distance around the circle
TRANSCRIPT
Distance around the circle
2 p r = Cor
dp = C
Find the circumference.
in. 6.28 pC m 33pin.8.89C
Portion of the circumference
P
A
B3602
mAB
r
ABlength
p
60º50º
3602
mAB
r
ABlength
p
360
50
52
p
length
p38.1
360
60
2
82.3
rp
647.3r
647.32 pC
A = p r2
ANSWERS WILL BE IN SQUARE UNITS
6.8
Find the area.
If S has a circumference of 10p inches, find the area of the circle to the nearest hundredth.
C = 2pr
10p = 2pr
5 = r
A = pr2
A = p 52
A = p 25
A = 78.54 in2
Find the area of the shaded region.
188.49in2
6 in
2 in
A = p22A = p82
A = p4
A = 12.57 in2
A = p64
A = 201.06 in2
A shaded = A –A= 201.06 - 12.57 =
SECTOR: region bounded by two radii of the circle and their intercepted arc
R
O
Q
Area of a Sector
2
sec
360
Area of tor RQ mRQ
rp
60°
120 °Q
R
Q
R
218.85cm
251.31cm
2
60
(6) 360
Area
p
2
sec
360
Area of tor QR mQR
rp
2
sec
360
Area of tor QR mQR
rp
2
120
(7) 360
Area
p
A SEGMENT is a region bounded by a chord and its intercepted arc
A segment is a minor segment if the intercepted arc is less than
180 degrees
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 hp
R
Q290 1
(12) (12)*(12)360 2
p
113.10 72
241.10yd