105 32 1 2 3 4 52.5 16 36.5 105 warm-up find the measures of angles 1 – 4
TRANSCRIPT
10532
1
23
4
52.5
16
36.5
105
Warm-up Find the measures of angles 1 – 4.
Homework Review
1) D 1) C
2) A 2) B
3) B 3) A
4) A 4) C
5) A 5) A
6) C
7) D
8) B
EOCT Practice
Question of the Day
GPS GeometryDay 44 (10-12-12)
UNIT QUESTION: What special properties are found with the parts of a circle?Standard: MM2G1, MM2G2
Today’s Question:How do we the area of a sector using proportions?Standard: MM2G3.b
1) Find the circumference and area of a circle with a diameter 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
64
A = r2
ANSWERS WILL BE IN SQUARE UNITS
6.8
Find the area.
If S has a circumference of 10 inches, find the area of the circle to the nearest hundredth.
C = 2r10 = 2r
5 = r
A = r2
A = 52A = 25A =
in2
Find the area of the shaded region.
188.49in2
8 in
2 in
A = 22A = 82
A = 4
A = 12.57in2
A = 64
A = 201.06in2
A shaded = A – A= 201.06 - 12.57 =
SECTOR: region bounded by two radii of the circle and their
intercepted arcR
O
Q
Area of a Sector
2
sec
360
Area of tor RQ mRQ
r
60°
120 °
6 cm
7 cm
Q
R
Q
R
218.85cm
251.31cm
2
60
(6) 360
Area
2
sec
360
Area of tor QR mQR
r
2
sec
360
Area of tor QR mQR
r
2
120
(7) 360
Area
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 h
R
Q290 1
(12) (12)*(12)360 2
113.10 72
241.10yd
