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Schaum's Outline of Geometry, 5th Edition: 665 Solved Problems + 25 Videos
Schaum's Outline of Geometry, 5th Edition: 665 Solved Problems + 25 Videos
Christopher Thomas, Barnett Rich
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Tough Test Questions? Missed Lectures? Not Enough Time?
Fortunately, theres Schaums. This allinonepackage includes more than 650 fully solved problems, examples, and practice exercises to sharpen your problemsolving skills. Plus, you will have access to 25 detailed videos featuring Math instructors who explain how to solve the most commonly tested problemsits just like having your own virtual tutor! Youll find everything you need to build confidence, skills, and knowledge for the highest score possible.
More than 40 million students have trusted Schaums to help them succeed in the classroom and on exams. Schaums is the key to faster learning and higher grades in every subject. Each Outline presents all the essential course information in an easytofollow, topicbytopic format. You also get hundreds of examples, solved problems, and practice exercises to test your skills.
This Schaums Outline gives you
 665 fully solved problems
 Concise explanations of all geometry concepts
 Support for all major textbooks for geometry courses
Fully compatible with your classroom text, Schaums highlights all the important facts you need to know. Use Schaums to shorten your study timeand get your best test scores!
Fortunately, theres Schaums. This allinonepackage includes more than 650 fully solved problems, examples, and practice exercises to sharpen your problemsolving skills. Plus, you will have access to 25 detailed videos featuring Math instructors who explain how to solve the most commonly tested problemsits just like having your own virtual tutor! Youll find everything you need to build confidence, skills, and knowledge for the highest score possible.
More than 40 million students have trusted Schaums to help them succeed in the classroom and on exams. Schaums is the key to faster learning and higher grades in every subject. Each Outline presents all the essential course information in an easytofollow, topicbytopic format. You also get hundreds of examples, solved problems, and practice exercises to test your skills.
This Schaums Outline gives you
 665 fully solved problems
 Concise explanations of all geometry concepts
 Support for all major textbooks for geometry courses
Fully compatible with your classroom text, Schaums highlights all the important facts you need to know. Use Schaums to shorten your study timeand get your best test scores!
Категории:
Год:
2012
Издание:
5
Издательство:
McGrawHill
Язык:
english
Страницы:
336
ISBN 10:
0071795405
ISBN 13:
9780071795401
Серии:
Schaum's Outline Series
Файл:
EPUB, 25,66 MB
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CHAPTER 6 Circles 6.1 The Circle; Circle Relationships The following terms are associated with the circle. Although some have been defined previously, they are repeated here for ready reference. A circle is the set of all points in a plane that are at the same distance from a fixed point called the center. The symbol for circle is [image: Image]; for circles [image: Image]. The circumference of a circle is the distance around the circle. It contains 360°. A radius of a circle is a line segment joining the center to a point on the circle. Note: Since all radii of a given circle have the same length, we may at times use the word radius to mean the number that is “the length of the radius.” A central angle is an angle formed by two radii. An arc is a continuous part of a circle. The symbol for arc is ⌒. A semicircle is an arc measuring onehalf the circumference of a circle. A minor arc is an arc that is less than a semicircle. A major arc is an arc that is greater than a semicircle. Thus in Fig. 61, [image: Image] is a minor arc and [image: Image] is a major arc. Three letters are needed to indicate a major arc. [image: Image] Fig. 61 To intercept an arc is to cut off the arc. Thus in Fig. 61, ∠BAC and ∠BOC intercept [image: Image]. A chord of a circle is a line segment joining two points of the circumference. Thus in Fig. 62, [image: Image] is a chord. [image: Image] Fig. 62 A diameter of a circle is a chord through the center. A secant of a circle is a line that intersects the circle at two points. A tangent of a circle is a line that touches the circle at one and only one point no matter how far produced. Thus, [image: Image] is a diameter of circle O in Fig. 62, [image: Image] is a secant, and [image: Image] is a tangent to the circle at P. P is the point of contact or the point of tangency. An inscribed polygon is a polygon all of whose sides are chords of a circle. A circumscribed circle is a circle passing through each vertex of a polygon. Thus ΔABD, ΔBCD, and quadrilateral; ABCD are inscribed polygons of circle O in Fig. 63. Circle O is a circumscribed circle of quadrilateral ABCD. [image: Image] Fig. 63 A circumscribed polygon is a polygon all of whose sides are tangents to a circle. An inscribed circle is a circle to which all the sides of a polygon are tangents. Thus, ΔABC is a circumscribed polygon of circle O in Fig. 64. Circle O is an inscribed circle of ΔABC. Concentric circlesare circles that have the same center. [image: Image] Fig. 64 Thus, the two circles shown in Fig. 65 are concentric circles. [image: Image] is a tangent of the inner circle and a chord of the outer one. [image: Image] is a secant of the inner circle and a chord of the outer one. [image: Image] Fig. 65 Two circles are equal if their radii are equal in length; two circles are congruent if their radii are congruent. Two arcs are congruent if they have equal degree measure and length. We use the notation m[image: Image] to denote “measure of arc AC.” 6.1A Circle Principles PRINCIPLE 1: A diameter divides a circle into two equal parts. Thus, diameter [image: Image] divides circle O of Fig. 66 into two congruent semicircles, [image: Image] and [image: Image]. PRINCIPLE 2: If a chord divides a circle into two equal parts, then it is a diameter. (This is the converse of Principle 1.) Thus if [image: Image] ≐ [image: Image] in Fig. 66, then [image: Image] is a diameter. [image: Image] Fig. 66 PRINCIPLE 3: A point is outside, on, or inside a circle according to whether its distance from the center is greater than, equal to, or smaller than the radius. F is outside circle O in Fig. 66, since [image: Image] is greater in length than a radius. E is inside circle O since [image: Image] is smaller in length than a radius. A is on circle O since [image: Image] is a radius. PRINCIPLE 