Download Unit 3. Circles and spheres

Accelerated Math I
Unit 3
1
Unit 3. Circles and spheres
Primary Standards Addressed:
MA1G4. Students will understand the properties of circles.
a. Understand and use properties of chords, tangents, and secants as an application
of triangle similarity.
b. Understand and use properties of central, inscribed, and related angles.
c. Use the properties of circles to solve problems involving the length of an arc and
the area of a sector.
d. Justify measurements and relationships in circles using geometric and algebraic
properties.
MA1G5. Students will find and compare the measures of spheres.
a. Use and apply surface area and volume of a sphere.
b. Determine the effect on surface area and volume of changing the radius or
diameter of a sphere.
Other Standards Addressed:
From the Unit 2,
MA1G1. Students will investigate properties of geometric figures in the
coordinate plane.
a.
b.
c.
d.
e.
Determine the distance between two points.
Determine the distance between a point and a line.
Determine the midpoint of a segment.
Understand the distance formula as an application of the Pythagorean theorem.
Use the coordinate plane to investigate properties of and verify conjectures
related to triangles and quadrilaterals.
MA1G2. Students will understand and use the language of
mathematical argument and justification.
a. Use conjecture, inductive reasoning, deductive reasoning, counterexamples, and
indirect proof as appropriate.
b. Understand and use the relationships among a statement and its converse,
inverse, and contrapositive.
MA1G3. Students will discover, prove, and apply properties of
triangles, quadrilaterals, and other polygons.
a. Determine the sum of interior and exterior angles in a polygon.
b. Understand and use the triangle inequality, the side-angle inequality, and the
exterior-angle inequality.
c. Understand and use congruence postulates and theorems for triangles (SSS,
SAS, ASA, AAS, HL).
All the process standards:
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MA1P1. Students will solve problems (using appropriate technology).
a.
b.
c.
d.
Build new mathematical knowledge through problem solving.
Solve problems that arise in mathematics and in other contexts.
Apply and adapt a variety of appropriate strategies to solve problems.
Monitor and reflect on the process of mathematical problem solving.
MA1P2. Students will reason and evaluate mathematical arguments.
a.
b.
c.
d.
Recognize reasoning and proof as fundamental aspects of mathematics.
Make and investigate mathematical conjectures.
Develop and evaluate mathematical arguments and proofs.
Select and use various types of reasoning and methods of proof.
MA1P3. Students will communicate mathematically.
a. Organize and consolidate their mathematical thinking through communication.
b. Communicate their mathematical thinking coherently and clearly to peers,
teachers, and others.
c. Analyze and evaluate the mathematical thinking and strategies of others.
d. Use the language of mathematics to express mathematical ideas precisely.
MA1P4. Students will make connections among mathematical ideas
and to other disciplines.
1. Recognize and use connections among mathematical ideas.
2. Understand how mathematical ideas interconnect and build on one another to
produce a coherent whole.
3. Recognize and apply mathematics in contexts outside of mathematics.
MA1P5. Students will represent mathematics in multiple ways.
a. Create and use representations to organize, record, and communicate
mathematical ideas.
b. Select, apply, and translate among mathematical representations to solve
problems.
c. Use representations to model and interpret physical, social, and mathematical
phenomena.
This material for unit 3 starts with studying terms with their precise definitions and
simplifying a real context mathematically, based on students’ knowledge from middle
school. Students’ tasks are focused on investigating properties and relationships among
circles, lines, and angles formed by circles and lines. By completing tasks in this unit,
students prove theorems and properties about circles, lines, and angles formed by circles
and lines. However, this material does not include enough problems to determine what
the students have learned because it focuses on students’ core tasks during classes but not
on practice tasks that might be given as homework. Thus, more application problems are
required to deepen conceptual understanding, encourage retention, and test what they
learned in this unit. The followings are specific terms with definitions, theorems and
properties that students will study.
Terms and definitions:
A circle is the set of all points in a plane that are equidistant (the radius) from a given
point (the center).
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A chord is a segment that joins two points on a circle.
A diameter is a chord that passes through the center of a circle.
An arc is the curve between two points on a circle.
If a circle is divided into two unequal arcs, the shorter arc is called the minor arc and the
longer arc is called the major arc.
If a circle is divided into two equal arcs, each arc is called a semicircle.
A secant is a line that intersects a circle in two points.
A tangent line to a circle intersects the circle at only one point.
A central angle of a circle is an angle whose vertex is the center of the circle and whose
measure is less than or equal to 180.
An inscribed angle is an angle whose vertex is on a circle and whose sides contain
chords of the circle.
A sector of a circle is a region bounded by two radii and an arc of the circle.

Properties, theorems, and corollaries:
 Chords for the same central angle in one or congruent circles have the same
length.
 Arcs for the same central angle in one or congruent circles have the same length.
 The lengths of arcs are proportional to the sizes of the central angles, but lengths
of chords are not proportional to the sizes of the central angles subtended by the
chords.
 If the radius of a circle is perpendicular to a chord, then the radius bisects the
chord and the arc for the central angle of the chord.
 The relationships between the radius of a circle (r) and the distance between a line
and the circle explain the position of the line and the circle.
 A tangent line of a circle is perpendicular to the radius whose one point ends at
the tangent point of the circle and the line.
 The measure of any inscribed angle for a certain arc is always half the measure of
the central angle for the complementary arc (the rest of the circle).
 Intersecting chord theorem (see §7)
 Intersecting secant theorem (see §8)
 The sum of opposite angles of an inscribed quadrilateral in a circle is always
180.
 The size of an inscribed angle in a semicircle is always 90.
 The angle between a chord and a tangent line, which passes through one end point
of the chord, and the inscribed angle for the chord have the same size.

 Carpenter’s construction of a circle by using inscribed angle theorem.

