Chapter Form 2

1. Warm-up 1) Find the circumference and area of a circle with a diamater of 10 in. 2) What would be the radius of a circle if it’s area is ? C = 31.42 in., A = 78..54 in. 2 radius = 8

2. Homework Review 2) 2 4) AB=16, DE=17 6) AB=32, DE=24 8) 6 10) RT=20, TV=16 12) RT=36, TV=27 14) 12 16) 18 28) 14.2 ft.

3. Math II Day 34 (9-29-09) UNIT QUESTION: What special properties are found with the parts of a circle? Standard: MM2G1, MM2G2 Today’s Question: How do we find arc length and area of sectors using proportions? Standard: MM2G3.c

4. 6.7 Circumference and Arc Length

5. Circumference Distance around the circle

6. Circumference 2  r = C or d  = C

7. 14.3 in. 33 m 103.7 m

8. Arc Length Portion of the circumference P A B

9. Find the arc length Find the circumference 3.82 m 60 º 50º 5cm 4.36 cm 22.92 m

10. Practice! Wkbk 6.7

11. Area of a circle A =  r 2 “Apple Pie are Square” ANSWERS WILL BE IN SQUARE UNITS 6.8

12. 7.6 cm 181.5 square cm 78.54 square in 10 in

13. If  S has a circumference of 10  inches, find the area of the circle to the nearest hundredth. C = 2  r 10  = 2  r 5 = r A =  r 2 A =  5 2 A =  25 A =  in 2

14. Find the area of the shaded region. 188.49 in 2 6 in 2 in A =  2 2 A =  8 2 A =  4 A = 12.57  in 2 A =  64 A = 201.06  in 2 A shaded = A – A= 201.06 - 12.57 =

15. SECTOR: region bounded by two radii of the circle and their intercepted arc R O Q Area of a Sector

16. 60° 120 ° 6 cm 7 cm Q R Q R

17. A SEGMENT is a region bounded by a chord and its intercepted arc A segment is a minor segment if the intercepted arc is less than 180 degrees Area of minor segment = (Area of sector) – (Area of triangle)

18. Area of minor segment = (Area of sector) – (Area of triangle) 12 yd R Q