4: Radii of the same or congruent circles are congruent. Thus in circle O of Fig. 67, [image: Image] ≐ [image: Image]. [image: Image] Fig. 67 PRINCIPLE 5: Diameters of the same or congruent circles are congruent. Thus in circle O of Fig. 67, [image: Image] ≐ [image: Image]. PRINCIPLE 6: In the same or congruent circles, congruent central angles have congruent arcs. Thus in circle O of Fig. 68, if ∠1 ≐ ∠2, then [image: Image] ≐ [image: Image]. [image: Image] Fig. 68 PRINCIPLE 7: In the same or congruent circles, congruent arcs have congruent central angles. Thus in circle O of Fig. 68, if [image: Image] ≐ [image: Image], then ∠1 ≐ ∠2. (Principles 6 and 7 are converses of each other.) PRINCIPLE 8: In the same or congruent circles, congruent chords have congruent arcs. Thus in circle O of Fig. 69, if [image: Image] ≐ [image: Image], then [image: Image] ≐ [image: Image]. [image: Image] Fig. 69 PRINCIPLE 9: In the same or congruent circles, congruent arcs have congruent chords. Thus in circle O of Fig. 69, if [image: Image] ≐ [image: Image], then [image: Image] ≐ [image: Image]. (Principles 8 and 9 are converses of each other.) PRINCIPLE 10: A diameter perpendicular to a chord bisects the chord and its arcs. Thus in circle O of Fig. 610, if [image: Image] [image: Image] [image: Image], then [image: Image] bisects [image: Image], [image: Image], and [image: Image]. A proof of this principle is given in Chapter 16. [image: Image] Fig. 610 PRINCIPLE 11: A perpendicular bisector of a chord passes through the center of the circle. Thus in circle O of Fig. 611, if [image: Image] is the perpendicular bisector of [image: Image], then [image: Image] passes through center O. [image: Image] Fig. 611 PRINCIPLE 12: In the same or congruent circles, congruent chords are equally distant from the center. Thus in circle O of Fig. 612, [image: Image] ≐ [image: Image], if [image: Image] [image: Image] [image: Image], and if [image: Image][image: Image] [image: Image], then [image: Image] ≐ [image: Image]. [image: Image] Fig. 612 PRINCIPLE 13: In the same or congruent circles, chords that are equally distant from the center are congruent. Thus in circle O of Fig. 612, if [image: Image] ≐ [image: Image], if [image: Image] [image: Image] [image: Image], and if [image: Image] [image: Image] [image: Image], then [image: Image] ≐ [image: Image]. (Principles 12 and 13 are converses of each other.) SOLVED PROBLEMS 6.1 Matching test of circle vocabulary Match each part of Fig. 613 on the left with one of the names on the right: [image: Image] [image: Image] Fig. 613 Solutions (a) 1 (b) 6 (c) 7 (d) 9 (e) 8 (f) 4 (g) 3 (h) 5 (i) 2 (j) 13 (k) 12 (l) 10 (m) 11 6.2 Applying principles 4 and 5 In Fig. 614, (a) what kind of triangle is OCD; (b) what kind of quadrilateral is ABCD? (c) In Fig. 615 if circle O = circle Q, what kind of quadrilateral is OAQB? Solutions Radii or diameters of the same or equal circles have equal lengths. (a) Since [image: Image] ≐ [image: Image], ΔOCD is isosceles. (b) Since diagonals [image: Image] and [image: Image] are equal in length and bisect each other, ABCD is a rectangle. (c) Since the circles are equal, [image: Image] ≐ [image: Image] ≐ [image: Image] ≐ [image: Image]and OAQB is a rhombus. [image: Image] Fig. 614 [image: Image] Fig. 615 6.3 Proving a circle problem [image: Image] PROOF: [image: Image] 6.4 Proving a circle problem stated in words Prove that if a radius bisects a chord, then it is perpendicular to the chord. Solutions [image: Image] PROOF: [image: Image] 6.2 Tangents The length of a tangent from a point to a circle is the length of the segment of the tangent from the given point to the point of tangency. Thus, PA is the length of the tangent from P to circle O in Fig. 616. [image: Image] Fig. 616 6.2A Tangent Principles PRINCIPLE 1: A tangent is perpendicular to the radius drawn to the point of contact. Thus if [image: Image] is a tangent to circle O at P in Fig. 617, and [image: Image] is drawn, then [image: Image] [image: Image] [image: Image]. [image: Image] Fig. 617 PRINCIPLE 2: A line is tangent to a circle if it is perpendicular to a radius at its outer end. Thus if [image: Image] [image: Image] radius [image: Image] at P of Fig. 617, then [image: Image] is tangent to circle O. PRINCIPLE 3: A line passes through the center of a circle if it is perpendicular to a tangent at its point of contact. Thus if [image: Image] is tangent to circle O at P in Fig. 618, and [image: Image] [image: Image] [image: Image] at P, then [image: Image] extended will pass through the center O. [image: Image] Fig. 618 PRINCIPLE 4: Tangents to a circle from an outside point are congruent. Thus if [image: Image] and [image: Image] are tangent to circle O at P and Q (Fig. 619), then [image: Image] ≐ [image: Image]. [image: Image] Fig. 619 PRINCIPLE 5: The segment from the center of a circle to an outside point bisects the angle between the tangents from the point to the circle. Thus [image: Image] bisects ∠PAQ in Fig. 619 if [image: Image] and [image: Image] are tangents to circle O. 6.2B Two Circles in Varying Relative Positions The line of centers of two circles is the line joining their centers. Thus, [image: Image] is the line of centers of circles O and O′ in Fig. 620. [image: Image] Fig. 620 Circles Tangent Externally Circles O and O′ in Fig. 621 are tangent externally at P. [image: Image] is the common internal tangent of both circles. The line of centers [image: Image] passes through P, is perpendicular to [image: Image], and is equal in length to the sum of the radii, R + r. Also [image: Image] bisects each of the common external tangents, [image: Image] and [image: Image] [image: Image] Fig. 621 Circles Tangent Internally Circles O and O′ in Fig. 622 are tangent internally at P. [image: Image] is the common external tangent of both circles. The line of centers [image: Image] if extended passes through P, is perpendicular to [image: Image], and is equal in length to the difference of the radii, R–r. [image: Image] Fig. 622 Overlapping Circles Circles O and O′ in Fig. 623 overlap. Their common chord is [image: Image]. If the circles are unequal, their (equal) common external tangents [image: Image] and [image: Image] meet at P. The line of centers [image: Image] ′ is the perpendicular bisector of [image: Image] and, if extended, passes through P. [image: Image] Fig. 623 Circles Outside Each Other Circles O and O′in Fig. 624 are entirely outside each other. The common internal tangents, [image: Image] and [image: Image] meet at P. If the circles are unequal, their common external tangents, [image: Image] and [image: Image] if extended, meet at P′. The line of centers [image: