1. FeatureLesson Geometry Lesson Main 1. Find the area of a trapezoid with bases 3 cm and 19 cm and height 9 cm. 2. Find the area of a trapezoid in a coordinate plane with vertices at (1, 1), (1, 6), (5, 9), and (5, 1). Find the area of each figure in Exercises 3–5. Leave your answers in simplest radical form. 3. trapezoid ABCD 4. kite with diagonals 20 m and 10 2 m long 5. rhombus MNOP 99 cm 2 26 square units 94.5 3 in. 2 100 2 m 2 840 mm 2 Lesson 10-2 Areas of Trapezoids, Rhombuses, and Kites Lesson Quiz 10-3

2. FeatureLesson Geometry Lesson Main 1.2. 3. 4. a hexagon with sides of 4 in. 5. an octagon with sides of 2 3 cm (For help, go to Lesson 8-2.) Lesson 10-3 Areas of Regular Polygons Find the area of each regular polygon. If your answer involves a radical, leave it in simplest radical form. Find the perimeter of the regular polygon. Check Skills You’ll Need 10-3

3. FeatureLesson Geometry Lesson Main Lesson 10-3 Areas of Regular Polygons Solutions 1212 1212 10 2 1. The triangle is equilateral and equiangular, so each of its angles is 60°. The altitude divides the triangle into two 30°-60°-90° triangles. Since the short leg of the 30°-60°-90° triangle is 5cm, the long leg, which is the altitude of the equilateral triangle, is 5 3 cm. The base is 10 cm and the height is 5 3 cm. The area A = bh = (10)(5 3) = 25 3 cm 2. 2. The diagonal is 10 cm and divides the square into two 45°-45°-90° triangles. The legs are each, or 5 2 ft. The base is 5 2 and the height is 5 2. The area A = bh = (5 2)(5 2) = (25)(2) = 50 ft 2. Check Skills You’ll Need 10-3

4. FeatureLesson Geometry Lesson Main Lesson 10-3 Areas of Regular Polygons 20 3 100 3 100 3 3 1212 Solutions (continued) 10 3 20 3 20 3 1212 Check Skills You’ll Need 3. The triangle is equilateral and equiangular, so each of its angles is 60°. The altitude divides the triangle into two 30°-60°-90° triangles. The altitude is 10 m. Since the long leg of the 30°-60°-90° triangle is 10m the short leg is m and the hypotenuse is m. Thus the base of the equilateral triangle is m. The area A = bh = ( ) (10) =, or m 2. 4. The perimeter of a polygon is the sum of the lengths of its sides. A regular hexagon has six sides of the same length. Since each side has length 4 in., the perimeter is = (6)(4) = 24 in. 5. The perimeter of a polygon is the sum of the lengths of its sides. A regular octagon has eight sides of the same length. Since each side has length 2 3 cm, the perimeter is (8)(2 3) = 16 3 cm. 10-3

5. FeatureLesson Geometry Lesson Main Lesson 10-3 Areas of Regular Polygons Notes 10-3 The center of a regular polygon is equidistant from the vertices. The radius of a regular polygon is the distance from the center to a vertex. The apothem is the distance from the center to a side. A central angle of a regular polygon has its vertex at the center, and its sides pass through consecutive vertices. Each central angle measure of a regular n-gon is

6. FeatureLesson Geometry Lesson Main Lesson 10-3 Areas of Regular Polygons Notes 10-3 Regular pentagon DEFGH has a center C, apothem BC, and central angle DCE.

7. FeatureLesson Geometry Lesson Main Lesson 10-3 Areas of Regular Polygons Notes 10-3 To find the area of a regular n-gon with side length s and apothem a, divide it into n congruent isosceles triangles. The perimeter is P = ns. area of each triangle: total area of the polygon:

8. FeatureLesson Geometry Lesson Main Lesson 10-3 Areas of Regular Polygons Notes 10-3

9. FeatureLesson Geometry Lesson Main 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 Areas of Regular Polygons 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 Quick Check Additional Examples 10-3 Finding Angle Measures

10. FeatureLesson Geometry Lesson Main 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 = apArea of a regular polygon 1212 Lesson 10-3 Areas of Regular Polygons Quick Check Additional Examples 10-3 Finding the Area of a Regular Polygon

11. FeatureLesson Geometry Lesson Main 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 Areas of Regular Polygons Additional Examples 10-3 Real-World Connection

12. FeatureLesson Geometry Lesson Main Step 2: Find the perimeter p. p = nsFind 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 Areas of Regular Polygons Additional Examples 10-3

13. FeatureLesson Geometry Lesson Main To the nearest 10 ft 2, the area is 1560 ft 2. (continued) Lesson 10-3 Areas of Regular Polygons Step 3: Find the area A. A = apArea of a regular polygon A (21.7)(144)Substitute 21.7 for a and 144 for p. A 1562.4Simplify. 1212 1212 Quick Check Additional Examples 10-3

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15. FeatureLesson Geometry Lesson Main

16. FeatureLesson Geometry Lesson Main 1. Find m 1. 2. Find m 2. 3. Find m 3. 4. Find the area of a regular 9-sided figure with a 9.6-cm apothem and 7-cm side. For Exercises 5 and 6, find the area of each regular polygon. Leave your answer in simplest radical form. 5.6. 36 18 72 302.4 cm 2 48 3 m 2 6 3 in. 2 Lesson 10-3 Areas of Regular Polygons Use the portion of the regular decagon for Exercises 1–3. Lesson Quiz 10-3

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18. FeatureLesson Geometry Lesson Main