 Finding the center of a circle given three points on the circle or a tangent line with
the tangent point and another point on the circle.
 Arc lengths and areas of sectors
 Surface areas and volume of spheres
Tasks for students
According to classroom situations or students’ abilities, it is desirable to modify the
following tasks for your students.
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Sunrise on the First Day of a New Year Learning Task
Sun rise
A boat sails
Sun
Chord
The horizon
Arc
Figure 1. Sunrise
In some countries in Asia, many people visit the seashore on the east side of their
countries the first day of every new year. Watching the gorgeous scene of the sun rising
up from the horizon over the ocean, they wish good luck on their new year.
As the sun rises above the horizon, the horizon cuts the sun in different positions. By
simplifying this scene, we can mathematically think of the relationships between lines
and circles and the angles formed by these lines and parts of the circle. We can use a
circle to represent the sun and a line to represent the horizon.
1. A circle is not a perfect representation of the sun. Why not?
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2. Using the simplified diagram, how many ways is it possible for the sun and the
horizon to intersect? Draw an example of each and explain how they differ.
3. A tangent is a line that intersects a circle in one point while a secant intersects a circle
in two points. Do any of your drawings in #2 have a tangent or a secant? If so,
identify them. Is it possible for a line to intersect a circle in 3 points? 4 points?
4. When a line intersects a circle in two points, it creates a chord. A chord is a segment
whose endpoints lie on a circle.
a.
b.
c.
d.
How does a chord differ from a secant?
How many chords can be in a circle?
What is the longest chord in a circle? How do you know?
How is a chord’s distance from the center of a circle related to the size of the
chord?
e. Mary made the following conjecture: If two chords are the same distance from
the center of the circle, the chords are congruent. Do you agree or disagree?
Support your answer mathematically. Is the converse of this conjecture true?
f. Ralph was looking at the figure to the right. He made the following conjecture: A
radius perpendicular to a chord bisects the chord. Prove his
conjecture is true. Remember, if we can prove something is
always true it can be named a theorem. Comment(s):
B
This is a very simple sample proof. There are many ways
students may choose to write their proof. They will need to use
the skills and theorems they learned in Unit 2. Encourage them
E
C
to truly prove this conjecture by writing a complete proof.
Solution(s):
1. BCE and BCD are right angles....(Given.)
2. BCE = BCD....(All right angles are congruent.)
3. Segment BE  segment BD...(All radii of a circle are congruent.)
4. Both triangles share the side OH...
Segment BC  segment BC (Reflexive Property)
5.  BEC   BDC (by Hypotenuse-Leg Theorem)
6.Segment EC  segment CD (CPCTC)
7.Therefore, C is the midpoint of segment ED.
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F
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g. Is 
the converse of the statement in part f true?
5. Think back to the sun rise. As the sun rises you see a portion of its outer
circumference. A portion of a circle is called an arc. An arc has two endpoints that lie
of the circle
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h. Describe what happens to the arc of the outer circumference of the sun as the sun
rises. How do the arcs of a sunset differ from the arcs of the sunrise?
i. If a circle is divided into two unequal arcs, the shorter arc is called the minor arc
and the longer arc is called the major arc. If a circle is divided into two equal
arcs, each arc is called a semicircle. Use these words to describe the arcs of the
sunrise. What has to be true for the arc to be a semicircle?
j. What has to be true for an arc to be a semi-circle?
6. Find
§2. Terms and definitions
Q. For each of the following geometric objects,
Step1. Sketch an example of each (like in the picture above). Use the circle below for
your sketches.
1. a chord.. (You may add one more chord)
2. a sector..
3. arcs.. (Distinguish the major arc from the minor one in your sketch.)
4. a secant line..
5. a tangent line..
For the same arc,
6. an inscribed angle..
7. a central angle..
O
chord
Figure 2
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Step2. Find the definitions and write them down in your notes by completing each of the
following sentences:
1. A chord of a circle is a segment …
2. A sector of a circle is a region formed by …
3. An arc of a circle is that part of the circle …
(Distinguish the major arc from the minor one in your sketch.)
4. A secant line is a line that intersects a circle …
5. A tangent line is a line that intersects a circle …
For the same arc,
6. An inscribed angle is an angle formed by …
7. A central angle is an angle formed by ...
Solution:
A chord is a segment that joins two points on a circle.
A diameter is a chord that passes through the center of a circle.
An arc is the curve between two points on a circle.
A secant is a line that intersects a circle in two points.
A tangent line is a line that intersects a circle at only one point.
A central angle of a circle is an angle whose vertex is the center of the circle and whose
measure is less than or equal to 180.
An inscribed angle is an angle whose vertex is on a circle and whose sides contain
chords of the circle.
A sector of a circle is a region bounded by two radii and an arc of the circle.

§3. Chords, and arcs in a circle or in congruent circles
Q. What conditions determine a unique circle on a plane?
Given a point on a plane and the length of the radius from that point, how many different
circles can we draw?
a fixed point
radius
Figure 3
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Solution:
A fixed point as the center of a circle and the radius are the conditions.
A fixed point determines the position of the circle in a plane while the radius
determines the size of the circle. Only one circle can be drawn with these givens.
How about a sector? In order for two sectors to be congruent in a circle or congruent
circles, what conditions are required?
Solution:
The same radius and the same central angle between two radii are the conditions.
D
O
A
C
B
Figure 4
The length of the arc AB is l and the length of the arc AC is 3l . Measure the angles,
AOB and AOC .
Compare the measures of these two angles.
Are they in the same proportions as the length of their arcs?


Solution:

Students will find AOC =3* AOB .By adding more activities, they will also find the
proportional relationships between arcs and their angles.

Comparethe lengthsof chords AB and AC ?
Are they in the same proportions as the measures of their angles?
Solution:
Use conditions among three sides to be a triangle. If they are in proportional


relationships, ABC cannot be a triangle (if AC =3 AB and BC =2 AB , then
AC = AB + BC ; therefore NOT a triangle as AB + BC > AC for a triangle). Thus, the
lengths of chords are not proportional to the measures of their central angles.