Image] passes through P and P′. Also, AB = CD and EF = GH. [image: Image] Fig. 624 SOLVED PROBLEMS 6.5 Triangles and quadrilaterals having tangent sides Points P, Q, and R in Fig. 625 are points of tangency. [image: Image] Fig. 625 (a) In Fig. 625(a), if AP = OP, what kind of triangle is OPA? (b) In Fig. 625(b), if AP = PQ, what kind of triangle is APQ? (c) In Fig. 625(b), if AP = OP, what kind of quadrilateral is OPAQ? (d) In Fig. 625(c), if [image: Image][image: Image] [image: Image], what kind of quadrilateral is PABR? Solutions (a) [image: Image] is tangent to the circle at P; then by Principle 1, ∠OPA is a right angle. Also, AP = OP. Hence, ΔOAP is an isosceles right triangle. (b) [image: Image] and [image: Image] are tangents from a point to the circle; hence by Principle 4, AP = AQ. Also, AP = PQ. Then ΔAPQ is an equilateral triangle. (c) By Principle 4, AP = AQ. Also, [image: Image] and [image: Image] are ≐ radii. And AP = OP. By Principle 1, ∠APO is a rt. ∠. Then AP = AQ = OP = OQ; hence, OPAQ is a rhombus with a right angle, or a square. (d) By Principle 1, [image: Image] [image: Image] [image: Image] and [image: Image] [image: Image] [image: Image]. Then [image: Image]  [image: Image], since both are [image: Image] to [image: Image]. By Principle 1, [image: Image] [image: Image] [image: Image] also, [image: Image] [image: Image] [image: Image] (Given). Then [image: Image]  [image: Image], since both are [image: Image] to [image: Image] Hence, PABR is a parallelogram with a right angle, or a rectangle. 6.6 Applying principle 1 (a) In Fig. 626(a), [image: Image] is a tangent. Find ∠A if m∠A: m∠O = 2:3. (b) In Fig. 626(b), [image: Image] and [image: Image] are tangents. Find m∠1 if m∠O = 140°. (c) In Fig. 626(c), [image: Image] and [image: Image] are tangents. Find m∠2 and m∠3 if ∠ OPD is trisected and [image: Image] is a diameter. Solutions (a) By Principle 1, m∠P = 90°. Then m∠A + m∠ O = 90°. If m∠A = 2x and m∠ O = 3x, then 5x = 90 and x = 18. Hence, m∠A = 36°. (b) By Principle 1, m∠P = m∠Q = 90°. Since m∠P + m∠Q + m∠ O + m∠A = 360°, m∠A + m∠O = 180°. Since m∠O = 140°, m∠A = 40°. By Principle 5, m∠ 1 = ½ m∠A = 20°. (c) By Principle 1, m∠DPQ = m∠PQC = 90°. Since m∠1 = 30°, m∠2 = 60°. Since ∠3 is an exterior angle of ΔPBQ, m∠3 = 90° + 60°= 150°. [image: Image] Fig. 626 6.7 Applying principle 4 (a) [image: Image], [image: Image] and [image: Image] in Fig. 627(a) are tangents. Find y. (b) Δ ABC in Fig. 627(b) is circumscribed. Find x. (c) Quadrilateral ABCD in Fig. 627(c) is circumscribed. Find x. Solutions (a) By Principle 4, AR = 6, and RB = y. Then RB = AB–AR = 14–6 = 8. Hence, y = RB = 8. (b) By Principle 4, PC = 8, QB = 4, and AP = AQ. Then AQ = AB–QB = 11. Hence, x = AP + PC = 11 + 8 = 19. (c) By Principle 4, AS = 10, CR = 5, and RD = SD. Then RD = CD–CR = 8. Hence, x = AS + SD = 10 + 8 = 18. [image: Image] Fig. 627 6.8 Finding the line of centers Two circles have radii of 9 and 4, respectively. Find the length of their line of centers (a) if the circles are tangent externally, (b) if the circles are tangent internally, (c) if the circles are concentric, (d) if the circles are 5 units apart. (See Fig. 628.) [image: Image] Fig. 628 Solutions Let R = radius of larger circle, r = radius of smaller circle. (a) Since R = 9 and r = 4, OO′ = R + r = 9 + 4 = 13. (b) Since R = 9 and r = 4, OO′ = R–r = 9–4 = 5. (c) Since the circles have the same center, their line of centers has zero length. (d) Since R = 9, r = 4, and d = 5, OO′ = R + d + r = 9 + 5 + 4 = 18. 6.9 Proving a tangent problem stated in words [image: Image] PROOF: [image: Image] 6.3 Measurement of Angles and Arcs in a Circle A central angle has the same number of degrees as the arc it intercepts. Thus, as shown in Fig. 629, a central angle which is a right angle intercepts a 90° arc; a 40° central angle intercepts a 40° arc, and a central angle which is a straight angle intercepts a semicircle of 180°. Since the numerical measures in degrees of both the central angle and its intercepted arc are the same, we may restate the above principle as follows: A central angle is measured by its intercepted arc. The symbol = may be used to mean “is measured by.” (Do not say that the central angle equals its intercepted arc. An angle cannot equal an arc.) [image: Image] Fig. 629 An inscribed angle is an angle whose vertex is on the circle and whose sides are chords. An angle inscribed in an arc has its vertex on the arc and its sides passing through the ends of the arc. Thus, ∠A in Fig. 630 is an inscribed angle whose sides are the chords [image: Image] and [image: Image]. Note that ∠A intercepts [image: Image] and is inscribed in [image: Image]. [image: Image] Fig. 630 6.3A AngleMeasurement Principles PRINCIPLE 1: A central angle is measured by its intercepted arc. PRINCIPLE 2: An inscribed angle is measured by onehalf its intercepted arc. A proof of this principle is given in Chapter 16. PRINCIPLE 3: In the same or congruent circles, congruent inscribed angles have congruent intercepted arcs. Thus in Fig. 631, if ∠1 ≐ ∠2, then [image: Image] ≐ [image: Image]. [image: Image] Fig. 631 PRINCIPLE 4: In the same or congruent circles, inscribed angles having congruent intercepted arcs are congruent. (This is the converse of Principle 3.) Thus in Fig. 631, if [image: Image] ≐ [image: Image], then ∠1 ≐ ∠2. PRINCIPLE 5: Angles inscribed in the same or congruent arcs are congruent. Thus in Fig. 632, if ∠C and ∠D are inscribed in [image: Image] then ∠C ≐ ∠D. [image: Image] Fig. 632 PRINCIPLE 6: An angle inscribed in a semicircle is a right angle. Thus in Fig. 633, since ∠C is inscribed in semicircle [image: Image], m∠C = 90°. [image: Image] Fig. 633 PRINCIPLE 7: Opposite angles of an inscribed quadrilateral are supplementary. Thus in Fig. 634, if ABCD is an inscribed quadrilateral, ∠ A is the supplement of ∠ C. [image: Image] Fig. 634 PRINCIPLE 8: Parallel lines intercept congruent arcs on a circle. Thus in Fig. 635, if [image: Image]  [image: Image], then [image: Image] ≐ [image: Image]. If tangent [image: Image] is parallel to [image: Image] then [image: Image] ≐ [image: Image]. [image: Image] Fig. 635 PRINCIPLE 9: An angle formed by a tangent and a chord is measured by onehalf its intercepted arc. PRINCIPLE 10: An angle formed by two intersecting chords is measured by onehalf the sum of the intercepted arcs. PRINCIPLE 11: An angle formed by two secants intersecting outside a circle is measured by onehalf the difference of the intercepted arcs. PRINCIPLE 12: An angle formed by a tangent and a secant intersecting outside a circle is measured by onehalf the difference of the intercepted arcs. PRINCIPLE 13: An angle formed by two tangents