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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