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
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.
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
Dis tance around the circle
2 π r = Cor
dπ = C
in. 6.28 π=C m 33πin.8.89≈C
Portion of the circumference
P
A
B3602
mAB
r
ABlength =π
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
A = π r2
ANSWERS WILL BE IN SQUARE UNITS
6.8
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
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 =
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=
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°=
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)
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
HomeworkMathematics Vol.2
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