 

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C
O
B
H
A
D
Figure 5
Theorem:
If radius OD is perpendicular to chord AB,
then OD bisects chord AB (i.e. length of segment AH = length of segment
HB) and arc AB (i.e. arc AD = arc DB) (see Figure 5).
Complete the steps in the following proof of this theorem:
(Solution: missing steps to be completed by student are given in blue.)
We first prove that  OAH and  OBH are congruent:
1.  OHA =  (OHB)....(Right Angles)
2. OA =(OB)...(radii forming Hypotenuse of each triangle)

3. Both triangles
sharethe side OH...(same Side)
 OAH   OBH (RHS)




Hence, AH = (HB) (Corresponding parts of congruent triangles are congruent)
 AOH =  BOH.
Further,

Thus, the corresponding arcs are congruent.

Is there any relationship between the center of the circle and the bisectors of chords?
Discuss this
 with others in your group.
Solution: All the bisectors of chords should pass through the center of the circle.
Note: This will be used later to find the center of a circle when three points on the circle
are given.
Chord properties

Chords equidistant from the center of a circle are equal.
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Equal chords are equidistant from the center.
The perpendicular bisector of a chord passes through the center of a circle.
A perpendicular line to a chord through the center of a circle bisects the chord.
The line through the center of the circle bisecting a chord is perpendicular to the
chord.
- (These last three are equivalent statements stemming from the uniqueness
of the perpendicular bisector.)
If a central angle and an inscribed angle of a circle are subtended by the same
chord and on the same side of the chord, then the central angle is twice the
inscribed angle.
If two angles are inscribed on the same chord and on the same side of the chord,
then they are equal.
If two angles are inscribed on the same chord and on opposite sides of the chord,
then they are supplemental.
For a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.
An inscribed angle subtended by a diameter is a right angle.
The diameter is the longest chord of the circle.
§4. Circles and lines
i
j
k
l
m
n
o
p
q
Figure 6. Positions of lines
Broadly speaking, there are two distinct cases: that a line does or does not intersect a
circle.
(In the above picture, the distinction of lines i, k, l, m, n, o, and p versus j and q)
In terms of the number of intersection points, we can divide the lines into three groups.
Q. Find and write the corresponding lines from the above picture.
1. No intersection point: j and q
2. One intersection point: i and p
3. Two intersection points: k, l, m, n, o, and p
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Secant lines of a circle are lines that intersect the circle in two points while tangent lines
intersect the circle in only one point.
Q. Find all the secant lines and all the tangent lines in the above picture.
Secant lines: k, l, m, n, o, and p
Tangent lines: i and p
§5. The radius of a circle and the distance between the center of the
circle and a line
We know how to find the distance between a point and a line.
Connect your knowledge to investigate the following.
In general, r represents the radius of a circle and d represents the perpendicular (or
shortest) distance between the center of the circle and a line. By comparing the lengths of
r and d, generalize the position relationships between a circle and lines.
i
j
m
k
l
r
n
o
d
p
q
Figure 7. The radius and the distance between the center of a circle and a line
Q. For what lines is d less than r? Specifically, given a circle and lines in a plane,
determine what length is greater than the other for each case. Refer to the above picture.
Use one of the notations of <, = or > between d and r in the following:
i) d ( ) r for a secant line,
ii) d ( ) r for a tangent line, and
iii) d ( ) r for the others.
Solution:
The answers are <, =, > (in this order).
Q. For i) above, how can you be sure that your answer is correct? Prove it by using the
Pythagorean theorem for a right triangle. Use the picture below.
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O
A
H
B
a secant line
Figure 8. The distance between a secant line and the center of a circle, and its radius
In the picture, the segment OA (r) is the radius of the circle, and the segment OH is the
distance (d) between the center of the circle and the secant line.
Apply Pythagorean theorem to the right triangle OAH.
 
2


 
2

 
2
Solution:
AH, OH, and OA in this order.
Which segment is longer than the other: OH(d) or OA(r) ? Why? Discuss this with
others in your group.
Solution:
OA(r)  OH (d). That is, the distance between a secant line and the center of a circle is

less than the radius.

§6. The orthogonality of tangent lines
Q. We know that tangent lines of a circle are the same distance away from the center as
the radius of the circle. So tangent lines should be perpendicular to the radius that ends at
the intersection point. Let’s prove this by discussing and answering the given questions.
(A group activity would be good for this, and it will be even better if the teacher’s
thorough proof follows after the group work.)
In the picture below, let the line l be a tangent line of the circle with the tangent
point A and center, B. Then the radius BA must be perpendicular to the tangent line
at A.
Proof:
Suppose that BA is not perpendicular to the tangent line. Then we can draw the
perpendicular line to the tangent line from the center, B. Name the intersection point of
the tangent line l and the perpendicular line through B as point C.
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B
l
A
C
Figure 9
(Note: The center point B has been offset in the figure above in order to make the angle
BCA appear to be a right angle as given by construction of the perpendicular through B
to l).
For the right triangle BCA, what is the distance between the tangent line l and the circle?
What side is the radius of the circle?
For right  BCA,
Which side is the hypotenuse, and which sides are the legs? Find each of them (do not
rely on “looks” but on the supposed construction of the perpendicular BC to the tangent
line).

Next, apply the Pythagorean theorem to  BCA.
2
2
2
     

Thus the segment BA { <, =, or
 > } the segment BC. Determine the correct notation
and circle it. For the circle and the tangent line l, what segment represents r and what
segment represents d? Any contradiction?
If d is less than r, then the line must be a secant line of the circle.
If there is inconsistency, where does it originate? Discuss this with others.
Solution:
2
2
2
BC  AC  BA and BA  BC .
That is, the distance is shorter than the radius. This means the line should be a secant line.
This is contradiction to the fact that the line l is a tangent line to a circle, and the
contradiction originates
 from what we assume: that the radius is not perpendicular to the
line.
     

