Transcript Circles - TeacherWeb

AREAS OF CIRCLES AND SECTORS
These regular polygons, inscribed in circles with radius r,
demonstrate that as the number of sides increases, the area of the
polygon approaches the value  r 2.
3-gon
4-gon
5-gon
6-gon
AREAS OF CIRCLES AND SECTORS
THEOREM
THEOREM 11.7 Area of a Circle
The area of a circle is  times the
square of the radius, or A =  r 2
r
Using the Area of a Circle
.
Find the area of P.
SOLUTION
8 in.
P
Use r = 8 in the area formula.
A = r2
=  • 82
= 64 
 201.06
So, the area is 64, or about 201.06, square inches.
Using the Area of a Circle
Find the diameter of
•
Z.
Z
SOLUTION
Area of
•
Z = 96 cm2
The diameter is twice the radius.
A = r2
96 =  r 2
96 = r 2

30.56  r 2
5.53  r
Find the square roots.
The diameter of the circle is about 2(5.53), or about 11.06, centimeters.
Using the Area of a Circle
The sector of a circle is the region bounded by
two radii of the circle and their intercepted arc.
A
P
r
B
In the diagram, sector APB is bounded by AP, BP, and AB.
Using the Area of a Circle
The following theorem gives a method for finding the area of a sector.
THEOREM
THEOREM 11.8
Area of a Sector
The ratio of the area A of a sector of
a circle to the area of the circle is
equal to the ratio of the measure of
the intercepted arc to 360°.
A
r2
=
mAB
360°
, or A =
mAB
360°
• r 2
A
P
A
B
Finding the Area of a Sector
Find the area of the sector shown at the right.
C
4 ft
80°
SOLUTION
P
Sector CPD intercepts an arc whose
measure is 80°. The radius is 4 feet.
A =
=
m CD
360°
• r2
80°
• • 42
360°
 11.17
Write the formula for the
area of a sector.
Substitute known values.
Use a calculator.
So, the area of the sector is about 11.17 square feet.
D
USING AREAS OF CIRCLES AND REGIONS
Finding the Area of a Region
Find the area of a the shaded region shown.
5m
SOLUTION
The diagram shows a regular hexagon inscribed in a circle
with radius 5 meters. The shaded region is the part of the
circle that is outside of the hexagon.
Area of
Area of
=
shaded region
circle
–
Area of
hexagon
USING AREAS OF CIRCLES AND REGIONS
Finding the Area of a Region
Area of
Area of
=
shaded region
circle
=
= • 5
2
= 25 –
–
r2
5
2
1 •
2
75
2
3
–
Area of
hexagon
–
1
aP
2
•
(6 • 5)
5m
The apothem of a hexagon is
1 • side length •
2
3
So, the area of the shaded region is 25  – 75
2
or about 13.59 square meters.
3,
3
Finding the Area of a Region
Complicated shapes may involve a number of regions.
P
P
Notice that the area of a portion of the ring is the
difference of the areas of two sectors.
Finding the Area of a Region
WOODWORKING You are cutting the front face of a clock out of wood,
as shown in the diagram. What is the area of the front of the case?
SOLUTION
The front of the case is formed by a rectangle
and a sector, with a circle removed. Note that
the intercepted arc of the sector is a semicircle.
Area = Area of rectangle + Area of sector – Area of circle
Finding the Area of a Region
WOODWORKING You are cutting the front face of a clock out of wood,
as shown in the diagram. What is the area of the front of the case?
Area = Area of rectangle + Area of sector – Area of circle
=
6
•
11
2
180°
+
360°
=
33 + 1
=
33 + 9  – 4
2
• •
9 – 
•
••
3
2
– •
1
2
•
4
(2)2
2
 34.57
The area of the front of the case is about 34.57 square inches.
