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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
Use r = 8 in the area formula.
Using the Area of a Circle
P
8 in.
SOLUTION
A = r 2
= • 8 2
= 64
201.06
So, the area is 64, or about 201.06, square inches.
Find the area of P..
Using the Area of a Circle
Find the diameter of Z.•
ZSOLUTION
The diameter of the circle is about 2(5.53), or about 11.06, centimeters.
Area of Z = 96 cm2•
Find the square roots.
The diameter is twice the radius.
A = r 2
96 = r 2
30.56 r 2
= r 296
5.53 r
P
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
Br
In the diagram, sector APB is bounded by AP, BP, and AB.
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°.
Using the Area of a Circle
The following theorem gives a method for finding the area of a sector.
A
r 2
= , or A = • r 2mAB
360°
mAB
360°
A
A
B
P
Finding the Area of a Sector
Find the area of the sector shown at the right.
P
C
D
4 ft80°SOLUTION
Sector CPD intercepts an arc whose measure is 80°. The radius is 4 feet.
m CD
360°A = • r
2
80°
360°= • • 4
2
11.17
So, the area of the sector is about 11.17 square feet.
Use a calculator.
Substitute known values.
Write the formula for the area of a sector.
USING AREAS OF CIRCLES AND REGIONS
Finding the Area of a Region
Find the area of a the shaded region shown.
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 shaded region = Area of
circleArea of hexagon–
SOLUTION
5 m
USING AREAS OF CIRCLES AND REGIONS
Finding the Area of a Region
Area of shaded region = Area of
circleArea of hexagon–
= • 5 2 – 1
252
3• • (6 • 5)The apothem of a hexagon is
• side length •12
3
= r 2 –
12
a P
or about 13.59 square meters.
So, the area of the shaded region is 25 – 752
3 ,
5 m
= 25 – 3752
Finding the Area of a Region
Complicated shapes may involve a number of regions.
Notice that the area of a portion of the ring is the difference of the areas of two sectors.
P P
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?
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 of rectangleArea = + Area of sector –
Finding the Area of a Region
Area Area of circleArea of rectangle= + Area of sector –
6 • 112
= + –180°360°
• • 32 • 12 • 4
2
= 33 + • • 9 – • (2)212
= 33 + – 492
34.57
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?
The area of the front of the case is about 34.57 square inches.
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.
inscribedangle
interceptedarc
THEOREM
USING INSCRIBED ANGLES
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.
m ADB = m AB12
mAB = 2m ADB
C
A
BD
Find the measure of the blue arc or angle.
Finding Measures of Arcs and Inscribed Angles
SOLUTION
N
P
M100°C
W
XY
Z
115°
C
RS
T Q
C
mQTS = 2m QRS = 2(90°) = 180°
mZWX = 2m ZYX = 2(115°) = 230°
M NMP = mNP = (100°) = 50°12
12
THEOREM
A
D
CB
USING INSCRIBED ANGLES
THEOREM 10.9
If two inscribed angles of a circle intercept the same arc, then the angles are congruent.
C D
Using the Measure of an Inscribed Angle
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?
THEATER DESIGN 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.
movie screenE G
F
SOLUTION
Using the Measure of an Inscribed Angle
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 POLYGONSTHEOREM 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.
C
A
B
B is a right angle if and only if AC is a diameter of the circle.
USING PROPERTIES OF INSCRIBED PLOYGONS
THEOREMS ABOUT INSCRIBED POLYGONSTHEOREM 10.11
A quadrilateral can be inscribed in a circle if and only if its opposite angles are supplementary.
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°.
.
C
E
D
F
G
Using an Inscribed Quadrilateral
In the diagram, ABCD is inscribed in P. Find the measure of each angle.
.P
B
C
D
A
2y °
3y °
5x °
3x °
PB
C
D
A
2y °
3y °
5x °
3x °
Using an Inscribed Quadrilateral
SOLUTION
ABCD is inscribed in a circle, so opposite angles are supplementary.
3x + 3y = 180 5x + 2y = 180
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.
y = 60 – 20 = 40 Substitute and solve for y.
5x + 2y = 180 Write second equation.
5x + 2(60 – x) = 180 Substitute 60 – x for y.
5x + 120 – 2x = 180 Distributive property
3x = 60 Subtract 120 from each side.
x = 20 Divide each side by 3.
PB
C
D
A
2y °
3y °
5x °
3x °
Using an Inscribed Quadrilateral
PB
C
D
A
2y °
3y °
5x °
3x °
x = 20 and y = 40,
so m A = 80°, m B = 60°,
m C = 100°, and m D = 120°.
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