intersecting outside a circle is measured by onehalf the difference of the intercepted arcs. Proofs of Principles 10 to 13 are given in Chapter 16. [image: Image] Fig. 636 6.3B Table of AngleMeasurement Principles [image: Image] [image: Image] SOLVED PROBLEMS 6.10 Applying principles 1 and 2 (a) In Fig. 637(a), if m∠y = 46°, find m∠x. (b) In Fig. 637(b), if m∠y = 112°, find m∠x. (c) In Fig. 637(c), if m∠x = 75°, find m[image: Image]. Solutions (a) ∠y ≐ [image: Image], so m[image: Image] = 46°. Then ∠x ≐ [image: Image][image: Image] = [image: Image] (46°) = 23°, so m∠x = 23°. (b) ∠y ≐ m[image: Image] so m[image: Image] = 112°. m[image: Image] = m([image: Image]–[image: Image]) = 180°–112° = 68°. Then ∠x ≐ [image: Image] [image: Image] = [image: Image](68°) = 34°, so m∠x = 34°. (c) ∠x ≐ [image: Image] [image: Image], so m[image: Image] = 150°. Then m[image: Image] = m([image: Image] [image: Image]) = 150°–60° = 90°. [image: Image] Fig. 637 6.11 Applying principles 3 to 8 Find x and y in each part of Fig. 638. [image: Image] Fig. 638 Solutions (a) Since m∠1 = m∠2, m[image: Image] = m[image: Image] = 50°. Since [image: Image] ≐ [image: Image], m∠y = m∠ABD = 65°. (b) ∠ABD and ∠x are inscribed in [image: Image]; hence, m∠x = m∠ABD = 40°. ABCD is an inscribed quadrilateral; hence, m∠y = 180°–m∠B = 95°. (c) Since ∠x is inscribed in a semicircle, m∠x = 90°. Since [image: Image]  [image: Image], m[image: Image] = m[image: Image] = 70°. 6.12 Applying principle 9 In each part of Fig. 639, CD is a tangent at P. (a) If m[image: Image] = 220° in part (a), find m∠x. (b) If m[image: Image] = 140° in part (b), find m∠x. (c) If m∠y = 75° in part (c), find m∠x. Solutions (a) ∠z ≐ [image: Image] [image: Image] = [image: Image](220°) = 110°. So m∠x = 180°–110° = 70°. (b) Since AB = AP, m[image: Image] = m[image: Image] = 140°. Then m[image: Image] = 360°–140°–140° = 80°. Since ∠x ≐ [image: Image] [image: Image] = 40°, m∠ x = 40°. (c) ∠y ≐ [image: Image] [image: Image], so m[image: Image] = 150°. Then m[image: Image] = 360°–100°–150° = 110°. Since ∠x ≐ [image: Image] [image: Image] = 55°, m∠x = 55°. [image: Image] Fig. 639 6.13 Applying Principle 10 (a) If m∠x = 95° in Fig. 640(a), find m[image: Image]. (b) If m[image: Image] = 80° in Fig. 640(b), find m∠x. (c) If m[image: Image] = 78° in Fig. 640(c), find m∠y. [image: Image] Fig. 640 Solutions (a) ∠x ≐ [image: Image]([image: Image] + [image: Image]); thus 95° = [image: Image](70° + m[image: Image]), so m[image: Image] = 120°. (b) [image: Image] Then m∠x = 180°–m∠z = 80°. (c) Because [image: Image]  [image: Image], m[image: Image] = m[image: Image] = 78°. Also, ∠z ≐ [image: Image]([image: Image] + [image: Image]) = 78°. Then m∠y = 180°–m ∠z = 102° 6.14 Applying principles 11 to 13 (a) If m∠x = 40° in Fig. 641(a), find m[image: Image]. (b) If m∠x = 67° in Fig. 641(b), find m[image: Image]. (c) If m∠x = 61° in Fig. 641(c), find m[image: Image]. Solutions (a) ∠x ≐ [image: Image]([image: Image]–[image: Image]), so 40° = [image: Image](200°–m[image: Image]) or m[image: Image] = 120°. (b) ∠x ≐ [image: Image]([image: Image]–[image: Image]), so 67° = [image: Image](200°–m[image: Image]) or m[image: Image] = 66°. Then m[image: Image] = 360°–200°–66° = 94°. (c) ∠x ≐ [image: Image]([image: Image]–[image: Image] and m[image: Image] = 360°–m[image: Image]. Then 61° = [image: Image][(360°m[image: Image])–m[image: Image]] = 180°–m[image: Image]. Thus m[image: Image] = 119°. [image: Image] Fig. 641 6.15 Using equations in two unknowns to find arcs In each part of Fig. 642, find x and y using equations in two unknowns. Solutions (a) By Principle 10, 70° = [image: Image](m[image: Image] + m[image: Image]) By Principle 11, 25° = [image: Image](m[image: Image]–m[image: Image]) If we add these two equations, we get m[image: Image] = 95°. If we subtract one from the other, we get m[image: Image] = 45°. (b) Since m[image: Image] + m[image: Image] = 360°, [image: Image]m[image: Image]+m[image: Image]) = 180° By Principle 13, [image: Image]m[image: Image]+m[image: Image]) = 62° If we add these two equations, we find that m[image: Image] = 242°. If we subtract one from the other, we get m[image: Image] = 118°. [image: Image] Fig. 642 6.16 Measuring angles and arcs in general Find x and y in each part of Fig. 643. Solutions (a) By Principle 2, 50° = [image: Image]m[image: Image] or [image: Image]m[image: Image] = 100°. Also, by Principle 9, 70° = [image: Image]m[image: Image] or m[image: Image] = 140°. Then m[image: Image] = 360°–m[image: Image]–m[image: Image] = 120°. By Principle 9, x = [image: Image]m[image: Image] = 60°. By Principle 13, y = [image: Image](m[image: Image]–m[image: Image]) = [image: Image](260°–100°) = 80°. (b) By Principle 1, m[image: Image] = 80°. Also, by Principle 8, m[image: Image] = m[image: Image] = 85°. Then m[image: Image] = 360°–m[image: Image]–m[image: Image]–m[image: Image] = 110°. By Principle 9, x = [image: Image] m [image: Image] = 55°. By Principle 12, y = [image: Image](m[image: Image]–m[image: Image] = [image: Image](195°–85°) = 55°. [image: Image] Fig. 643 6.17 Proving an angle measurement problem [image: Image] PROOF: [image: Image] 6.18 Proving an angle measurement problem stated in words Prove that parallel chords drawn at the ends of a diameter are equal in length. Solutions [image: Image] PROOF: [image: Image] SUPPLEMENTARY PROBLEMS 6.1. Provide the proofs requested in Fig. 644. (6.3) (a) [image: Image] (b) [image: Image] (c) [image: Image] [image: Image] Fig. 644 6.2. Provide the proofs requested in Fig. 645. Please refer to figure 645(a) for problems 6.2(a) and (b); to figure 645(b) for problems 6.2(c) and (d); and figure 645(c) for problems 6.2(e) and (f). (6.3) [image: Image] [image: Image] [image: Image] [image: Image] Fig. 645 6.3. Prove each of the following: (6.4) (a) If a radius bisects a chord, then it bisects its arcs. (b) If a diameter bisects the major arc of a chord, then it is perpendicular to the chord. (c) If a diameter is perpendicular to a chord, it bisects the chord and its arcs. 6.4. Prove each of the following: (6.4) (a) A radius through the point of intersection of two congruent chords bisects an angle formed by them. (b) If chords drawn from the ends of a diameter make congruent angles with the diameter, the chords are congruent. (c) In a circle, congruent chords are equally distant from the center of the circle. (d) In a circle, chords that are equally distant from the center are congruent. 