# Tangent Properties. Line m is a tangent to circle O. Point T is the point of tangency.

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2 Lesson 6.1 Tangent Properties (continued) Read the information about minor and major arcs on page 313 of your book, and study the example. Then read about internally tangent circles and externally tangent circles. Now read the example below. EXMPLE a. MN and MP are tangents to circle O. x y mnp 5 cm M x y mpqn N P 110 O Q b. D is tangent to both circle B and circle C. w mxt D w B T 80 C X c. Refer to the figure for part a. What type of quadrilateral is MNOP? Solution a. By the Tangent Conjecture, m MNO 90 and m MPO 90. By the Quadrilateral Sum Conjecture, x 360, so x 70. By the Tangent Segments Conjecture, MN MP, so y 5 cm. Because the measure of a minor arc equals the measure of its central angle, mnp 110. Subtract this measure from 360 to find the measure of the major arc, giving mpqn 50. b. By the Tangent Conjecture, m BD 90 and m CD 90. By the Quadrilateral Sum Conjecture, w 360, so w 100. The measure of minor arc T equals the measure of the central angle BT, so mt 100. Subtract this measure from 360 to find the measure of its major arc: mxt c. MNOP is a kite because MN MP and ON OP. 80 CHPTER 6 Discovering Geometry Condensed Lessons

4 Lesson 6. Chord Properties (continued) Investigation 3: Chords and the Center of the Circle Follow Step 1 in your book. The perpendicular to each chord divides the chord into two segments. How do the lengths of the segments compare? Use your findings to complete this conjecture. Perpendicular to a Chord Conjecture The perpendicular from the center of a circle to a chord is the of the chord. C-57 Now use your compass to compare the distances (measured along the perpendicular) from the center to the chords. How do the distances compare? Draw a circle of a different size and follow Step 1 in your book for the new circle. Compare the distances from the center of the circle to each chord. re your findings the same as they were for the first circle? Complete the conjecture. Chord Distance to Center Conjecture Two congruent chords in a circle are from the center of the circle. C-58 Investigation 4: Perpendicular Bisector of a Chord Construct a large circle and mark the center. Draw two nonparallel chords that are not diameters. Construct the perpendicular bisector of each chord and extend the bisectors until they intersect. Where is the point of intersection? Construct another circle that is larger or smaller than the first circle you drew. Draw two chords and construct their perpendicular bisectors. Where do the perpendicular bisectors intersect? Complete this conjecture. Perpendicular Bisector of a Chord Conjecture The perpendicular bisector of a chord passes through the of the circle. C-59 Now you can find the center of any circle and the vertex of the central angle of any arc. ll you have to do is construct the perpendicular bisectors of nonparallel chords. 8 CHPTER 6 Discovering Geometry Condensed Lessons

5 CONDENSED LESSON 6.3 rcs and ngles In this lesson you will Make conjectures about inscribed angles in a circle Investigate relationships among the angles in a cyclic quadrilateral Compare the arcs formed when two parallel lines intersect a circle In this lesson you ll discover properties of arcs and the angles associated with them. Investigation 1: Inscribed ngle Properties Refer to the diagram of circle O in your book. Notice that COR is a central angle, while CR is an inscribed angle. Both angles intercept CR. Recall that the measure of a minor arc is equal to the measure of its central angle. Find mcr by measuring COR. Then find the measure of CR. How does the measure of the inscribed angle compare with the measure of its intercepted arc? R Construct your own circle T. Draw an inscribed angle PQR and its corresponding central angle PTR. n example is shown at right. P T What is the measure of PR? What is the measure of PQR? How do the measures compare? Use your findings from this investigation to complete the conjecture. Q Inscribed ngle Conjecture The measure of an angle inscribed in a circle is the measure of its intercepted arc. C-60 Investigation : Inscribed ngles Intercepting the Same rc Follow Steps 1 and in your book. Note that selecting P and Q on the major arc creates two angles, PB and QB, that intercept the same minor arc. Draw another circle. Follow Step 4 in your book. Note that selecting P and Q on the minor arc creates two angles, PB and QB, that intercept the same major arc. P P Q B Q B In each case, how do the measures of the two inscribed angles compare? Use your findings to complete the conjecture. Inscribed ngles Intercepting rcs Conjecture Inscribed angles that intercept the same arc are. C-61 (continued) Discovering Geometry Condensed Lessons CHPTER 6 83

