Feature Lesson 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: 4 9 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
Transcript
FeatureLesson
GeometryGeometry
LessonMain
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 largertriangle is 48 ft2. 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 yd2 and 75 yd2. 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:49
1681
27 ft2
324 in.2
3 : 5; 3 : 5
$108
Lesson 10-4
Perimeters and Areas of Similar FiguresPerimeters and Areas of Similar Figures
Lesson Quiz
10-5
FeatureLesson
GeometryGeometry
LessonMain
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 m2
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 ft2
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Solutions
Lesson 10-5
Trigonometry and AreaTrigonometry and Area
Check Skills You’ll Need
10-5
FeatureLesson
GeometryGeometry
LessonMain
(For help, go to Lesson 10-3.)
Find the area of each regular polygon.
1. 3.2.
Lesson 10-5
Trigonometry and AreaTrigonometry and Area
Check Skills You’ll Need
Check Skills You’ll Need
10-5
36 m24536 in2 168 ft2
FeatureLesson
GeometryGeometry
LessonMain
FeatureLesson
GeometryGeometry
LessonMain
FeatureLesson
GeometryGeometry
LessonMain
Lesson 10-5
Trigonometry and AreaTrigonometry and Area
Notes
10-5
FeatureLesson
GeometryGeometry
LessonMain
Lesson 10-5
Trigonometry and AreaTrigonometry and Area
Notes
10-5
FeatureLesson
GeometryGeometry
LessonMain
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.1
2
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 AreaTrigonometry and Area
Additional Examples
36010
Because the polygon has 10 sides, m ACB = = 36.
and are radii, so CA = CB. Therefore, ACM BCM by the HLTheorem, soCA CB
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m ACM = m ACB = 18 and AM = AB = 6.
10-5
Finding Area
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(continued)
Now substitute into the area formula.
A = ap12
A = • • 12012
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 cm2.
Lesson 10-5
tan 18° = 6a Use the tangent ratio.
a = 6 tan 18° Solve for a.
Quick Check
Trigonometry and AreaTrigonometry and Area
Additional Examples
10-5
FeatureLesson
GeometryGeometry
LessonMain
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.1
2
Lesson 10-5
Trigonometry and AreaTrigonometry 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 HLTheorem, so
3605
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m ACM = m ACB = 36
10-5
Real-World Connection
FeatureLesson
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LessonMain
(continued)
So p = 5 • (2 • AM) = 10 • AM = 10 • 18(sin 36°) = 180(sin 36°).
Lesson 10-5
Trigonometry and AreaTrigonometry 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
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GeometryGeometry
LessonMain
(continued)
Finally, substitute into the area formula A = ap.12
A = • 18(cos 36°) • 180(sin 36°)12
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 ft2.
Lesson 10-5
Trigonometry and AreaTrigonometry and Area
Quick Check
Additional Examples
10-5
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LessonMain
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
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Theorem 9-1
Area = • 200 • 300 • sin 65°12
Substitute.
Area = 30,000 sin 65° Simplify.
Use a calculator27189.23361
The area of the park is approximately 27,200 ft2.
Lesson 10-5
Trigonometry and AreaTrigonometry and Area
Quick Check
Additional Examples
10-5
Real-World Connection
FeatureLesson
GeometryGeometry
LessonMain
Find the area of each figure. Give answers to the nearest unit.