6.5. Determine each of the following, assuming t, t′, and t′′ in Fig. 646 are tangents. (6.5) (a) If m∠A = 90° in Fig. 646(a), what kind of quadrilateral is PAQO? (b) If BR = RC in Fig. 646(b), what kind of triangle is ABC? (c) What kind of quadrilateral is PABQ in Fig. 646(c) if [image: Image] is a diameter? (d) What kind of triangle is AOB in Fig. 646(c)? [image: Image] Fig. 646 6.6. In circle O, radii [image: Image] and [image: Image] are drawn to the points of tangency of [image: Image] and [image: Image]. Find m∠AOB if m∠APB equals (a) 40°; (b) 120°; (c) 90°; (d) x°; (e) (180–x)°; (f) (90–x)°. (6.6) 6.7. Find each of the following (t and t′ in Fig. 647 are tangents). [image: Image] Fig. 647 In Fig. 647(a) (6.6) (a) If m∠POQ = 80°, find m∠PAQ. (b) If m∠PBO = 25°, find m∠1 and m∠PAQ. (c) If m∠PAQ = 72°, find m∠1 and m∠PBO. In Fig. 647(b) (d) If [image: Image] bisects ∠APQ, find m∠2. (e) If m ∠1 = 35°, find m∠2. (f) If PQ = QB, find m∠1. 6.8. In Fig. 648(a), ΔABC is circumscribed. (a) If y = 9, find x. (b) If x = 25, find y. (6.7) In Fig. 648(b), quadrilateral ABCD is circumscribed. (c) Find AB + CD. (d) Find perimeter of ABCD. In Fig. 648(c), quadrilateral ABCD is circumscribed. (e) If r = 10, find x. (f) If x = 25, find r. [image: Image] Fig. 648 6.9. If two circles have radii of 20 and 13, respectively, find their line of centers: (6.8) (a) If the circles are concentric (b) If the circles are 7 units apart (c) If the circles are tangent externally (d) If the circles are tangent internally 6.10. If the line of centers of two circles measures 30, what is the relation between the two circles: (6.8) (a) If their radii are 25 and 5? (b) If their radii are 35 and 5? (c) If their radii are 20 and 5? (d) If their radii are 25 and 10? 6.11. What is the relation between two circles if the length of their line of centers is (a) 0; (b) equal to the difference of their radii; (c) equal to the sum of their radii; (d) greater than the sum of their radii, (e) less than the difference of their radii and greater than 0; (f) greater than the difference and less than the sum of their radii? (6.8) 6.12. Prove each of the following: (6.9) (a) The line from the center of a circle to an outside point bisects the angle between the tangents from the point to the circle. (b) If two circles are tangent externally, their common internal tangent bisects a common external tangent. (c) If two circles are outside each other, their common internal tangents are congruent. (d) In a circumscribed quadrilateral, the sum of the lengths of the two opposite sides equals the sum of the lengths of the other two. 6.13 Find the number of degrees in a central angle which intercepts an arc of (a) 40°; (b) 90°; (c) 170°; (d) 180°; (e) 2x°; (f) (180–x)°; (g) (2x–2y)°. (6.10) 6.14. Find the number of degrees in an inscribed angle which intercepts an arc of (a) 40°; (b) 90°; (c) 170°; (d) 180°; (e) 260°; (f) 348°; (g) 2x°; (h) (180–x)°; (i) (2x–2y)°. (6.10) 6.15. Find the number of degrees in the arc intercepted by (6.10) (a) A central angle of 85° (b) An inscribed angle of 85° (c) A central angle of c° (d) An inscribed angle of i° (e) The central angle of a triangle formed by two radii and a chord equal to a radius (f) The smallest angle of an inscribed triangle whose angles intercept arcs in the ratio of 1:2:3 6.16. Find the number of degrees in each of the arcs intercepted by the angles of an inscribed triangle if the measures of these angles are in the ratio of (a) 1:2:3; (b) 2:3:4; (c) 5:6:7; (d) 1:4:5. 6.17. (a) If m[image: Image] = 40° in Fig. 649(a), find m∠x. (b) If m∠x = 165° in Fig. 649(a), find m[image: Image]. (c) If m∠y = 115° in Fig. 649(b), find m∠x. (d) If m∠x = 108° in Fig. 649(b), find m∠y. (e) If m[image: Image] = 105° in Fig. 649(c), find m∠x. (f) If m∠x = 96° in Fig. 649(c), find m[image: Image]. [image: Image] Fig. 649 6.18. If quadrilateral ABCD is inscribed in a circle in Fig. 650, find (6.11) (a) m∠A if m∠C = 45° (b) m∠B if m∠D = 90° (c) m∠C if m∠A = x° (d) m∠D if m∠B = (90–x)° (e) m∠A if m[image: Image] = 160° (f) m∠B if m[image: Image] = 200° (g) m∠C if m[image: Image] = 140° and m[image: Image] = 110° (h) m∠D if m∠D: m∠B = 2:3 [image: Image] Fig. 650 6.19. If BC and AD are the parallel sides of inscribed trapezoid ABCD in Fig. 651, find (a) m[image: Image] if m[image: Image] = 85° (b) m[image: Image] if m[image: Image] = y° (c) m[image: Image] if m[image: Image] = 60° and m[image: Image] = 80° (d) m[image: Image] if m[image: Image] + m[image: Image] = 170° (e) m∠A if m∠D = 72° (f) m∠A if m∠C = 130° (g) m∠B if m∠C = 145° (h) m∠B if m[image: Image] = 90°andm[image: Image] = 84° [image: Image] Fig. 651 6.20. A diameter is parallel to a chord. Find the number of degrees in an arc between the diameter and chord if the chord intercepts (a) a minor arc of 80°; (b) a major arc of 300°. 6.21. Find x and y in each part of Fig. 652. [image: Image] Fig. 652 6.22. Find the number of degrees in the angle formed by a tangent and a chord drawn to the point of tangency if the intercepted arc has measure (a) 38°; (b) 90°; (c) 138°; (d) 180°; (e) 250°; (f) 334°; (g) x°; (h) (360–x)°; (i) (2x + 2y)°. (6.12) 6.23. Find the number of degrees in the arc intercepted by an angle formed by a tangent and a chord drawn to the point of tangency if the angle measures (a) 55°; (b) 67[image: Image]°; (c) 90°; (d) 135°; (e) (90–x)°; (f) (180–x)°; (g) (x–y)°; (h) 3[image: Image]x°. (6.12) 6.24. Find the number of degrees in the acute angle formed by a tangent through one vertex and an adjacent side of an inscribed (a) square; (b) equilateral triangle; (c) regular hexagon; (d) regular decagon. (6.12) 6.25. Find x and y in each part of Fig. 653 (t and t′ are tangents). [image: Image] Fig. 653 6.26. If [image: Image] and [image: Image] are chords intersecting in a circle as shown in Fig. 654, find (6.13) (a) m∠x if m[image: Image] = 90° and m[image: Image] = 60° (b) m∠x if m[image: Image] and m[image: Image] each equals 75° (c) m∠x if m[image: Image] + m[image: Image] = 230° (d) m∠x if m[image: Image] + m[image: Image] = 160° (e) m[image: Image] + m[image: Image] if m∠x = 70° (f) m[image: Image] + m[image: Image] if m∠x = 65° (g) m[image: Image] if m∠x = 60° and m∠AD = 160° (h) m[image: Image] if m∠y = 72° and m[image: Image] = 2m[image: Image] [image: Image] Fig. 654 6.27. If [image: Image] and [image: Image] are diagonals of an inscribed quadrilateral ABCD as shown in Fig. 655, find (6.13) (a) m∠1 if m[image: Image] = 95° and m[image: Image] = 75° (b) m∠1 if m[image: Image] = 88° and m[image: Image] = 66° (c) m∠1 if m[image: Image] and m[image: Image] = 100° (d) m∠1 if m[image: Image]:m[image: Image]:m[image: Image]:m[image: Image] = 1:2:3:4 (e) m∠2 if m[image: Image] + m[image: Image] = m[image: Image] + m[image: Image] (f) m∠2 if m[image: Image]  m[image: Image] and m[image: Image] = 70° (g) m∠2 if [image: Image] is a diameter and m[image: Image] = 