 # FeatureLesson Geometry Lesson Main 1.For the similar rectangles, give the ratios (smaller to larger) of the perimeters and of the areas. 2.The triangles.

## Presentation on theme: "FeatureLesson Geometry Lesson Main 1.For the similar rectangles, give the ratios (smaller to larger) of the perimeters and of the areas. 2.The triangles."— Presentation transcript:

FeatureLesson Geometry Lesson Main 1.For the similar rectangles, give the ratios (smaller to larger) of the perimeters and of the areas. 2.The triangles are similar. The area of the larger triangle is 48 ft 2. Find the area of the smaller triangle. 3.The similarity ratio of two regular octagons is 5 : 9. The area of the smaller octagon is 100 in. 2 Find the area of the larger octagon. 4.The areas of two equilateral triangles are 27 yd 2 and 75 yd 2. Find their similarity ratio and the ratio of their perimeters. 5.Mulch to cover an 8-ft by 16-ft rectangular garden costs \$48. At the same rate, what would be the cost of mulch to cover a 12-ft by 24-ft rectangular garden? perimeters: ; areas: 4949 16 81 27 ft 2 324 in. 2 3 : 5; 3 : 5 \$108 Lesson 10-4 Perimeters and Areas of Similar Figures Lesson Quiz 10-5

FeatureLesson Geometry Lesson Main 1.The perimeter is 4(6) = 24 m. The area A of a regular polygon is half the apothem a times the perimeter p: A = ap = (3)(24) = 36 m 2 2.The perimeter is 6(42) = 252 in. The area A of a regular polygon is half the apothem a times the perimeter p: A = ap = (36)(252) = 4536 in. 2 3.The perimeter is 6(8) = 48 ft. The area A of a regular polygon is half the apothem a times the perimeter p: A = ap = (7)(48) = 168 ft 2 1212 1212 1212 1212 1212 1212 Solutions Lesson 10-5 Trigonometry and Area Check Skills You’ll Need 10-5

FeatureLesson Geometry Lesson Main (For help, go to Lesson 10-3.) Find the area of each regular polygon. 1.3.2. Lesson 10-5 Trigonometry and Area Check Skills You’ll Need 10-5 36 m 2 4536 in 2 168 ft 2

FeatureLesson Geometry Lesson Main

FeatureLesson Geometry Lesson Main

FeatureLesson Geometry Lesson Main Lesson 10-5 Trigonometry and Area Notes 10-5

FeatureLesson Geometry Lesson Main Lesson 10-5 Trigonometry and Area Notes 10-5

FeatureLesson Geometry Lesson Main 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 Trigonometry and Area Additional Examples 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. 10-5 Finding Area

FeatureLesson Geometry Lesson Main (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. Quick Check Trigonometry and Area Additional Examples 10-5

FeatureLesson Geometry Lesson Main 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 Trigonometry and Area Additional Examples 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 10-5 Real-World Connection

FeatureLesson Geometry Lesson Main (continued) So p = 5 (2 AM) = 10 AM = 10 18(sin 36°) = 180(sin 36°). Lesson 10-5 Trigonometry and Area 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. Additional Examples 10-5

FeatureLesson Geometry Lesson Main (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 Trigonometry and Area Quick Check Additional Examples 10-5

FeatureLesson Geometry Lesson Main 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 Trigonometry and Area Quick Check Additional Examples 10-5 Real-World Connection

FeatureLesson Geometry Lesson Main Find the area of each figure. Give answers to the nearest unit. 1.regular hexagon with perimeter 90 ft 2.regular pentagon with radius 12 m 3.regular polygon with 12 sides of length 1 in. 4.5. 585 ft 2 342 m 2 11 in 2 490 mm 2 70 yd 2 Lesson 10-5 Trigonometry and Area Lesson Quiz 10-5

Download ppt "FeatureLesson Geometry Lesson Main 1.For the similar rectangles, give the ratios (smaller to larger) of the perimeters and of the areas. 2.The triangles."

Similar presentations