Now let’s think about the inverse.
If a line is perpendicular to the radius and it passes through only one point on the
circle, then the line is a tangent line of the circle.
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O
E'
E
Figure 10
Proof:
Suppose the segment OE is perpendicular to the line that passes through the point E, but
the line is not a tangent line of the circle. Then in what situation are the line and the
circle? We know that three different situations are possible for the line and the circle.
Since the line intersects the circle at least at the point E, it should be either a tangent line
or a secant line. Since we are trying to prove that it must be a tangent line, we shall
assume that the line is not a tangent line; thus it must be a secant line and therefore, it has
another intersection point E`. We now need to show that the assumption (that the line is
not a tangent line) leads to a contradiction and thus, is a false assumption:
Let’s investigate  OEE`:
Does OE  OE' ? If so, what kind of triangle is  OEE`?
What do you know about the measures of  OEE` and  OE`E?
Are they same?
 condition states that OE is perpendicular to the line through E, so what is
The given


 OEE`?
 find here? 
What kind of contradiction do you
What do you think caused the contradiction?




Solution:
OE  OE' since both are radii of the circle. Thus OEE' is an isosceles triangle and
OEE ' OE ' E . From the given condition, OEE' 90 . That is,
OEE'OE'E  90 .
It is not possible for two interior angles of a triangle to both be 90º since the sum of three
 . Thus we have a contradiction. Our
interior angles of a triangle is always 180

assumption (that the line was not a tangent)
must be false as it leads to this contradiction,
which can only be overcome if point E` is the same as point E and, therefore, is tangent to
the circle.

Tangent properties

The line drawn perpendicular to the end point of a radius (on the circle) is a
tangent to the circle.
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A line drawn perpendicular to a tangent at the point of contact with a circle passes
through the center of the circle.
Tangents drawn from a point outside the circle are equal in length.
Two tangents can always be drawn from a point outside of the circle.
Task for Students: Making Curved Designs Using Triangles
In this activity, you will use cardboard triangles moving between two fixed points
(pushpins) to create curved designs by tracing the vertex (between the 2 pins) as its two
adjacent sides stay in contact with the pushpins (see Figure 11 below). Your challenge is
to find out what kinds of curves you can create in this manner and to make designs with
those curves based on your particular triangle. You will then compare your designs with
others in your class. Later on in this unit we shall use properties of circles and chords to
explain why it is possible to draw the curves you obtain in this manner.
Materials needed:
 9” x 12” cardboard rectangles, at least ¼” thick for each pair of students.
 Sheets of plane US-Letter size copy paper (several for each pair of students).
 Various cardboard triangles of different shapes (right, acute and obtuse), one for
each pair of students.
 A pair of plastic-topped pushpins for each pair of students.
 Fine point felt-tipped pens or ball-point pens (one for each pair of students).
You will work in pairs for this activity:
 Place a clean sheet of copy paper on your cardboard rectangle.
 Stick the pushpins through the copy paper into the thick cardboard rectangle about
2 to 3 inches apart somewhere close to the middle of the rectangle.
 Slide one vertex of your triangle between the 2 pins until the sides touch the pins.
(If the vertex moves beyond the edge of the copy paper, move the pushpins closer
together.)
 Mark a point on the paper at the vertex of your triangle.
 Rotate your triangle through a small angle, keeping the sides in contact with the
pushpins. Mark the new position of your vertex on the paper.
 Continue to rotate your triangle, keeping the sides in contact with the pushpins
and marking each new position of the vertex on your paper until the vertex
reaches each of the two pins in turn.
 Join the points that you marked with a smooth curve (If you can keep your pen at
the vertex while you rotate the triangle and maintain sides in contact with the two
pins at the same time, you will be able to draw the curves themselves.)
 Repeat all of the above steps using a different vertex of your triangle between the
pins. Also change the direction through which you push the vertex through the
pins so that you make curves on the other side of the two pins.
 Change the paper and readjust the positions and distance between the pins to
begin a new set of curves.
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V'
V
V''
V'''
V''''
B
A
Figure 11: Drawing curves tracing vertex V of a triangle between 2 pushpins, A & B
Q.
What kinds of curves did you create? How many different curves could you make
with your triangle (keeping the position of the pins the same)?
What mathematical properties make these curves possible?
Draw a segment between the pin positions in each of your designs. What is this
segment relative to each of the curves that you created in that design?
Solutions: All curves should be arcs of circles. Students should have been able to make
three different arcs on each side of the pins if they have triangles with 3 different angles
(2 different arcs for isosceles triangles, one arc for equilateral triangles).
Inscribed angles in an arc are all congruent, or all angles subtended by a chord of
a circle, on one side of the chord, are congruent.
A chord of the circles containing each of the arcs.
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§6. Central angle theorem
Your curves in the design activity you just completed are the result of a special property
of chords, arcs and the inscribed angles. The property is derived from the Central Angle
Theorem. The central angle theorem states that the measure of any inscribed angle for
a certain major arc is always half the measure of the central angle for the
complementary minor arc.
P
O
Figure 11
Now you are ready to prove this. Complete the following three steps and discuss your
conclusion.
Q.
Step I. Let's find the relationship between two angles,  OAB and  COB, where AC is
a diameter of the circle with center O, and B is any other point on the circle.

A

O
B
C
Figure 12
1. Find the angle whose size is the same as the size of  OAB.
Solution: OBA
2. What kind of triangle is  OAB?
Solution: it is an isosceles triangle since OA  OB as the radius.

 OAB and  COB?
3. What can you tell about the relationship between





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Accelerated Math I
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Solution: COB  OAB  OBA  2  OAB (exterior angle of triangle is equal to sum
of non-adjacent interior angles)
Step II. Let's find the relationship between the angles,  IHG and  IOG.

H


O
G
I
J
Figure 13
1. In Figure 14, point J is the opposite end of the diameter from H through O. What is the
relationship between  JOG and  JHG? (see Step I above) Similarly, what is the
relationship between  IOJ and  IHJ?


Solution: IOJ  2IHJ & JOG  2JHG




2.  IHG =  (
)+  (
).
 IOG =  (
)+  (
).

IHG  IHJ  JHG
Solution:


IOG  IOJ  JOG


3. Represent the size of  IHG with the size of  IOG.

IOG  IOJ  JOG
 2  IHJ  2  JHG


Solution:
 2  (IHJ  JHG)
 2  IHG
Step III. In the following diagram (fig. 15), let's find the relationship between the angles,
 KNM and  KOM. Point L is the opposite end of the diameter from N through O.