80° (h) m∠2 if ABCD is a rectangle and mm[image: Image] = 70° [image: Image] Fig. 655 6.28. Find x and y in each part of Fig. 656. (6.13) [image: Image] Fig. 656 6.29. If [image: Image] and [image: Image] are intersecting secants as shown in Fig. 657, find (6.14) (a) m∠A if m[image: Image] = 100° and m[image: Image] = 40° (b) m∠A if m[image: Image]–mm[image: Image] = 74° (c) m∠A if m[image: Image] = mm[image: Image] + 40° (d) m∠A if m[image: Image]: m[image: Image]: m[image: Image]:m[image: Image] = 1:4:3:2 (e) m[image: Image] if m[image: Image] = 160° and m∠A = 20° (f) m[image: Image] if m[image: Image] = 60° and m∠A = 35° (g) m[image: Image]–m[image: Image] if m∠A = 47° (h) m[image: Image] if m[image: Image] = 3m[image: Image] and m∠A = 25° [image: Image] Fig. 657 6.30. If tangent [image: Image] and secant [image: Image] intersect as shown in Fig. 658, find (6.14) (a) m∠A if m[image: Image] = 150° and m[image: Image] = 60° (b) m∠A if m[image: Image] = 200° and m[image: Image] = 110° (c) m∠A if m[image: Image] = 120° and m[image: Image] = 70° (d) m∠A if m[image: Image]–m[image: Image] = 73° (e) m∠A if m[image: Image]:m[image: Image]:m[image: Image] = 1:4:7 (f) m[image: Image] if m[image: Image] = 220° and m∠A = 40° (g) m[image: Image] if m[image: Image] = 55° and m∠A = 30° (h) m[image: Image] if m[image: Image] = 3m[image: Image] and m∠A = 45° (i) m[image: Image] if m[image: Image] = 100° and m∠A = 50° [image: Image] Fig. 658 6.31. If [image: Image] and [image: Image] are intersecting tangents as shown in Fig. 659, find (6.14) (a) m∠A if m[image: Image] = 200° (b) m∠A if m[image: Image] = 95° (c) m∠A if m[image: Image] = x° (d) m∠A if m[image: Image] = (90–x)° (e) m∠A if m[image: Image] = 3m[image: Image] (f) m∠A if m[image: Image] = m[image: Image] + 50° (g) m∠A if m[image: Image]–= m[image: Image] = 84° (h) m∠A if m[image: Image]:m[image: Image] = 5:1 (i) m∠A if m[image: Image]:m[image: Image] = 7:3; (j) m∠A if m[image: Image] = 5m[image: Image]–60° (k) m[image: Image] if m∠A = 35° (l) m[image: Image] if m∠A = y° (m) m[image: Image] if m∠A = 60° (n) m[image: Image] if m∠A = x° (o) m[image: Image] if m[image: Image] [image: Image]m[image: Image] [image: Image] Fig. 659 6.32. Find x and y in each part of Fig. 660 (t and t′ are tangents). (6.14) 6.33. If [image: Image] and [image: Image] are intersecting secants as shown in Fig. 661, find [image: Image] Fig. 660 (a) m[image: Image] if m∠1 = 80° and m∠A = 40° (b) m[image: Image] if m∠1 + m∠A = 150° (c) m[image: Image] if ∠1 and ∠A are supplementary (d) m[image: Image] if m∠1 = 95° and m∠A = 45° (e) m[image: Image] if m∠1–m∠A = 22[image: Image]° (f) m[image: Image] if m[image: Image] + m[image: Image] = 190° and m∠A = 50° [image: Image] Fig. 661 6.34. Find x and y in each part of Fig. 662 (t and t′ are tangents). (6.16) [image: Image] Fig. 662 6.35. If ABC is an inscribed triangle as shown in Fig. 663, find (6.16) [image: Image] Fig. 663 (a) m∠A if m[image: Image] = 110° and m[image: Image] = 200° (b) m∠A if [image: Image][image: Image][image: Image] and m[image: Image] = 102° (c) m∠A if [image: Image] is a diameter and m[image: Image] = 80° (d) m∠A if m[image: Image]:m[image: Image]:m[image: Image] = 3:1:2 (e) m∠A in [image: Image] is a diameter and m[image: Image]:m[image: Image] = 5:4 (f) m∠B if m[image: Image] = 208° (g) m∠B if m[image: Image] + m[image: Image] = 3m[image: Image] (h) m∠B if m[image: Image] = 75° and m[image: Image] = 2m[image: Image] (i) m∠[image: Image] if [image: Image][image: Image][image: Image] and m[image: Image] = 5m[image: Image] (j) m[image: Image] if m∠A:m∠B:m∠C = 5:4:3 6.36. If ABCP is an inscribed quadrilateral, [image: Image] a tangent, and m[image: Image] a secant in Fig. 664, find (6.16) (a) m∠1 if m[image: Image] = 94° and m[image: Image] = 54° (b) m∠2 if [image: Image] is a diameter (c) m∠3 if m[image: Image] = 250° (d) m∠3 if m∠ABC = 120° (e) m∠4 if m[image: Image] = 130° and m[image: Image] = 50° (f) m∠4 if [image: Image]  [image: Image] and m[image: Image] = 74° (g) m[image: Image] if [image: Image]  [image: Image] and m∠6 = 42° (h) m[image: Image] if [image: Image] is a diameter and m∠5 = 35° (i) m[image: Image] if [image: Image] [image: Image] [image: Image] and m∠2 = 57° (j) m[image: Image] if [image: Image] and [image: Image] are diameters and m∠5 = 41° (k) m[image: Image] if m∠1 = 95° and m[image: Image] = 95° (l) m∠CPA if m∠3 = 79° [image: Image] Fig. 664 6.37. Find x and y in each part of Fig. 665 (t and t′ are tangents). (6.16) [image: Image] Fig. 665 6.38. Find x and y in each part of Fig. 666. [image: Image] Fig. 666 6.39. Provide the proofs requested in Fig. 667. (6.17) [image: Image] [image: Image] [image: Image] [image: Image] Fig. 667 6.40. Prove each of the following: (6.18) (a) The base angles of an inscribed trapezoid are congruent. (b) A parallelogram inscribed in a circle is a rectangle. (c) In a circle, parallel chords intercept equal arcs. (d) Diagonals drawn from a vertex of a regular inscribed pentagon trisect the vertex angle. (e) If a tangent through a vertex of an inscribed triangle is parallel to its opposite side, the triangle is isosceles. CHAPTER 3 Congruent Triangles 3.1 Congruent Triangles Congruent figures are figures that have the same size and the same shape; they are the exact duplicates of each other. Such figures can be moved on top of one another so that their corresponding parts line up exactly. For example, two circles having the same radius are congruent circles. Congruent triangles are triangles that have the same size and the same shape. If two triangles are congruent, their corresponding sides and angles must be congruent. Thus, congruent triangles ABC and A′B′C′ in Fig. 31 have congruent corresponding sides ([image: Image], and [image: Image]) and congruent corresponding angles (∠A ≐ ∠A′, ∠B ≐ ∠B’, and ∠ C ≐ ∠ C’). [image: Image] Fig. 31 (Read ΔABC ≐ ΔA′B′C′ as “Triangle ABC is congruent to triangle Aprime, Bprime, Cprime.”) Note in the congruent triangles how corresponding equal parts may be located. Corresponding sides lie opposite congruent angles, and corresponding angles lie opposite congruent sides. 