Geometry - Semester 2
Mrs. Day-Blattner
3/14/2017 Area regular polygons continued.
Agenda 3/14/2017
1. Bulletin
2. Turn in page 67- 68. Make sure your name is on the
sheet.
3. From Area of a polygon to area of a circle, starting at page
71
4. Arcs and chords
5. homework
Regular polygons an all be inscribed in a circle.
What happens to the area of the regular polygon
compared to the area of the circle as the number of
sides increases?
For a regular polygon inscribed in a circle, as n, then
number of sides, gets very large what happens to
perimeter and apothem?
radius
From area of a regular polygon to area of a circle:
Area of any regular polygon with n sides.
AREAregular polygon= ½(ns)(apothem)
When n gets very large
ns approaches the circumference of the
circle, 2πr and the apothem approaches r,
the radius of the circle
From area of a regular polygon to area of a circle:
Eventually - at the limit when n is infinitely large
AREAregular polygon= ½(2πr)(r)
AREAregular polygon= πr2 = area of the circle
Approach #2 page 71
A circle can be decomposed into a set of thin, concentric rings as shown on the
left in the following diagram. If we unroll and stack those rings we can approximate
a triangle as shown in the figure on the right.
Approach #2 page 71
7. How might we describe the height of this “triangle” relative to the circle?
height?
Approach #2 page 71
8. How might we describe the length of the base of this “triangle” relative to the
circle?
Height
approx.
radius
base ?
Approach #2 page 71
9. As the rings get narrower and narrower, the triangular shape gets closer and
closer to an exact triangle with the same area as the circle. What does our
diagram suggest for the area of a circle?
Height
approx.
radius
base approx. circumference
Area triangle = ½ b h = ½ r. 2πr = πr2
For large numbers of rings the area of the triangle
approaches the area of the circle (informal limit argument).
Height
approx.
radius
base approx. circumference
Approach #3 (page 72)
Draw a circle on colored paper (with a radius at
least 10cm), then cut up your circle into equal
segments as small yet accurately as you can.
We have “decomposed the circle into a set of
congruent sectors.”
We have “decomposed the circle into a set of
congruent sectors.”
Now stick the sectors back together to
approximate a parallelogram as shown:
10. How might we describe the height of the
“parallelogram” relative to the circle?
Lesson 11. (7.5) From polygons to circles
Approach #3 (page 72)
Height
approx.
radius
10. How might we describe the height of the
“parallelogram” relative to the circle?
height approximates to the radius of the circle
if we cut very small segments
Lesson 11. (7.5) From polygons to circles
Approach #3 (page 72)
11. How might we describe the base of
the“parallelogram” relative to the circle?
Lesson 11. (7.5) From polygons to circles
Approach #3 (page 72)
base approx. ½ circumference
11. How might we describe the base of
the“parallelogram” relative to the circle?
base is half the circumference of the circle
Lesson 11. (7.5) From polygons to circles
Approach #3 (page 72)
12. As we decompose the circle into more and
more sectors the “parallelogram” shape gets
closer and closer to an exact parallelogram
with the same area as the circle. What would
the diagram suggest for the area of a circle?
Lesson 11. (7.5) From polygons to circles
Approach #3 (page 72)
12. area parallelogram = base x height
= ½ circumference x radius
Circumference = 2 r
Lesson 11. (7.5) From polygons to circles
Approach #3 (page 72)
12. area parallelogram = ½ ( 2 r ) r
=
r2
for large numbers of sectors the area of the parallelogram
approaches the area of the circle (informal limit
argument).
Finding the radius, area and circumference of a
Circle page 73
1. Radius = 1m
Area = πr2 =
Circumference = 2πr =
Finding the radius, area and circumference of a
Circle page 73
1. Radius = 1m
Area = π m2
Circumference = 2π m
Page 73 cont.
2. Radius =
Area = 9πft2
=
πr2
Circumference = 2πr
Page 73 cont.
2. Radius = 3ft
Area = 9πft2
(sq.rt of 9)
9= r2
Circumference = 6πft
(d = 2x3ft)
3. Radius = r
Area = πr2
Circumference = 8π yd = 2πr
(r = ½ 8yd)
Continue, finish this page for homework.
3. Radius = 4yd
Area = 16πyd2
Circumference = 8πyd
(r = ½ 8yd )
4. Radius = 1m
Area = 3.14 m2
(π = 3.14, r x r = 1)
Circumference = 6.28 m
(d = 2 x r )
5. Radius = 7 miles
Area = 49πmiles2
(72 = 49)
Circumference = 14π miles
(2x 7 =14)
6. Radius = 40.5 in
Area = 1640.25π in2
(40.52 = 1640.25)
Circumference = 81π in (2r = 81, r = 40.5)
7. Which of the polygons below would have an
area and perimeter closest to the circle it is
inscribed in? Why?
We can see that the last polygon which has the greatest
number of sides (8) has an area and perimeter that are
approaching the area and circumference of the circle itself.
This is the idea of an “informal limit.”
8. Find the perimeter and area of an equilateral
triangle with apothem of 7cm.
Perimeter = 3 x 14√3 cm = 42√3 cm
cm
Area = ½ (perimeter)(apothem)
14
cm
14
√3
= ½ (42√3 cm) (7cm)
7√3cm
7cm
7√3cm
= 147√3cm2
9. Find the perimeter and area of this regular
hexagon with sidelength 8in.
8in
8in
4in
4√3in
Perimeter = 6(8in) = 48in
Area = ½ (perimeter)(apothem)
= ½ (48in) (4√3in)
= 96√3in2
Arcs and Chords
B
Opening Exercise
Given circle A with line segment
BC perpendicular to line
segment DE
FA = 6
AC = 10
Find BF and DE. Explain your
work.