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Accelerated Math I
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N
O
L
M
K
Figure 14
1. Compare the size of angles,  MNL and  MOL (see Step I above). Similarly, what
relationship can you tell about the sizes of angles,  KNL and  KOL? Specifically,
 MOL = x*(  MNL), and  KOL = y*(  KNL). What are the values of x and y? Is x
equal to y?
 MNL, andx = y =2
Solution:  MOL = 2


 -  (MOL)

2. KOM =  KOL
 KNM =  KNL -  (MNL)

relationship between the angles,  KOM and  KNM.
Describe the




  KOL  MOL
KOM


 2  KNL  2  MNL

Solution:
 2  (KNL  MNL)
 2  KNM

Q. Why do we need three steps to prove the central angle theorem? Discuss this with
others. Also discuss how this theorem explains the curves that you produced in your

design task.
A GSP simulation would be very helpful to generalize the theorem regardless of the
position of an inscribed angle (see accompanying GSP file, Fig. 15s).
§7. Intersecting chord theorem
This theorem states that a×b is always equal to c×d no matter where the chords are
(the products of the two parts of each chord are equal).
By adding two segments in the picture (see figure 16), you can create two triangles:
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c
a
b
d
Figure 15
Q. Prove these triangles are similar. What condition for the similarity do you use to
prove it? By establishing proportional equations for corresponding sides of the similar
triangles, you can make the same conclusion as the theorem.
Solution: By using AAA similarity, we can show the two triangles on opposite sides of
the intersection point are similar. Because corresponding sides of similar triangles are in
proportion, we can set up the proportional equation, a :c  d :b . Since the product of
inners is equal to that of outers, ab  cd .
§8. Intersecting secant theorem


B
A
P
C
D
Figure 16
The theorem states that when two secant lines intersect each other at the point P
outside a circle, the products of the segments from point P to the intersection points
on the circle for each secant are equal. That is, PA  PB  PC  PD.
Q. Prove this theorem by adding two segments BC and AD to figure 17.

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Find pairs of angles of the same size, and determine what triangles are similar to each
other. Set up proportionality equations for ratios of corresponding sides of these triangles.

Solution:
Consider PAD and PCB.
PBC  PDA
BCD  DAB
Thus, PAD  PCB and P is common.

 PAD  PCB.
PA : PC  PD : PB

Hence, PA  PB  PC  PD

§9. A property between opposite angles of an inscribed quadrilateral in
a circle
A
B
O
D
C
Figure 17
Q.
1. What central angle is corresponding to its inscribed angle,  ABC?
Solution: AOC for the minor arc AC
 angle,  ADC?
2. What central angle is corresponding to its inscribed

Solution: AOC for the major arc AC

3. From 1 and 2, what is the sum of two central angles?
Solution: from Q1 and Q2, the sum of two central angles is 360º.

So what do you conclude about the sum of two inscribed angles  ABC and  ADC?

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Accelerated Math I
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1
1
AOC mirnor AOC major
2
2
1
  AOC mirnorAOC major
Solution: 2
.
1
  360
2
 180
§10. The size of an inscribed angle in a semicircle

A
A'
A"
B
C
O
Figure 18
Q.
What is the size of the inscribed angle,  BAC in the above circle?
The points B and C are the end points of the diameter of the circle, so the central angle,
 BOC of the arc of the lower half of the circle is always 180. Thus, the size of any
inscribed angle in a semicircle is constant.


1
BAC  BOC
Solution:
2
 90


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§11. Theorem: The angle formed by a chord and a tangent line in a
circle is equal to the inscribed angle subtended by the chord.
F
O
l
E
G
H
Figure 19
In figure 20, point E is the opposite end of the diameter from F through O. Line l is
tangent to the circle at H. Point G is on l.  FHG is the angle chord HF makes with the
tangent line, l.
Q. Prove this theorem.
First, using theorem §10 above,

 EHF= 90 so  (OHE)+  OHF = 90=  FHG +  OHF (angle of radius and tangent)
Thus,  FHG =  (OHE).

Next, Triangle OEH is a/an (isosceles) triangle.
Thus, (OHE) 
=  OEH.  



Hence,  FHG =  (OEH).
However,  OEH=  FEH= an inscribed angle of chord FH.




§12.
Carpenter’s
construction of a semicircle.


Q. A carpenter wants to construct a semi-circular arch above a doorframe that currently
has a pointed top frame as in the following diagram:
A
B
In order to create her semicircle she uses a large square piece of
wood with two large nails hammered into the top corners of the
doorframe at points A and B. Experiment with a piece of
cardboard with a square corner and two pushpins pinned into
another (larger) thick piece of cardboard (see diagram belowright) to see if you can draw her semi-circle above the
doorframe. Why does this work? What theorem(s) above did
you use? Why doesn’t the carpenter just use a nail at the center
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Accelerated Math I
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of her semicircle and a piece of string to draw her semicircle? Discuss this activity with
others in your group. How is this similar to your design
task with the cardboard triangle?
By connecting the dots determined by the various
positions of the corner of the square as it slides between
the nails A and B, we can find a semi-circle whose
diameter is the segment AB.
A
B
Discuss the mathematics behind this activity with
others. Hint, consider the constant size of inscribed
angles in a semicircle.
Figure 21
P
C
A
B
Figure 20
To draw a circle in the same manner as the Carpenter's Method with GSP, begin with two
points A and B. Construct a line through A that includes the free point P. Construct the
perpendicular through point B to the line AP. Label the intersection of the two lines C.
Trace point C while you move the point P around point A.
Measure  ACB. What kind of angle is this? Does the size of the angle change along
with the position of the point P? Why or why not? Select points A, C and B (in that order)
and select Arc through 3 points from the Construct menu. What kind of arc appeared?