3.1A Basic Principles of Congruent Triangles PRINCIPLE 1: If two triangles are congruent, then their corresponding parts are congruent. (Corresponding parts of congruent triangles are congruent.) Thus if ΔABC ≐ ∠A′B′C′ in Fig. 32, then ∠A ≐ ∠A′, ∠B ≐ ∠B′, ∠ C ≐ ∠ C′, a = A′, b = b’, and c = C′. [image: Image] Fig. 32 Methods of Proving that Triangles are Congruent PRINCIPLE 2: (SideAngleSide, SAS) If two sides and the included angle of one triangle are congruent to the corresponding parts of another, then the triangles are congruent. Thus if b = b’, c = c’, and ∠A = ∠A’ in Fig. 33, then ΔABC = ΔA′B′C′. [image: Image] Fig. 33 PRINCIPLE 3: (AngleSideAngle, ASA) If two angles and the included side of one triangle are congruent to the corresponding parts of another, then the triangles are congruent. Thus if ∠A ≐ ∠A′, ∠ C ≐ ∠ C’, and b = b’ in Fig. 34, then ΔABC ≐ Δ A′B′C′. [image: Image] Fig. 34 PRINCIPLE 4: (SideSideSide, SSS) If three sides of one triangle are congruent to three sides of another, then the triangles are congruent. Thus if a = A′, b = b’, and c = c’ in Fig. 35, then ΔABC ≐ ∠A′B′C. [image: Image] Fig. 35 SOLVED PROBLEMS 3.1 Selecting congruent triangles From each set of three triangles in Fig. 36 select the congruent triangles and state the congruency principle that is involved. Solutions (a) ΔI ≐ ΔII, by SAS. In ΔIII, the right angle is not between 3 and 4. (b) ΔII ≐ ΔIII, by ASA. In ΔI, side 10 is not between 70° and 30°. (c) ΔI ≐ ΔII ≐ ΔIII by SSS. [image: Image] Fig. 36 3.2 Determining the reason for congruency of triangles In each part of Fig. 37, ΔI can be proved congruent of ΔII. Make a diagram showing the equal parts of both triangles and state the congruency principle that is involved. [image: Image] Fig. 37 Solutions (a) AC is a common side of both [image: Image] [Fig. 38(a)]. ΔI ≐ ΔII by ASA. (b) ∠1 and ∠2 are vertical angles [Fig. 38(b)]. ΔI ≐ ΔII by SAS. (c) BD is a common side of both [image: Image] [Fig. 38(c)]. ΔI ≐ ΔII by SSS. [image: Image] Fig. 38 3.3 Finding parts needed to prove triangles congruent State the additional parts needed to prove ΔI ≐ ΔII in the given figure by the given congruency principle. [image: Image] Fig. 39 (a) In Fig. 39(a) by SSS. (b) In Fig. 39(a) by SAS. (c) In Fig. 39(b) by ASA. (d) In Fig. 39(c) by ASA. (e) In Fig. 39(c) by SAS. Solutions (a) If [image: Image] ≐ [image: Image], then ΔI ≐ ΔII by SSS. (b) If ∠1 ≐ ∠4, then ΔI ≐ ΔII by SAS. (c) If [image: Image] ≐ [image: Image], then ΔI ≐ ΔII by ASA. (d) If ∠2 ≐ ∠3, then ΔI ≐ ΔII by ASA. (e) If [image: Image] ≐ [image: Image], then ΔI ≐ ΔII by SAS. 3.4 Selecting corresponding parts of congruent triangles In each part of Fig. 310, the equal parts needed to prove ΔI = ΔII are marked. List the remaining parts that are congruent. [image: Image] Fig. 310 Solutions Congruent corresponding sides lie opposite congruent angles. Congruent corresponding angles lie opposite congruent sides. (a) Opposite 45°, [image: Image] ≐ [image: Image], Opposite 80°, [image: Image] ≐ [image: Image]. Opposite the side of length 12; ∠C ≐ ∠ D. (b) Opposite [image: Image] and [image: Image] ∠3 ≐ ∠2. Opposite [image: Image] and [image: Image], ∠1 ≐ ∠4. Opposite common side [image: Image], ∠A ≐ ∠C. (c) Opposite [image: Image] and [image: Image], ∠2 ≐ ∠3. Opposite [image: Image] and [image: Image], ∠1 ≐ ∠4. Opposite ∠5 and ∠6, [image: Image] ≐ [image: Image]. 3.5 Applying algebra to congruent triangles In each part of Fig. 311, find x and y. [image: Image] Fig. 311 Solutions (a) Since ΔI ≐ ΔII, by SSS, corresponding angles are congruent. Hence, 2x = 24 or x = 12, and 3y = 60 or y = 20. (b) Since ΔI ≐ ΔII, by SSS, corresponding angles are congruent. Hence, x + 20 = 26 or x = 6, and y –5 = 42 or y = 47. (c) Since ΔI ≐ ΔII, by ASA, corresponding are congruent. Then 2x = 3y + 8 and x = 2y. Substituting 2y for x in the first of these equations, we obtain 2(2y) = 3y + 8 or y = 8. Then x = 2y = 16. 3.6 Proving a congruency problem [image: Image] PROOF: [image: Image] 3.7 Proving a congruency problem stated in words Prove that if the opposite sides of a quadrilateral are equal and a diagonal is drawn, equal angles are formed between the diagonal and the sides. Solution If the opposite sides of a quadrilateral are congruent and a diagonal is drawn, congruent angles are formed between the diagonal and the sides. [image: Image] PROOF: [image: Image] 3.2 Isosceles and Equilateral Triangles 3.2A Principles of Isosceles and Equilateral Triangles PRINCIPLE 1: If two sides of a triangle are congruent, the angles opposite these sides are congruent. (Base angles of an isosceles triangle are congruent.) Thus in ΔABC in Fig. 312, if [image: Image] ≐ [image: Image], then ∠A ≐ ∠ C. A proof of Principle 1 is given in Chapter 16. [image: Image] Fig. 312 PRINCIPLE 2: If two angles of a triangle are congruent, the sides opposite these angles are congruent. Thus in ΔABC in Fig. 313, if ∠A ≐ ∠C, then [image: Image] ≐ [image: Image]. [image: Image] Fig. 313 Principle 2 is the converse of Principle 1. A proof of Principle 2 is given in Chapter 16. PRINCIPLE 3: An equilateral triangle is equiangular. Thus in ΔABC in Fig. 314, if [image: Image] ≐ [image: Image] ≐ [image: Image], then ∠A ≐ ∠B ≐ ∠ C. [image: Image] Fig. 314 Principle 3 is a corollary of Principle 1. A corollary of a theorem is another theorem whose statement and proof follow readily from the theorem. PRINCIPLE 4: An equiangular triangle is equilateral. Thus in ΔABC in Fig. 315, if ∠A ≐ ∠B ≐ ∠ C, then [image: Image] ≐ [image: Image] ≐ [image: Image]. Principle 4 is the converse of Principle 3 and a corollary of Principle 2. [image: Image] Fig. 315 SOLVED PROBLEMS 3.8 Applying principles 1 and 3 In each part of Fig. 316, name the congruent angles that are opposite congruent sides of a triangle. [image: Image] Fig. 316 Solutions (a) Since [image: Image] ≐ [image: Image], ∠A ≐ ∠B. (b) Since [image: Image] ≐ [image: Image], ∠1 ≐ ∠2. Since [image: Image] ≐ [image: Image], ∠3 ≐ ∠4. (c) Since [image: Image] ≐ [image: Image] ≐ [image: Image], ∠A ≐ ∠1 ≐ ∠3. Since [image: Image] ≐ [image: Image], ∠2 ≐ ∠ D. (d) Since [image: Image] ≐ [image: Image] ≐ [image: Image], ∠A ≐ ∠ACB ≐ ∠ABC. Since [image: Image] ≐ [image: Image] ≐ [image: Image], ∠A ≐ ∠ D ≐ ∠ E. 3.9 Applying principles 2 and 4 In each part of Fig. 317, name the congruent sides that are opposite congruent angles of a triangle. [image: Image] Fig. 317 Solutions (a) Since m∠A = 55°, ∠A ≐ ∠ D. Hence, [image: Image] ≐ [image: Image]. (b) Since ∠A ≐ ∠1, [image: Image] ≐ [image: Image]. Since ∠2 ≐ ∠ C, [image: Image] ≐ [image: Image]. (c) Since ∠1 ≐ ∠3, [image: Image] ≐ [image: Image]. Since ∠2 ≐ ∠4 ≐ ∠ D, [image: Image] ≐ [image: Image] ≐ [image: Image]. (d) Since ∠A Δ ∠1 ≐ ∠4, [image: Image] ≐ [image: Image] ≐ [image: Image]. Since ∠2 ≐ ∠ C, [image: Image] ≐ [image: Image]. 