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Hello. Properties of Circles. General Information. Content. EDD 5161E Educational Communication and Technology. Group Project Presentation Package. Next page. Main Menu. Instructors:. Dr. LEE Fong Lok. Mr. TAM Tat Sang. Students :. Shek Ting ( 98041130).
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Hello Properties of Circles General Information Content
EDD 5161E Educational Communication and Technology Group Project Presentation Package Next page Main Menu
Instructors: Dr. LEE Fong Lok Mr. TAM Tat Sang Students : Shek Ting ( 98041130) Cheung Chun Pong ( 98040140 ) Chow Ming Hing ( 98040630 ) Next page Previous page Main Menu
Target Audience: Form 4 students (band 1) Type of Software: Lecturing in one lesson Previous page Main Menu
Properties of Circle Content Review Angle in semi-circle Angle in the same segment Exercise Main Menu
a O b Review What is the relationship between the a and b? Hint : O is the centre. b = 2a at centre twice at circumference Content
C A B O Angle in Semi-circle Given: In a circle , O is centre. diameter AB is AOB = 180o 90o ACB = Next page Content
90 90 75 75 105 105 60 60 120 120 C 45 45 135 135 C’ Protractor Protractor 30 30 150 150 15 15 165 165 Made in China Made in China 0 0 180 180 B A O ACB = AC ’B=90o Next page Previous page Content
C 30 A B Example 1 AB is a diameter. Find CAB. Solutions: BCA = 90 ( in semi-circle ) CAB + ABC + BCA = 180 ( s sum of ) CAB + 30 + 90 = 180 CAB = 60 Next page Previous page Content
C 6 8 A B Example 2 Given that AC = 6 and BC = 8. Find the radius of the circle. Solutions: Since BCA = 90, AB is a diameter. ( the converse of in semi-circle ) AB2 = AC2 + BC2 = 62 + 82 AB = 10 radius = 10 2 = 5 Content Previous page
C C’ B A Angles in same Segment Given: AB is not a diameter, just a chord. IsACB = AC’B ? Next page Content
C C’ C’ C’ C’ C’ C’ C’ C’ C’ C’ C’ C’ B A We try to rotate the DABC’ Next page Previous page Content
C C’ O B A Proof O is the centre of circle AOB = 2 ACB AOB = 2 AC’B (at centre twice at circumference) ACB= AC’B Next page Previous page Content
Q R x 95 T 40 P S Example 3 Find x. Solutions: PQT = 180QTPTPQ ( sum of ) = 1809540 = 45 Besides,x = PQT ( in same segment ) = 45 Next page Previous page Content
Q R 45 T 55 y P S Example 4 Given that TP = TS. Find y. Solutions: QPR = QSR = 55. ( in same segment ) Since TP = TS, TPS = TSP = y. TPS + TSP + QPT + SQP = 180 ( sum of ) y + y + 55 + 45 = 180 2y = 80 Content y = 40 Previous page
Exercises Work Harder Question 1 Ans 1 Ans 2 Question 2 Ans 3 Question 3 Ans 4 Question 4 Ans 5 Question 5 Content
Exercise One Find x . 90 35 55 45 x 65 35
Exercise Two AB=8 and Radius=5. Find x . B 6 8 A 4 x 5 8 C 5 10
Exercise Three Find x . 40 x 100 55 30 50 30 70
Exercise Four Find x . 60 60 50 x 45 140 65 100
Exercise Five Find x . 90 55 40 45 x 50 40
Exercise One Find x . 90 35 55 45 x 65 35 Exercises
Exercise Two AB=8 and Radius=5. Find x . B 6 8 A 4 x 5 8 C 5 10 Exercises
Exercise Three Find x . 40 x 100 55 30 50 30 70 Exercises
Exercise Four Find x . 60 60 50 x 45 140 65 100 Exercises
Exercise Five Find x . 90 55 40 45 x 50 40 Exercises