Sunrise on the first day of a new year



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. Determine the distance between two points.

b. Determine the distance between a point and a line.

c. Determine the midpoint of a segment.

d. Understand the distance formula as an application of the Pythagorean theorem.

e. 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:

MA1P1. Students will solve problems (using appropriate technology).

a. Build new mathematical knowledge through problem solving.

b. Solve problems that arise in mathematics and in other contexts.

c. Apply and adapt a variety of appropriate strategies to solve problems.

d. Monitor and reflect on the process of mathematical problem solving.

MA1P2. Students will reason and evaluate mathematical arguments.

a. Recognize reasoning and proof as fundamental aspects of mathematics.

b. Make and investigate mathematical conjectures.

c. Develop and evaluate mathematical arguments and proofs.

d. 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).

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 [pic].

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 [pic].

• The size of an inscribed angle in a semicircle is always [pic].

• 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.

Sunrise on the First Day of a New Year Learning Task

[pic]

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.

[pic]

1. A circle is not a perfect representation of the sun. Why not?

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. How does a chord differ from a secant?

b. How many chords can be in a circle?

c. What is the longest chord in a circle? How do you know?

d. 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):

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 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. [pic] BEC [pic] [pic] BDC (by Hypotenuse-Leg Theorem)

6. Segment EC ( segment CD (CPCTC)

7. Therefore, C is the midpoint of segment ED.

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

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..

[pic]

Figure 2

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 [pic].

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?

[pic]

Figure 3

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.

[pic]

Figure 4

The length of the arc AB is [pic] and the length of the arc AC is [pic]. Measure the angles, [pic] and[pic].

Compare the measures of these two angles.

Are they in the same proportions as the length of their arcs?

Solution:

Students will find [pic]=3*[pic].By adding more activities, they will also find the proportional relationships between arcs and their angles.

Compare the lengths of chords [pic] and [pic]?

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, [pic] cannot be a triangle (if [pic]=3[pic] and [pic]=2[pic], then [pic]=[pic]+[pic]; therefore NOT a triangle as [pic]+[pic] > [pic] for a triangle). Thus, the lengths of chords are not proportional to the measures of their central angles.

[pic]

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 [pic]OAH and [pic]OBH are congruent:

1. [pic]OHA = [pic](OHB)....(Right Angles)

2. [pic] =(OB)...(radii forming Hypotenuse of each triangle)

3. Both triangles share the side OH...(same Side)

[pic]OAH [pic] [pic]OBH (RHS)

Hence, [pic] = (HB) (Corresponding parts of congruent triangles are congruent)

Further, [pic]AOH = [pic]BOH.

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.

• 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

[pic]

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

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.

[pic]

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 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.

[pic]

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.

[pic]

Solution:

AH, OH, and OA in this order.

Which segment is longer than the other: [pic]? Why? Discuss this with others in your group.

Solution:

[pic]. 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.

[pic]

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[pic]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 [pic]BCA.

[pic]

Thus the segment BA { } 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:

[pic] and [pic].

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.

[pic]

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 [pic]OEE`:

Does [pic]? If so, what kind of triangle is [pic]OEE`?

What do you know about the measures of [pic]OEE` and [pic]OE`E?

Are they same?

The given condition states that OE is perpendicular to the line through E, so what is [pic]OEE`?

What kind of contradiction do you find here?

What do you think caused the contradiction?

Solution:

[pic] since both are radii of the circle. Thus [pic] is an isosceles triangle and [pic]. From the given condition, [pic]. That is, [pic].

It is not possible for two interior angles of a triangle to both be 90º since the sum of three interior angles of a triangle is always [pic]. Thus we have a contradiction. Our 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.

• 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.

[pic]

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.

§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.

[pic]

Figure 12

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, [pic]OAB and [pic]COB, where AC is a diameter of the circle with center O, and B is any other point on the circle.

[pic]

Figure 13

1. Find the angle whose size is the same as the size of [pic]OAB.

Solution: [pic]

2. What kind of triangle is [pic]OAB?

Solution: it is an isosceles triangle since [pic] as the radius.

3. What can you tell about the relationship between [pic]OAB and [pic]COB?

Solution: [pic] (exterior angle of triangle is equal to sum of non-adjacent interior angles)

Step II. Let's find the relationship between the angles, [pic]IHG and [pic]IOG.

[pic]

Figure 14

1. In Figure 14, point J is the opposite end of the diameter from H through O. What is the relationship between [pic]JOG and [pic]JHG? (see Step I above) Similarly, what is the relationship between [pic]IOJ and [pic]IHJ?

Solution: [pic]

2. [pic]IHG = [pic]( )+ [pic]( ).

[pic]IOG = [pic]( )+ [pic]( ).

Solution: [pic]

3. Represent the size of [pic]IHG with the size of [pic]IOG.

Solution:[pic]

Step III. In the following diagram (fig. 15), let's find the relationship between the angles, [pic]KNM and [pic]KOM. Point L is the opposite end of the diameter from N through O.

[pic]

Figure 15

1. Compare the size of angles, [pic]MNL and [pic]MOL (see Step I above). Similarly, what relationship can you tell about the sizes of angles, [pic]KNL and [pic]KOL? Specifically, [pic]MOL = x*([pic]MNL), and [pic]KOL = y*([pic]KNL). What are the values of x and y? Is x equal to y?

Solution: [pic]MOL = 2[pic]MNL, and x = y =2

2. [pic]KOM = [pic]KOL - [pic](MOL)

[pic]KNM = [pic]KNL - [pic](MNL)

Describe the relationship between the angles, [pic]KOM and [pic]KNM.

Solution: [pic]

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:

[pic]

Figure 16

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, [pic]. Since the product of inners is equal to that of outers, [pic].

