area circle§or
DESCRIPTION
ChapterForm 2TRANSCRIPT
![Page 1: Area circle§or](https://reader033.vdocument.in/reader033/viewer/2022052413/559c264f1a28ab482c8b4633/html5/thumbnails/1.jpg)
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
64π
![Page 2: Area circle§or](https://reader033.vdocument.in/reader033/viewer/2022052413/559c264f1a28ab482c8b4633/html5/thumbnails/2.jpg)
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.
![Page 3: Area circle§or](https://reader033.vdocument.in/reader033/viewer/2022052413/559c264f1a28ab482c8b4633/html5/thumbnails/3.jpg)
Math IIDay 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
![Page 4: Area circle§or](https://reader033.vdocument.in/reader033/viewer/2022052413/559c264f1a28ab482c8b4633/html5/thumbnails/4.jpg)
![Page 5: Area circle§or](https://reader033.vdocument.in/reader033/viewer/2022052413/559c264f1a28ab482c8b4633/html5/thumbnails/5.jpg)
Dis tance around the circle
![Page 6: Area circle§or](https://reader033.vdocument.in/reader033/viewer/2022052413/559c264f1a28ab482c8b4633/html5/thumbnails/6.jpg)
2 π r = Cor
dπ = C
![Page 7: Area circle§or](https://reader033.vdocument.in/reader033/viewer/2022052413/559c264f1a28ab482c8b4633/html5/thumbnails/7.jpg)
in. 6.28 π=C m 33πin.8.89≈C
![Page 8: Area circle§or](https://reader033.vdocument.in/reader033/viewer/2022052413/559c264f1a28ab482c8b4633/html5/thumbnails/8.jpg)
Portion of the circumference
P
A
B3602
mAB
r
ABlength =π
![Page 9: Area circle§or](https://reader033.vdocument.in/reader033/viewer/2022052413/559c264f1a28ab482c8b4633/html5/thumbnails/9.jpg)
3.82 m60º50º
5cm
3602
mAB
r
ABlength =π
360
50
52=
⋅πlength
π38.1
360
60
2
82.3 =⋅ rπ
647.3≈r647.32 ⋅= πC
![Page 10: Area circle§or](https://reader033.vdocument.in/reader033/viewer/2022052413/559c264f1a28ab482c8b4633/html5/thumbnails/10.jpg)
![Page 11: Area circle§or](https://reader033.vdocument.in/reader033/viewer/2022052413/559c264f1a28ab482c8b4633/html5/thumbnails/11.jpg)
A = π r2
ANSWERS WILL BE IN SQUARE UNITS
6.8
![Page 12: Area circle§or](https://reader033.vdocument.in/reader033/viewer/2022052413/559c264f1a28ab482c8b4633/html5/thumbnails/12.jpg)
![Page 13: Area circle§or](https://reader033.vdocument.in/reader033/viewer/2022052413/559c264f1a28ab482c8b4633/html5/thumbnails/13.jpg)
If S has a circumference of 10π inches, find the area of the circle to the nearest hundredth.
C = 2πr10π = 2πr
5 = r
A = πr2
A = π 52
A = π 25A = 78.54 in2
![Page 14: Area circle§or](https://reader033.vdocument.in/reader033/viewer/2022052413/559c264f1a28ab482c8b4633/html5/thumbnails/14.jpg)
Find the area of the shaded region.
188.49in2
6 in
2 in
A = π22A = π82
A = π4
A = 12.57 in2
A = π64
A = 201.06 in2
A shaded = A – A= 201.06 - 12.57 =
![Page 15: Area circle§or](https://reader033.vdocument.in/reader033/viewer/2022052413/559c264f1a28ab482c8b4633/html5/thumbnails/15.jpg)
S E C T O R : region bounded by two radii of the circle and their intercepted
arcR
O
Q
A rea of a Sector
2
sec
360
Area of tor RQ mRQ
rπ=
![Page 16: Area circle§or](https://reader033.vdocument.in/reader033/viewer/2022052413/559c264f1a28ab482c8b4633/html5/thumbnails/16.jpg)
60°
120 °
6 cm
7 cm
Q
R
Q
R
218.85cm≈
251.31cm≈
2
60
(6) 360
Area
π°=
2
sec
360
Area of tor QR mQR
rπ=
2
sec
360
Area of tor QR mQR
rπ=
2
120
(7) 360
Area
π°=
![Page 17: Area circle§or](https://reader033.vdocument.in/reader033/viewer/2022052413/559c264f1a28ab482c8b4633/html5/thumbnails/17.jpg)
A S E G M E N T is a r e g io n b o u n d e d b y a c h o r d a n d
i t s in t e r c e p t e d a r c
A segment is a m in o r s e g m e n t i f t h e
in t e r c e p t e d a r c i s le s s t h a n 18 0
d e g r e e s
Area of minor segment =
(Area of sector) – (Area of triangle)
![Page 18: Area circle§or](https://reader033.vdocument.in/reader033/viewer/2022052413/559c264f1a28ab482c8b4633/html5/thumbnails/18.jpg)
Area of minor segment =
(Area of sector) – (Area of triangle)
12 yd
»2 1
*360 2
mRQr b hπ
−
R
Q290 1
(12) (12)*(12)360 2
π −
113.10 72−
241.10yd
![Page 19: Area circle§or](https://reader033.vdocument.in/reader033/viewer/2022052413/559c264f1a28ab482c8b4633/html5/thumbnails/19.jpg)
HomeworkMathematics Vol.2
Book page No. Book page No.