7 CONDENSED LESSON 6.4 Proving Circle Conjectures In this lesson you will Prove the Inscribed ngle Conjecture Prove other conjectures related to circles In Lesson 6.3, you discovered the Inscribed ngle Conjecture. The measure of an inscribed angle in a circle equals half the measure of its intercepted arc. In this lesson you will prove this conjecture. First, note that there are three ways an inscribed angle can relate to the circle s center. Case 1 The circle s center is on the angle. Case The center is outside the angle. Case 3 The center is inside the angle. These are the only possibilities, so if you can prove the conjecture is true for each case, you will have proved the conjecture is always true. Case 1: The circle s center is on the inscribed angle. Look at the drawing for Case 1 on page 330 of your book. The dashed radius O has been added to form OB. To prove the conjecture is true for this case, you need to show that BC 1 mc that is, x 1 mc. Notice the following things: OC is the central angle for C. (How are z and mc related?) OC is an exterior angle for OB. (How are x, y, and z related?) OB is isosceles. (How are x and y related?) Combine these observations to write a flowchart proof for Case 1. Then, compare your proof with the one in your book. Case : The circle s center is outside the inscribed angle. You can use Case 1 to help you prove Case. Look at the drawing for Case on page 331 of your book. dashed diameter, DB, has been added. To prove the conjecture is true for this case, you must show that m BC 1 mc q that is, x. Think about the following things: How are x, y, and z related? How are p, q, and mdc related? Using Case 1, how is y related to p? How is z related to p q? (continued) Discovering Geometry Condensed Lessons CHPTER 6 85

8 Lesson 6.4 Proving Circle Conjectures (continued) Combine these observations to write a flowchart proof for Case. Then compare your proof with the one in your book. Case 3: The circle s center is inside the inscribed angle. Use the given information and the marked figure in Example below to write a flowchart proof for this case. You may want to use your flowchart from Case as a guide for this proof. Write your proof before you look at the solution to Example. Then compare your proof with the one given. EXMPLE Prove Case 3 of the Inscribed ngle Conjecture: The measure of an inscribed angle in a circle equals half the measure of its intercepted arc when the center of the circle is inside the angle. Given: Circle O with inscribed angle BC whose sides lie on either side of diameter BD B y x O z C D Show: m BC 1 mc,or x 1 mc Solution 1 y _ 1 md Proof of Case 1 4 rc addition md mdc mc 7 _ 1 (md mdc) _ 1 mc Multiplication property of equality _ 1 md _ 1 mdc 1 _ mc Distributive property z _ 1 mdc Proof of Case 1 _ 1 _ 1 md mdc x Substitution 5 8 x _ 1 mc 3 y z x ngle addition 6 x m BC Given 9 Substitution _ 1 m BC mc Substitution EXMPLE B Write a paragraph proof or a flowchart proof of the conjecture: If a kite is inscribed in a circle, then the nonvertex angles are right angles. Given: Circle with inscribed kite BCD where B D and BC CD B Show: B and D are right angles C D Solution By the Kite ngles Conjecture, B D. By the Cyclic Quadrilateral Conjecture, B and D are supplementary. Because B and D are both equal and supplementary, they must have measures of 90. Therefore, B and D are right angles. 86 CHPTER 6 Discovering Geometry Condensed Lessons

9 CONDENSED LESSON 6.5 The Circumference/Diameter Ratio In this lesson you will Discover the relationship between the diameter and the circumference of a circle The distance around a polygon is called the perimeter. The distance around a circle is called the circumference. Think of a can of three tennis balls. Which do you think is greater, the height of the can or the circumference of the can? The height is about three tennis-ball diameters, which is about the same as three can diameters. If you have a tennis-ball can, wrap a string around the can to measure the circumference, and then compare the measurement with the height. You should find that the circumference of the can is slightly greater than the height. That is, the circumference is a little more than three diameters. In the next investigation you will discover the relationship between the diameter and circumference of any circle. Investigation: Taste of Pi For this investigation you will need several round objects, along with either a metric measuring tape or a string and a meterstick. Wrap the measuring tape or string around each object to measure its circumference. Then measure the diameter of each object. Make all measurements to the nearest millimeter (tenth of a centimeter). The table below includes some sample measurements. Copy the table and add your findings to it. Object Circumference (C) Diameter (d) Ratio C d Quarter 7.8 cm.5 cm Compact disc 37.7 cm 1.0 cm Mug 5.9 cm 8. cm Small plate 47.4 cm 15.1 cm Calculate the ratio C d for each object. You should find that, in each case, the ratio is a little more than 3. In fact, the ratio C d is exactly the same number for every circle. This constant ratio is denoted by the Greek letter (pi). So, C d. If you solve this equation for C, you get the formula for the circumference of a circle in terms of its diameter. Because the diameter of a circle is twice the radius, you can also write a formula for the circumference in terms of the radius. Circumference Conjecture If C is the circumference and d is the diameter of a circle, then there is a number such that C. If d r where r is the radius, then C. C-65 (continued) Discovering Geometry Condensed Lessons CHPTER 6 87