Solution: (If available, do the same activity with GSP.)
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Since BCAP regardless of the position of the point P, ACB is always 90. Thus, the
arc ACB is the semicircle.
Theorem 10 states that inscribed angles in a semicircle are always 90.
The carpenter used the theorem to construct a circle like the above picture, but the angle


of the corner of a piece of cardboard does not have to be 90. Infact, this activity can be
generalized with any angle size instead of 90 and two fixed points as ending points of a
 above)
chord instead of a diameter. (See the Triangle Design activity
§13. Finding the center of 
a circle

Figure 21
Here is a circle.
Now we want to find the center of the circle. You can use a compass and a ruler (or GSP
if available). What principles do you use to find it with three points on the circle?
Q.
Select three points on the circle. How many chords can you make?
Construct two of them.
What is the relationship between the center and chords in a circle? Use it to find the
center. (Hint, remember properties of bisectors of chords in §3.) Recall the archeological
method for finding the circumference of a piece of broken pottery from Unit 2. How
does this method relate to that problem?
Solution:
Three chords can be drawn from three points on the circle, and perpendicular
bisectors of chords in a circle pass through the center of the circle. Thus, by
constructing two bisectors of any two chords, students can find the center. The
intersection point of two perpendicular bisectors is the center of the circle.
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Accelerated Math I
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Figure 22
Let’s think about another condition to find the center.
Suppose that you know a tangent line and one more point on the circle like the above
picture. What will you use? Find it by using a compass and a ruler.
Discuss your way with others.
Solution: Use §6 and §3.
Properties or theorems for circles





A circle’s circumference and radius are proportional.
The enclosed area and the square of its radius are proportional.
The constants of proportionality are 2  and  , respectively.
The circle centered at the origin with radius 1 is called the unit circle.
Through any three points, not all on the same line, there lies a unique circle.


§14. Arc lengths and areas of sectors
In our previous investigations we found that arc lengths in a circle are proportional to the
size of corresponding central angles in the same circle. Using this, develop the formulas
to find the arc length and areas of sectors.
Q.
What is the area of a circle whose radius is r? ………..(*)
What is the length of circumference of the circle?................(**)
Solution:
The area of a circle whose radius is r is r 2 , and the length of circumference of the circle
is 2 r .


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Accelerated Math I
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O
r

B
A
Figure 23
Let S be the area of the sector OAB, and l be the length of the arc AB (the minor one)
In the picture, find the area of the sector OAB, and the length of arc AB.
You can establish the proportional expression,
 : 360  l : the circumference. Here the circumference is the answer of (**) above.
From the above proportional equation, represent l with the expression of r and  .
Thus, l = (
).

 : 360  l : 2r

360

l

2

r

Solution:

2r
l
or r if  is measured in radians

360
Similarly,  : 360  S : the area of circle . Here the circle area is the answer of (*).
Represent the area of the sector with the expression of r and  .

Hence, S = (
).

Solution:
360 S  r 2

1 2
S
or r  if  is measured in radians
360
2
r 2
By using the above formula for l , find the analogy between the formula for the area of a
triangle and the formula to find the area of a sector.


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Accelerated Math I
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A

r
r
r
versus
h
C
B
a
l
S=(
)
Area=
1
2
ah
Figure 24
Solution:
For  measured in radians :
1
S  r 2, and l  r
2
1
1
S  rr   rl
2
2

Thus,
1
1
rl is the area of the sector and ah is the area of the triangle.
2
2
Q.
Suppose that you make a pendulum swing like the picture below.
The length of the string of the pendulum is 6 inch and the length of the arc AB is 12 inch.
 Find the size of APB .


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Accelerated Math I
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29
P
r
B
A
l
Figure 25


Hint: In the diagram above r is the radius of your arc and has length 6 inches. What
would the circumference of the full circle be in terms of π? The arc swing has length 12
inches. How does the ratio of this arc swing to the circumference of the full circle relate
to the angle APB?
Solution:
If the angle is measured in radians rather than degrees, the solution is straightforward:
From l  r
l=12 and r=6
360
360
Hence,   2 radians. As 1 radian =
deg rees,  
deg rees
2

Q.

Did you ever think of the container
shapes of drinks? Many of their shapes are
cylindrical. See the pictures below.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
Figure 26. Vending machines in Japan
Have you ever seen drink containers of the following shapes?
Note: In fact, there are non-cylindrical containers for some reasons. Rectangular shaped
paper containers for milk or juice are examples. One reason is that rectangular shaped
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Accelerated Math I
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paper containers help people recognize whether milk or juice have gone bad since the
containers swell when they have gone bad.
Koke
or
fepsi
Figure 27
Beverage companies do not use those containers. Do you know why?
Suppose you are a CEO in a beverage company. Then it is natural that you want to invest
less money to make your containers. For the same amount of beverage, what containers
have less surface area (and thus less material)? This is important to save money, as the
cost depends on the total amount of material needed to make the container. In fact,
beverage companies are spending big money to pay for materials to make containers.
For containers of the same height, we just need to compare the area of each different
shaped base to determine the container that will hold the most beverage (why?).
In the following diagram, the perimeter (or circumference) of each base has been fixed at
4  . Use your area formulas for each shape to determine the area of each base.

Note: Teachers can provide the fixed perimeters as 24 cm instead of 4  asking students
to use calculators to compute the areas.
Let's fix the circumference for the circle and perimeters for the square
and the equilateral triangle as 4š

2š r=4š
r=2
The length of one side =š
Figure 28
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The length of one side =
4
3
š
Accelerated Math I
Unit 3

31
I hope they knew that the square has the biggest area for the same perimeter among
rectangles (Since x  y  2  where x is the width and y is the height of a rectangle, the
area must be S  xy  x(2   x)  2 x  x 2 , where does the parabola have its vertex? Yes,
when x   ) and the same argument is held for an equilateral triangle. See the following
picture. Do all the following triangles have the same area? Why?

How about the perimeters? (I recommend GSP software to simulate the following.)

QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
Area
Area
ABC = 6.17 cm2
A'BC = 6.17 cm2
m PB+m CP+m BC = 13.88 cm
m AB+m CA+m BC = 11.32 cm
A''
A'
A'''
A
B
A'''' P
A'''''
C
Area
PBC = 6.17 cm2
Figure 29
Now, we just consider a circle, a square, and an equilateral triangle for the same
circumference or perimeter.
Q.
For the circumference 4 
The area of the circle = (
)
The area of the square = ( )
The area
of the equilateral triangle = ( )

Which one has the biggest area?
You know that the volume is found by multiplying the height to the area of the base. So
for the same height, the biggest area gives the biggest volume.
Discuss your investigation with others in your group.
Solution:
The area of the circle = 4 
The area of the square =  2


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Accelerated Math I
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The area of the equilateral triangle =
4 3 2

9
Which one has the biggest area? Since 4   and
biggest area.
4 3
 1, the circle has the
9

Alternate version using a fixed perimeter of 24 cm:

In the following diagram, the perimeter (or circumference) of each base has been fixed at
24 centimeters. Use your area formulas for each shape to determine the area of each base.
Circumference AB = 24.0 cm
A
Radius AB = 3.82 cm
Perimeter of square = 24.0 cm
Perimeter of Triangle = 24.0 cm
B
Side length = 6.00 cm
Side length = 8.00 cm
Alternate Figure 31
Q.
For a circumference or perimeter of 24 cm:
The area of the circle = (π(3.82)2 sq. cm.)
The area of the square = (36 sq. cm.)
The area of the equilateral triangle = (16√3 sq. cm.)
Which one has the biggest area? (using a calculator: Circle = 45.84, square = 36,
triangle = 27.71 sq. cm.)
You know that the volume is found by multiplying the height to the area of the base. So
for the same height, the biggest area gives the biggest volume. If the height of each
container is 12 cm, find the volume for each different shaped container. Find the surface
area of each container not counting the area of the top and bottom of the container
(remember the perimeter of each is 24 cm). If you now add in the area of the top and
bottom of each container, compare total surface area to total volume of each container.
Do the soft drink companies use the best-shaped container for saving material costs?
Discuss your investigation with others in your group.
Solutions: Volume of Cylinder: 550 cm3, square prism: 432 cm3, triangular prism: 332.5
cm3. Surface area of each can without ends is 12x24=288 cm2. Surface area of cans with
ends added: Cylinder: 333.84, Square prism: 324, Triangular prism: 315.71 sq. cm. To
determine best cost-benefit, take ratio of volume to surface area for each can: Cylinder:
1.65; for square prism: 1.33; for triangular prism: 1.05, so Cylinder still has best volume
to materials ratio.
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§15. Surface areas and volumes of spheres
Remember the formulas to find the surface area and the volume of a sphere,
4
S  4 r 2 , and V  r 3
3
(*I assume that students at this level learn the formulas at the previous level. otherwise,
the following activity would be helpful for students to get the formulas).

(If possible, I recommend that students do experiment with containers and water rather
than just a theoretical approach).
We are going to compare the volumes of the following three different shaped containers
by pouring water into the cylinder that contains each.
r
Figure 30. A cone and a sphere with the same diameter and height as the containing
cylinder
In the picture, the sphere and the cone touch the interior surface of the cylinder and thus
have the same diameter and height as the cylinder.
By performing experiments, we can find the ratio among volumes of these three
containers although it will not be precise.
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QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
Figure 31. An image from Google
For example, suppose we pour the water into the cylinder that contains the sphere like the
left picture below.
Once the water is full in the cylinder, then we pull out the sphere from the cylinder
(without spilling any of the water!). The empty cylinder will look like the picture on the
right.
r
h
1
h
3
Figure 32
In theory, the water in the cylinder now fills one third of the height of the cylinder.
Q. How much did the sphere occupy the interior of the cylinder? Solution:
2
3
We can apply a similar method to find the volume of the cone. In the case of the cone,
when it is removed, the water fills two thirds of the height of the cylinder.

2
h
3
h
Figure 33
Q. How much of the interior volume of the cylinder did the cone occupy?
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Accelerated Math I
Unit 3
Solution:
35
1
3
By comparing the volume of water that each container can maximally hold, now you
know the ratios for the volumes. That is, Vcone :Vsphere :Vcylindar    :   :  

Remember that the volume of a solid body is the amount of "space" it occupies. Thus, the
volume of a cylinder is computed by multiplying its height to the area of its base since it
is the solid that congruent circles are accumulated in perpendicular direction to the base.
 in the figure 6 is
Hence, the volume of the cylinder
S  Base(a circle with radius r) height
 r 2  h
Or you can get it in a different way.
Let’s consider that you slice the cylinder like the picture below.
Solution: Vcone:Vsphere :Vcylindar   1  :  2  :  3 

Figure 34
The height of the cylinder is h and the radius of the circle base is r. Let’s consider that
you can slice the cylinder infinitely many times. Then each sector is close to the triangle.
That is, y is close to the half of the circumference of the circle.
Q. Find x and y, and the volume of the cylinder by computing the right rectangular solid.
Solution:
x  r, and y  r
V  r 2 h
Now go back to the topic of volumes of a cone and a sphere.
The volume of a cone is one third of the volume of a cylinder and the volume of a sphere

is two thirds of the volume of cylinder in the figure 6.
Write down each formula
Vcylinder 
Vcone 
(***)
Vsphere 

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Unit 3
36
Vcylinder  r 2 h
1
Solution: Vcone  r 2 h
3
2
Vsphere  r 2 h
3
In the figure 30, h  2r

Vcylinder  r 2 h  2 r 3
1
2
 r 2 h  2r 3
3
3
4
Vsphere  r 3
3
Next, let’s find the surface area of a sphere.
The idea is shown in the picture below.
 V
cone

