theorem 9-1: The area of a triangle is one half the product of the lengths of two sides and the sine of the included angle. Area = side length side length sine of included angle 1212 Theorem 9-1 Area = 200 300 sin 65° 1212 Substitute. Area = 30,000 sin 65°Simplify. Use a calculator 27189.23361 The area of the park is approximately 27,200 ft 2. Lesson 10-5

21. Kennedy, D., Charles, R., Hall, B., Bass, L., Johnson, A. (2009) Geometry Prentice Hall Mathematics. Power Point made by: Robert Orloski Jerome High School.