11 CONDENSED LESSON 6.6 round the World In this lesson you will Solve application problems involving radius, diameter, and circumference Now that you have learned about and the formula for circumference, you can solve many real-world problems involving circles. The example in your book shows the solution to a problem related to the novel round the World in Eighty Days, by Jules Verne. Read this example carefully. Then read the examples below. EXMPLE The wheel on Devin s unicycle has a diameter of 7 inches. 7 in. a. How far will the unicycle travel in 100 revolutions of the wheel? Give your answer to the nearest whole foot. b. Devin rides his unicycle miles to school each day. bout how many revolutions does the wheel make during his trip? (580 feet 1 mile) Solution a. In one revolution, the wheel covers a distance equal to its circumference. C d The formula for circumference. C (7) Substitute 7 for d. The unicycle travels 7 inches in one revolution. So, in 100 revolutions, it covers (100) inches, or about 707 feet. b. Two miles is 10,560 feet, or 16,70 inches. The wheel makes one revolution every 7 inches, so it makes 16,70 7, or about 1494 revolutions during Devin s trip to school. (continued) Discovering Geometry Condensed Lessons CHPTER 6 89

12 Lesson 6.6 round the World (continued) EXMPLE B n audio CD spins at a rate of 00 rotations per minute. If the radius of a CD is 6 cm, how far does a point on the outer edge travel during the playing of a 57-minute CD? 6 cm Solution The point makes 00 rotations in 1 minute. So, in 57 minutes, it will make 00 57, or 11,400 rotations. The distance the point travels in one rotation is equal to the circumference of the CD. Use the circumference formula to find this distance. C r The formula for circumference. C (6) Substitute 6 for r. The point travels 1 cm in one rotation. So, in 11,400 rotations, it will travel 11,400 1, or 49,770 cm. This is equal to about 4.3 kilometers. EXMPLE C The city of Rutledge wants to build a circular walking track that is a quarter-mile long. To keep vehicles off the track, the city plans to build a fence around the track. The fenced area will be square and the fence, at its closest point, will be 5 feet from the edge of the circular track. How many feet of fencing will be needed? Give your answer to the nearest foot and ignore any gates or openings. (1 mile 580 feet) Solution Refer to the sketch at right. Track 5 d 5 Fence First find the diameter of the track. The circumference of the track is 580 4, or 130 feet. Using C d, solve for d to get d C. So,d 130, or d 40. feet. Each side of the square fence is 10 feet longer than the diameter of the track. Therefore, each side of the square must be about 430. feet. The amount of fencing needed is about feet, or to the nearest foot, 171 feet. 90 CHPTER 6 Discovering Geometry Condensed Lessons

13 CONDENSED LESSON 6.7 rc Length In this lesson you will Learn the difference between arc length and arc measure Find a method for calculating arc length Solve problems involving arc length In the figure at right, B and CD have the same measure (45 ) because their central angles, PB and CPD, are the same angle. However, the lengths of B and CD are clearly different CD is longer than B. In this lesson you ll learn how to find the length of an arc. Read Example in your book. It illustrates that if an arc has measure x, x then the arc is 36 0 of the circle. For example, each of the 45 arcs in 45 the figure at right is 3 6, 0 or 1, 8 of a complete circle. Think about how the fractions in Example are related to the length of the arc. If you travel a quarter of the way around a circle, you cover 1 4 of its circumference. If you travel halfway around, you cover 1 the circumference. The length of an arc, or arc length, is a fraction of the circumference of its circle. While arc measure is expressed in degrees, arc length is expressed in a unit of distance. P 45 B C D Investigation: Finding the rcs Look at the figures in Step 1 in your book. Find the fraction of the circle each arc makes up. You should get the following results. B is 1 4 of circle T CED 1 is of circle O GH is , 0 or 1, 8 of circle P Now find the circumference of each circle using the formula C d or C r. For example, because circle P has radius 36 feet, its circumference is (36), or 7 feet. Use the fractions you found in Step 1 and the circumference of each circle to find the length of each arc. For example, GH 7 makes up 1 8 of circle P, and circle P has a circumference of 7. So, Length of GH 7 1 (7 ) 8 8 The length of GH is 8 feet. Generalize this method for finding the length of an arc as a conjecture. rc Length Conjecture The length of an arc equals the circumference times the arc measure divided by. C-66 (continued) Discovering Geometry Condensed Lessons CHPTER 6 91

14 Lesson 6.7 rc Length (continued) Examples B and C in your book show how you can apply the rc Length Conjecture. Read these examples carefully. Then read the example below. EXMPLE The length of SM is 6 centimeters. What is the radius of circle P? S M 0 P Solution m SM 0, so msm 40 by the Inscribed ngle Conjecture. This means that SM 40 is 3, or 1, 9 of the circumference of the circle. 60 rc length SM 1 9 C Use the rc Length Conjecture C Substitute 6 for the arc length. 54 C Multiply both sides by 9. The circumference is 54 centimeters. Use the circumference formula to find the radius. C r 54 r The circumference formula. Substitute 7 for the arc length. 7 r Divide both sides by. The radius is 7 centimeters. 9 CHPTER 6 Discovering Geometry Condensed Lessons

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