Jim Smith JCHS Jim Smith JCHS Arc Length, Sectors, Sections Arc Length, Sectors, Sections spi.3.2.B spi.3.2.B spi.3.2.L spi.3.2.L
Jim Smith JCHSJim Smith JCHS
Arc Length, Sectors, SectionsArc Length, Sectors, Sections
spi.3.2.Bspi.3.2.Bspi.3.2.Lspi.3.2.L
77
C = 2C = 2ππrrC = 2C = 2ππ77C = 14C = 14ππ
99
A = A = ππr²r²A = A = ππ9²9²A = A = 8181ππ
AA
BB
The length of AB represents a The length of AB represents a fractional part of the circle’s fractional part of the circle’s
circumference. If the mAB Is 90°, circumference. If the mAB Is 90°, then the length of AB is 90/360then the length of AB is 90/360thth
(1/4(1/4thth) of the circumference.) of the circumference.
AA
BB66
Find the length of ABFind the length of AB
C = 2 C = 2 ππ r rC = 2 C = 2 ππ 6 6C = 12C = 12ππ
Length of AB = Length of AB = ¼¼·12·12ππ = =
33ππ
90/36090/360 = ¼ = ¼
AA
BB
60°60°
88
C = 2C = 2ππrrC = 16C = 16ππ
AB = 60/360 of AB = 60/360 of 1616ππAB = 1/6 · 16AB = 1/6 · 16ππAB = 1 · 16AB = 1 · 16ππ 6 16 1AB = 8AB = 8ππ 33
XX YY
ZZ
120°120°
Find the length of Find the length of XYZ XYZ
99
XYZ = 240 of XYZ = 240 of 1818ππ 360360XYZ = 2 · 18XYZ = 2 · 18ππ 3 13 1XYZ = 12XYZ = 12ππ
360°360°- 120° - 120° 240°240°
C = 2C = 2ππr C = 18r C = 18ππ
Sectors are a fractional Sectors are a fractional part of a circle’s areapart of a circle’s area
Find the shaded Find the shaded areaarea
88 A = A = ππr² A = 64r² A = 64ππ
Sector area Sector area ==¼ of 64¼ of 64ππ
6464π π = 16= 16ππ 44
90 of circle’s 90 of circle’s areaarea360360
60°60°
60 of circle’s area60 of circle’s area360360
1212Area = Area = ππr² A = 144r² A = 144ππ
Sector area =Sector area =
A = 1 of 144A = 1 of 144ππ 66144144ππ = 24 = 24ππ 66
SectionSectionssLet’s talk Let’s talk
pizzapizza
AREA OF SECTIONAREA OF SECTION = = AREA OF SECTOR – AREA OF SECTOR – AREA OF AREA OF TRIANGLETRIANGLE
¼ ¼ ππ r² - r² - ½ bh½ bh
Area of sectionArea of section = = area of sector – area of sector – area of area of triangletriangle ¼ ¼ ππ r² - r² - ½ bh½ bh
1010A OF = ½∙10∙10=A OF = ½∙10∙10= 5050
A OF SECTION = A OF SECTION =
2525ππ - 50 - 50A of circle = A of circle = 100100ππ
A OF = ¼ 100A OF = ¼ 100ππ == 2525ππ