D
F
6
A
10
C
E
Arc and Chords
B
Find BF and DE. Explain your
work.
AB = AD = AE = 10 All radii
4
D
6
10
AF + BF = 10
6 + BF = 10
BF = 4
E
F
A
10
C
10
Arcs and Chords
Find BF and DE. Explain your
work.
By Pythagorean theorem
2
2
10 = 6 + FE
2
102 - 62 = FE2
100 - 36 = 64 = FE2
FE = sq. rt 64 = 8
FE = FD
therefore DE = 2(8) =16
B
4
D
6
10
E
F
A
10
C
10
Exercises
54°
1. Given circle A with mBC = 54°
and angle CDB is congruent to
angle DBE, find measure of
arc DE. Explain your work.
B
C
A
E
D
Lesson 12 cont. Exercises
54°
1. Central angle has same
measure as subtended arc.
Angle CAB = 54°
Inscribed angles are half the
central angle that intercepts the
same arc.
Angle CDB = 54°/2 = 27° and we
are told Angle DBE is congruent
to this angle.
B
C
27°
54°
A
27°
D
E
Lesson 12 cont. Exercises
Angle CDB = 54°/2 = 27°
B
C
Given angle CDB is congruent to
angle DBE,
mDE = 54°
because it must be double the
angle DBE (27°) which is an
inscribed angle intercepting arc
DE.
A
E
D
54°
2. If two arcs in a circle have the same measure, what
can you say about the quadrilateral formed by the four
endpoints?
...
A
A quadrilateral with one pair of opposite sides that are
parallel and the other pair congruent is an isoceles
trapezoid.
We always end up with one pair
of congruent sides
And one pair of parallel sides
(Isosceles trapezoid)
Could get a rectangle - try it
out...
A
3. Find the angle measure of CD and ED.
C
B
A
65°
D
E
3. Find the angle measure of CD and ED.
C
angleCBD = 65°
2 parallel lines cut by a
transversal, alternate
interior angles are
congruent
130°
B 65°
130°
A
65°
mCD = 2x65°= 130°
D
E
3. Find the angle measure of CD and ED.
BD is a diameter of the
circle so
C
50°
mCD = 2x65°= 130°
B 65°
130°
A
mBE + mED = 180°
130°
So mED = 50°
D
E
50°
Lesson Summary - Highlight and use to answer
questions that follow.
Theorems:
Congruent chords have congruent arcs
Congruent arcs have congruent chords
Arcs between parallel chords are congruent
Homework
Page 73 finish - questions 4 - 9 (see
slides above to check your work)
Page 78 - 79 questions 1 - 4 (see slides
below to check work)
Problem Set
1. a) mCE = 70°
Central angle is twice
the measure of an
inscribed angle
intercepting the same
arc
B
C
35°
70°
E
D
Problem Set
1. b) mBD = 70°
Arcs between parallel
chords are congruent
B
70°
C
35°
70°
E
D
Problem Set
1. c) mED =
180° - 140° = 40°
If F is the center of
the circle then
measure of arcCEDB
must be 180°
B
70°
F
C
35°
70°
E
40°
D
Problem Set
2. In circle A, BC is a
diameter, mCE = mED
and measure of angle
CAE = 32°
Congruent chords must
have congruent central
angles intercepting them
D
E
C
B
32°
32°
A
70°
Problem Set
2. a) Find measure of
angle CAD
= 32° + 32°
= 64°
D
E
C
B
32°
32°
A
70°
Problem Set
2. b) Find measure of angle
D
ADC = 58°
DA=CA both radii
E
Angle ADC is one of the C
base angles of an isosceles
triangle = ½ (180° - 64°) =
58°
B
64°
A
Problem Set
3. In circle A, BC is a
diameter, 2mCE = mED
B
and BC is parallel to DE.
A
C
D
E
Problem Set
3. Arcs between parallel
chords are congruent, so
B
mBD =mCE
mED = 2mCE, given
2mCE +mCE+mCE= 180°
4mCE = 180°
A
C
D
E
Problem Set
4mCE = 180°
mCE = 180° divide by 4
B
A
mCE = 45°
C
45°
E
D
Problem Set
mCE = 45°
Measure of the inscribed
angle CDE would be half
this arc
= ½ (45°) = 22.5°
B
A
C
45°
E
22.5°
D
Problem Set
4. A tangent line has been drawn on the
diagram that is parallel to chord DE, and the
diameter BC has to be perpendicular to this
tangent line, so we know the diameter is the
perpendicular bisector of chord DE. If diameter
bisects the chord, it will also bisect the arc
DCE, so we know arcs DC and EC are
congruent.
Problem Set
4. a) Given mCE = 68° and the fact that CE
and DE are congruent
mCD = 68°
b) measure of angle DBE = 68°
(angle DBC is an inscribed angle intercepting
arc of measure 68°, so it is 34° and so is angle
CBE, so the total measure of angle DBE is 68°)
Problem Set
4. c) Find m∠DCE =
Inscribed angle DBE is 68° and so the central
angle intercepting that same arc is 2(68°)=
136°
For a circle the sum of the major arc and the
minor arc have to be 360°, so the arc DBE
must be 360° - 136° = 224°
Problem Set
4. c) Find m∠DCE =
arc DBE must be 360° - 136° = 224°
The central angle intercepting this arc will be
twice the inscribed angle we are looking for.
Hence m∠DCE = ½ (224°) = 112°