Download Lesson Plan - epawelka-math

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
Elizabeth Pawelka
Geometric Probability
4/11/12
Geometry
Lesson Plans
Section 9-5: Trigonometry and Area
4/18/12
Pre-Quiz Warm-up (5 mins)
Find value of each variable to the nearest tenth.
 Check your sides with the Pythagorean Theorem.
 Check angles with Triangle Angle Sum Theorem/Complementary Angles
Ask for any questions on sections 9-1/9-2
Quiz on sections 9-1/9-2 (25 mins)
Post-Quiz Seat-work (15 mins)


“When a Ruler Isn’t Enough” worksheet
Reteaching 9-3: p. 109 # 1, 3, 4, 5
p.1
Elizabeth Pawelka
Geometric Probability
4/11/12
p.2
Homework Review (10 mins) – ask for any questions on homework
Homework (H)
 p. 484, #1 – 21, 23, 28, 33, 34
Homework (R)
 p. 484, #1 – 21, 23, 33
Statement of Objectives (5 mins)

The student will be able to use trigonometry to find area of polygons.
Teacher Input (25 mins)
Area of a regular polygon = ½ ap
Use trigonometry to find apothem:
p = 5 * 8 = 40cm
Find measure of central angle: 360/5 = 72
m∠XCY = ½ m∠XCZ = 36.
XY = ½ XZ = 4
tan(36) = 4/a
a = 4/tan(36)
a = 5.5
A = ½ (5.5)(40) = 110.11055
Find area of regular octagon with perimeter of 80.
Elizabeth Pawelka
Geometric Probability
4/11/12
side = 80/8 = 10
angle = ½(360/8) = 22.5
tan(22.5) = 5/a
a = 5/tan(22.5)
a = 12.1
A = ½(12.1)(80) = 482.8 in2
Find area of these regular polygons
p.3
Elizabeth Pawelka
have a, need perimeter: 2x(6)
tan(30) = x/6
x = 6 tan(30)
x = 3.46
p = 2(3.46)*6 = 41.57
A = ½(6)(41.57) = 124.71 cm2
Geometric Probability
4/11/12
p.4
p = 10 * 12 = 120
need a: central angle = 360/10 = 36
36/2 = 18
tan(18) = 6/a
a = 6/tan(18) = 18.466
A = ½(18.466)(120) = 1107.966 = 1108 cm2
Elizabeth Pawelka
Geometric Probability
Area of Triangles
What if you only had m∠A and lengths b and c? Find h:
Area of Triangle = ½ b*h
h
c
h = c (sin A)
sin A =
Area = ½ bc(sin A)
Theorem 9-1: Area of a Triangle Given SAS
Area of ΔABC = ½ bc(sin A)
Find area of this triangle to the nearest hundredth
A = ½ 5*9*sin(34) = 12.58 in2
Find area of these triangles to nearest tenth:
4/11/12
p.5
Elizabeth Pawelka
A = ½ 7*7*sin(31) = 12.6 ft2
Geometric Probability
A = ½ (7*10*sin(44)) = 24.3 m2
4/11/12
p.6
A = ½ * 8*6*sin(61) = 20.99 =
21 in2
Extension: If area of a triangle = ½ b*h and area of a parallelogram = b*h, what is the area of a
parallelogram given SAS?
Area of a parallelogram = bc(sin A):
Find area of this parallelogram to the nearest hundredth
A = 10*8*sin(63) = 71.28 m2
(Show Heron’s formula – p. 353: Challenge)
Find area of these regular polygons to nearest tenth
Elizabeth Pawelka
Geometric Probability
4/11/12
p.7
A dodecagon with perimeter = 108
½(central angle) = ½(45) = 22.5
cos(22.5) = a/10
a = 10*cos(22.5)
sin(22.5) = ½ b/10
½ b = 10sin(22.5)
b = 20sin(22.5)
p = 8*20sin(22.5)
A = ½(16.8)108 = 906.9 cm
A = ½ (10*cos(22.5)) * (8*20sin(22.5)) = 282.8 m
Closure (5 mins)


Today you learned to use trigonometry to find area of polygons.
Tomorrow we’ll review for the Chapter Test on Friday.
Homework (H)
 p. 500, # 1 – 18, 20, 21, 23, 25, 27
Homework (R)
 p. 500, # 1 – 18, 20, 21, 25
Related documents