Figure 35

Suppose we can cut the sphere as lots of pyramids (or cones) whose vertices meet at the
center of the sphere. The volume of the sphere is approximately same as the sum of
V1 V2 V3 V4  Vn if n is big enough. The volume of a pyramid can be found by a
similar method to the way we found the volume of a cone, but using a container with a
square base rather than a cylinder (i.e. a square prism). The volume of the pyramid turns
out to be one-third the volume of the square prism with the same height. Let base1 be the
area of the base of one small pyramid, base2 the area of the base of a second pyramid, etc.
Thus, Vsphere  1 base1 height  1 base2  height  1 base3  height  1 base4  height   1 basen  height
3
3
3
3
3
Q. What is the height in the above equation? How can we consider that the value of the
is fixed?
Solution: The height is the radius of the sphere since we assumed that we cut the sphere
into many pyramids whose vertices meet at the center of the sphere.
height
The second draft
Accelerated Math I
Unit 3
37
1
1
1
1
base1 height  base2  height  base3  height  base4  height 
3
3
3
3
1
 (base1  base 2  base3   basen )  height
3
1
  (the sum of the areas of all the bases)  height
3
Vsphere 

1
basen  height
3
What is the sum of areas of all the bases? Isn’t that the surface area of the sphere?

Solution: The surface area of the sphere.
1
Vsphere  the surface area of the sphere  height
where the height is the radius of the sphere.
3
1
 the surface area of the sphere r
3

Let S be the surface area of the sphere.
1
Vsphere  Sr
3
3V
 S  sphere
r
What was the volume of the sphere from (***)? By plugging that value to Vsphere , you can
derive the formula to find the surface area of the sphere whose radius is r.

2
V  r 3h, and h  2r
3
Solution:
2
4
V  r 2 2r   r 3
3
3

Q. Write down the formula to find the surface area of a sphere with the radius r.
Solution:

3V
4
From S  sphere , and V  r 3
r
3
4

3 r 3 
3

S
 4 r 2
r

The area of a zone of a sphere
The curved area of a circular portion on the surface of a sphere (called a “zone of one
base”) can be determined from the plane intersecting the sphere to form the
circumference of the circular region (its base). The surface area of this zone is given by
the product of the circumference of the sphere (2πr, where r is the radius of the sphere)
with the perpendicular distance of the plane to the center of the circular region on the
surface of the sphere (h).
The second draft
Accelerated Math I
Unit 3
38
h
r
Figure 38: Zone formed by the intersection of one plane with a sphere (3-D view and 2-D
cross section). Surface area of zone= 2πrh.
Q. In the case where the plane passes through the center of the sphere, h=r and thus the
zone is a hemisphere. Based on the formula for the area of a zone, find the surface area of
a hemisphere. Is this formula consistent with the formula you found for the surface area
of a sphere with radius r?
Solution:
For a hemisphere, surface area = 2πrh, where h=r; thus surface area = 2πr2.
A sphere consists of two hemispheres; thus surface area of a sphere = 2*2πr2 = 4πr2.
This formula is consistent with the one above.
§16. Culminating Task: Find the surface area of the earth covered by
one low-orbit communication satellite.
The modern world relies on orbiting communications satellites for world-wide
communications. It is your job to determine how much of the surface of the earth will be
covered by one satellite in a low-earth orbit at a height of 485 miles (or 780 km). The
polar radius of the earth is approximately 6,357 km. Draw a scale drawing using 1
millimeter to represent 10 km. Figure 39 is approximately to scale, using one pixel as 10
km. in GSP.
The second draft
Accelerated Math I
Unit 3
39
m AE = 635.68 pixels
ES = 78 pixels
S
E
A
Figure 39: Scale drawing representing the position of a satellite, S above the surface of
the earth.
You may use this scale drawing to complete your task, but first measure segments AE
and ES to determine your scale in millimeters. You can also complete the task without
relying on scaled measurements. Following are some hints to help you!
Hints: Signals from the satellite travel in straight lines in all directions. Only those
hitting the surface of the earth can be used for communicating with earth-based stations.
Recall the properties of tangents from a point outside of a circle. Can you use those
properties to determine the extreme points on the arc of the earth’s surface with which the
satellite can communicate? Once you have those points, can you determine the diameter
of the circular region on the earth’s surface covered by this particular satellite? Can you
use that information to determine the value of h in the formula for the surface area of
your circular region on a sphere (zone of one base)? This low-earth orbit satellite orbits
the earth from pole to pole once every 100 minutes. Calculate the speed of the satellite in
meters/second.
In your group, discuss your strategies for solving this task before you begin. What data
do you have? What data do you need to calculate? How will you go about finding that
data?
Your group should be prepared to give a report to the class, showing your constructions
and measurements as well as any formulas and calculations that you used.
The second draft
Accelerated Math I
Unit 3
40
The data used in this task were based on the Iridium Satellite Constellation and were
obtained via Wikipedia, the free on-line encyclopedia:
http://en.wikipedia.org/wiki/Iridium_%28satellite%29#The_constellation
A possible solution (students may come up with other approaches to the problem):
T1
m AE = 635.68 pixels
ES = 78 pixels
m T 1 T 2 = 578 pixels
AT 1 ST 1 GSAGT1
S
So
E
G
AS
AT 1
=
AT 1
AG
But AT 1 is the radius of
the earth (AE, known).
AS = AE+ES (known).
GE=AE-AG, thus the
exact measurement of
GE can be calculated
from the above
equations:
Midpoint of AS
GE=AE-(
T2
A
m AE-
AE2
AE+ES
m AE2
m AE+ES
)
= 69 pixels
GE = 69 pixels
2 š m AEGE = 277411
 
2 š m AE+ES10
pix2
4 š  m AE 2 = 5.08 10 6 pix2
6 pixels
= 7473
Figure 40
Based on the calculations given in Figure 40, the area of the circular region on the earth’s
surface (2πrh square kilometers) will be approximately 27,741,100 square kilometers.
The surface area of the earth is approximately 508 million square kilometers.
The speed of the satellite is given by the circumference of the orbit (in meters) divided by
6000 seconds: 2π*mAS*10,000/6000  7,473 meters/second, which is approximately 7.5
kilometers/second or 27,000 kilometers/hour!

The second draft