§8. Intersecting secant theorem

[pic]

Figure 17

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, [pic].

Q. Prove this theorem by adding two segments BC and AD to figure 17.

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 [pic].

[pic]

[pic]

[pic]

§9. A property between opposite angles of an inscribed quadrilateral in a circle

[pic]

Figure 18

Q.

1. What central angle is corresponding to its inscribed angle, [pic]ABC?

Solution: [pic]for the minor arc AC

2. What central angle is corresponding to its inscribed angle, [pic]ADC?

Solution: [pic]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 [pic]ABC and [pic]ADC?

Solution: [pic].

§10. The size of an inscribed angle in a semicircle

[pic]

Figure 19

Q.

What is the size of the inscribed angle, [pic]BAC in the above circle?

The points B and C are the end points of the diameter of the circle, so the central angle, [pic]BOC of the arc of the lower half of the circle is always [pic]. Thus, the size of any inscribed angle in a semicircle is constant.

Solution: [pic]

§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.

[pic]

Figure 20

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. [pic]FHG is the angle chord HF makes with the tangent line, l.

Q. Prove this theorem.

First, using theorem §10 above,

[pic]EHF=[pic] so[pic](OHE)+[pic]OHF = [pic]=[pic]FHG + [pic]OHF (angle of radius and tangent)

Thus, [pic]FHG = [pic](OHE).

Next, Triangle OEH is a/an (isosceles) triangle.

Thus, [pic](OHE) =[pic]OEH.

Hence, [pic]FHG = [pic](OEH).

However, [pic]OEH=[pic]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:

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 below-right) 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 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.

Discuss the mathematics behind this activity with others. Hint, consider the constant size of inscribed angles in a semicircle.

Figure 21

[pic]

Figure 22

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 [pic]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.)

Since [pic] regardless of the position of the point P, [pic] is always [pic]. Thus, the arc ACB is the semicircle.

Theorem 10 states that inscribed angles in a semicircle are always [pic].

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 [pic]. In fact, this activity can be generalized with any angle size instead of [pic] and two fixed points as ending points of a chord instead of a diameter. (See the Triangle Design activity above)

§13. Finding the center of a circle

[pic]

Figure 23

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.

[pic]

Figure 24

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 [pic] and [pic], 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 [pic], and the length of circumference of the circle is [pic].

[pic]

Figure 25

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,

[pic]. Here the circumference is the answer of (**) above.

From the above proportional equation, represent [pic] with the expression of r and [pic].

Thus, [pic] = ( ).

Solution: [pic]

Similarly, [pic]. Here the circle area is the answer of (*).

Represent the area of the sector with the expression of r and [pic].

Hence, S = ( ).

Solution: [pic]

By using the above formula for [pic], find the analogy between the formula for the area of a triangle and the formula to find the area of a sector.

[pic]

Figure 26

Solution:

[pic]

Thus, [pic] is the area of the sector and [pic] is the area of the triangle.

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 [pic].

[pic]

Figure 27

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 [pic]

l=12 and r=6

Hence, [pic] radians. As 1 radian = [pic]

Q.

Did you ever think of the container shapes of drinks? Many of their shapes are cylindrical. See the pictures below.

[pic] [pic]

Figure 28. 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 paper containers help people recognize whether milk or juice have gone bad since the containers swell when they have gone bad.

[pic]

Figure 29

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 [pic]. 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 [pic] asking students to use calculators to compute the areas.

[pic]

Figure 30

I hope they knew that the square has the biggest area for the same perimeter among rectangles (Since[pic] where x is the width and y is the height of a rectangle, the area must be [pic], where does the parabola have its vertex? Yes, when [pic]) 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.)

[pic][pic]

Figure 31

Now, we just consider a circle, a square, and an equilateral triangle for the same circumference or perimeter.

Q.

For the circumference [pic]

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 = [pic]

The area of the square = [pic]

The area of the equilateral triangle = [pic]

Which one has the biggest area? Since [pic], the circle has the biggest area.

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.

[pic]

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.

§15. Surface areas and volumes of spheres

Remember the formulas to find the surface area and the volume of a sphere, [pic]

(*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.

[pic]

Figure 32. 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.

[pic]

Figure 33. 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.

[pic] [pic]

Figure 34

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: [pic]

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.

[pic] [pic]

Figure 35

Q. How much of the interior volume of the cylinder did the cone occupy?

Solution: [pic]

By comparing the volume of water that each container can maximally hold, now you know the ratios for the volumes. That is, [pic]

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.

Hence, the volume of the cylinder in the figure 6 is

[pic]

Or you can get it in a different way.

Let’s consider that you slice the cylinder like the picture below.

Solution:[pic]

[pic] [pic]

Figure 36

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: [pic]

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

[pic] (***)

Solution: [pic]

In the figure 30, [pic]

[pic]

Next, let’s find the surface area of a sphere.

The idea is shown in the picture below.

[pic]

Figure 37

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 [pic] 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, [pic]

Q. What is the height in the above equation? How can we consider that the value of the height 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.

[pic]

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.

[pic] where the height is the radius of the sphere.

Let S be the surface area of the sphere.

[pic]

What was the volume of the sphere from (***)? By plugging that value to [pic], you can derive the formula to find the surface area of the sphere whose radius is r.

Solution: [pic]

Q. Write down the formula to find the surface area of a sphere with the radius r.

Solution:

[pic]

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).

[pic] [pic]

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.

[pic]

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 data used in this task were based on the Iridium Satellite Constellation and were obtained via Wikipedia, the free on-line encyclopedia:

A possible solution (students may come up with other approaches to the problem):

[pic]

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 [pic] 7,473 meters/second, which is approximately 7.5 kilometers/second or 27,000 kilometers/hour!

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