jim smith jchs arc length, sectors, sections spi.3.2.bspi.3.2.l

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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

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Page 1: Jim Smith JCHS Arc Length, Sectors, Sections spi.3.2.Bspi.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

Page 2: Jim Smith JCHS Arc Length, Sectors, Sections spi.3.2.Bspi.3.2.L

77

C = 2C = 2ππrrC = 2C = 2ππ77C = 14C = 14ππ

Page 3: Jim Smith JCHS Arc Length, Sectors, Sections spi.3.2.Bspi.3.2.L

99

A = A = ππr²r²A = A = ππ9²9²A = A = 8181ππ

Page 4: Jim Smith JCHS Arc Length, Sectors, Sections spi.3.2.Bspi.3.2.L

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.

Page 5: Jim Smith JCHS Arc Length, Sectors, Sections spi.3.2.Bspi.3.2.L

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

Page 6: Jim Smith JCHS Arc Length, Sectors, Sections spi.3.2.Bspi.3.2.L

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

Page 7: Jim Smith JCHS Arc Length, Sectors, Sections spi.3.2.Bspi.3.2.L

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

Page 8: Jim Smith JCHS Arc Length, Sectors, Sections spi.3.2.Bspi.3.2.L

Sectors are a fractional Sectors are a fractional part of a circle’s areapart of a circle’s area

Page 9: Jim Smith JCHS Arc Length, Sectors, Sections spi.3.2.Bspi.3.2.L

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

Page 10: Jim Smith JCHS Arc Length, Sectors, Sections spi.3.2.Bspi.3.2.L

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

Page 11: Jim Smith JCHS Arc Length, Sectors, Sections spi.3.2.Bspi.3.2.L

SectionSectionssLet’s talk Let’s talk

pizzapizza

Page 12: Jim Smith JCHS Arc Length, Sectors, Sections spi.3.2.Bspi.3.2.L

AREA OF SECTIONAREA OF SECTION = = AREA OF SECTOR – AREA OF SECTOR – AREA OF AREA OF TRIANGLETRIANGLE

¼ ¼ ππ r² - r² - ½ bh½ bh

Page 13: Jim Smith JCHS Arc Length, Sectors, Sections spi.3.2.Bspi.3.2.L

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