2
USING INSCRIBED ANGLES
An inscribed angle is an angle whose vertex is on a circle and
whose sides contain chords of the circle. The arc that lies in the
interior of an inscribed angle and has endpoints on the angle is
called the intercepted arc of the angle.
inscribed
angle
intercepted
arc
USING INSCRIBED ANGLES
THEOREM
THEOREM 10.8 Measure of an Inscribed Angle
If an angle is inscribed in a circle, then its measure is
half the measure of its intercepted arc.
A
m
ADB =
1
m AB
2
C
D
mAB = 2m
ADB
B
Finding Measures of Arcs and Inscribed Angles
Find the measure of the blue arc or angle.
W
R
S
C
C
Z
M
115°
T
N
Q
100°
C
X
Y
SOLUTION
mQTS = 2m QRS = 2(90°) = 180°
mZWX = 2m ZYX = 2(115°) = 230°
1
1
M NMP = mNP = (100°) = 50°
2
2
P
USING INSCRIBED ANGLES
THEOREM
A
THEOREM 10.9
If two inscribed angles of a circle
intercept the same arc, then the
angles are congruent.
B
C
D
C
D
Using the Measure of an Inscribed Angle
When you go to the movies, you want to be close to the
movie screen, but you don’t want to have to move your eyes too much to
see the edges of the picture. If E and G are the ends of the screen and you
are at F, m EFG is called your viewing angle.
THEATER DESIGN
E
movie screen
You decide that the middle of the sixth
row has the best viewing angle. If
someone is sitting there, where else can
you sit to have the same viewing angle?
F
G
Using the Measure of an Inscribed Angle
SOLUTION
Draw the circle that is determined by the
endpoints of the screen and the sixth row
center seat. Any other location on the
circle will have the same viewing angle.
USING PROPERTIES OF INSCRIBED POLYGONS
If all of the vertices of a polygon lie on a circle, the polygon
is inscribed in the circle and the circle is circumscribed
about the polygon. The polygon is an inscribed polygon
and the circle is a circumscribed circle.
USING PROPERTIES OF INSCRIBED POLYGONS
THEOREMS ABOUT INSCRIBED POLYGONS
THEOREM 10.10
If a right triangle is inscribed in a circle,
then the hypotenuse is a diameter of
the circle. Conversely, if one side of
an inscribed triangle is a diameter of the
circle, then the triangle is a right triangle
and the angle opposite the diameter is
the right angle.
B is a right angle if and only if
AC is a diameter of the circle.
A
B
C
USING PROPERTIES OF INSCRIBED PLOYGONS
THEOREMS ABOUT INSCRIBED POLYGONS
THEOREM 10.11
F
E
A quadrilateral can be inscribed in
a circle if and only if its opposite
angles are supplementary.
C
D
D, E, F, and G lie on some circle, . C, if and only if
m D + m F = 180° and m E + m G = 180°.
G
Using an Inscribed Quadrilateral
A
In the diagram, ABCD is inscribed in . P.
Find the measure of each angle.
2y °
D 3y °
P
3x °
5x °
C
B
Using an Inscribed Quadrilateral
SOLUTION
A
2y °
ABCD is inscribed in a circle, so opposite
angles are supplementary.
3x + 3y = 180
5x + 2y = 180
D 3y °
P
3x °
5x °
C
B
Using an Inscribed Quadrilateral
3x + 3y = 180
5x + 2y = 180
To solve this system of linear equations,
you can solve the first equation for y to get
y = 60 – x. Substitute this expression into
the second equation.
A
2y °
D 3y °
Write second equation.
5x + 2(60 – x) = 180
Substitute 60 – x for y.
5x + 120 – 2x = 180
Distributive property
x = 20
y = 60 – 20 = 40
3x °
5x °
C
5x + 2y = 180
3x = 60
P
Subtract 120 from each side.
Divide each side by 3.
Substitute and solve for y.
B
Using an Inscribed Quadrilateral
A
x = 20 and y = 40,
so m
m
A = 80°, m
2y °
B = 60°,
C = 100°, and m
D = 120°.
D 3y °
P
3x °
5x °
C
B