3.10 Applying isosceles triangle principles In each of Fig. 318(a) and (b), ΔI can be proved congruent to ΔII. Make a diagram showing the congruent parts of both triangles and state the congruency principle involved. [image: Image] Fig. 318 Solutions (a) Since [image: Image] ≐ [image: Image], ∠A ≐ ∠ C. ΔI ≐ ΔII by SAS [see Fig. 319(a)]. (b) Since [image: Image] ≐ [image: Image], ∠B ≐ ∠ C. ΔI ≐ ΔII by ASA [see Fig. 319(b)]. [image: Image] Fig. 319 3.11 Proving an isosceles triangle problem [image: Image] PROOF: [image: Image] 3.12 Proving an isosceles triangle problem stated in words Prove that the bisector of the vertex angle of an isosceles triangle is a median to the base. Solution The bisector of the vertex angle of an isosceles triangle is a median to the base. [image: Image] PROOF: [image: Image] SUPPLEMENTARY PROBLEMS 3.1. Select the congruent triangles in (a) Fig. 320, (b) Fig. 321, and (c) Fig. 322, and state the congruency principle in each case. (3.1) [image: Image] Fig. 320 [image: Image] Fig. 321 [image: Image] Fig. 322 3.2. In each figure below, ΔI can be proved congruent to ΔII. State the congruency principle involved. (3.2) [image: Image] [image: Image] 3.3. State the additional parts needed to prove ΔI ≐ ΔII in the given figure by the given congruency principle. (3.3) [image: Image] Fig. 323 (a) In Fig. 323(a) by SSS. (b) In Fig. 323(a) by SAS. (c) In Fig. 323(b) by ASA. (d) In Fig. 323(b) by SAS. (e) In Fig. 323(c) by SSS. (f) In Fig. 323(c) by SAS. 3.4. In each part of Fig. 324, the congruent parts needed to prove ΔI ≐ ΔII are marked. Name the remaining parts that are congruent. (3.4) [image: Image] Fig. 324 3.5. In each part of Fig. 325, find x and y. (3.5) [image: Image] Fig. 325 3.6. Prove each of the following. (3.6) [image: Image] [image: Image] [image: Image] Fig. 326 [image: Image] Fig. 327 [image: Image] [image: Image] [image: Image] Fig. 328 [image: Image] Fig. 329 3.7. Prove each of the following: (3.7) (a) If a line bisects an angle of a triangle and is perpendicular to the opposite side, then it bisects that side. (b) If the diagonals of a quadrilateral bisect each other, then its opposite sides are congruent. (c) If the base and a leg of one isosceles triangle are congruent to the base and a leg of another isosceles triangle, then their vertex angles are congruent. (d) Lines drawn from a point on the perpendicular bisector of a given line to the ends of the given line are congruent. (e) If the legs of one right triangle are congruent respectively to the legs of another, their hypotenuses are congruent. 3.8. In each part of Fig. 330, name the congruent angles that are opposite sides of a triangle. (3.8) [image: Image] Fig. 330 3.9. In each part of Fig. 331, name the congruent sides that are opposite congruent angles of a triangle. (3.9) [image: Image] Fig. 331 3.10. In each part of Fig. 332, two triangles are to be proved congruent. Make a diagram showing the congruent parts of both triangles and state the reason for congruency. (3.10) [image: Image] Fig. 332 3.11. In each part of Fig. 333, ΔI, ΔII, and ΔIII can be proved congruent. Make a diagram showing the congruent parts and state the reason for congruency. (3.10) [image: Image] Fig. 333 3.12. Prove each of the following: (3.11) [image: Image] [image: Image] [image: Image] Fig. 334 [image: Image] Fig. 335 3.13. Prove each of the following: (3.12) (a) The median to the base of an isosceles triangle bisects the vertex angle. (b) If the bisector of an angle of a triangle is also an altitude to the opposite side, then the other two sides of the triangle are congruent. (c) If a median to a side of a triangle is also an altitude to that side, then the triangle is isosceles. (d) In an isosceles triangle, the medians to the legs are congruent. (e) In an isosceles triangle, the bisectors of the base angles are congruent. BARNETT RICH held a doctor of philosophy degree (PhD) from Columbia University and a doctor of jurisprudence (JD) from New York University. He began his professional career at Townsend Harris Hall High School of New York City and was one of the prominent organizers of the High School of Music and Art where he served as the Administrative Assistant. Later he taught at CUNY and Columbia University and held the post of chairman of mathematics at Brooklyn Technical High School for 14 years. Among his many achievements are the 6 degrees that he earned and the 23 books that he wrote, among them Schaum’s Outlines of Elementary Algebra, Modern Elementary Algebra, and Review of Elementary Algebra. CHRISTOPHER THOMAS has a BS from University of Massachusetts at Amherst and a PhD from Tufts University, both in mathematics. He first taught as a Peace Corps volunteer at the Mozano Senior Secondary School in Ghana. Since then he has taught at Tufts University, Texas A&M University, and the Massachusetts College of Liberal Arts. He has written Schaum’s Outline of Math for the Liberal Arts as well as other books on calculus and trigonometry. [image: Image] Copyright © 2013 by The McGrawHill Companies, Inc. All rights reserved. Printed in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written permission of the publisher. ISBN: 9780071795418 MHID: 0071795413 The material in this eBook also appears in the print version of this title: ISBN: 9780071795401, MHID: 0071795405. 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THE WORK IS PROVIDED “AS IS.” McGRAWHILL AND ITS LICENSORS MAKE NO GUARANTEES OR WARRANTIES AS TO THE ACCURACY, ADEQUACY OR COMPLETENESS OF OR RESULTS TO BE OBTAINED FROM USING THE WORK, INCLUDING ANY INFORMATION THAT CAN BE ACCESSED THROUGH THE WORK VIA HYPERLINK OR OTHERWISE, AND EXPRESSLY DISCLAIM ANY WARRANTY, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO IMPLIED WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. McGrawHill and its licensors do not warrant or guarantee that the functions contained in the work will meet your requirements or that its operation will be uninterrupted or error free. Neither McGrawHill nor its licensors shall be liable to you or anyone else for any inaccuracy, error or omission, regardless of cause, in the work or for any damages resulting there from. McGrawHill has no responsibility for the content of any information accessed through the work. Under no circumstances shall McGrawHill and/or its licensors be liable for any indirect, incidental, special, punitive, consequential or similar damages that result from the use of or inability to use the work, even if any of them has been advised of the possibility of such damages. This limitation of liability shall apply to any claim or cause whatsoever whether such claim or cause arises in contract, tort or otherwise. CHAPTER 18 Transformations 18.1 Introduction to Transformations Two figures are congruent if one can be moved so that it exactly overlaps the other. A figure cut out of paper can be turned, slid, and flipped over to see if it matches up with another figure. If the figure is put on a graph, then these movements will change the coordinates of the points. A transformation is a way to describe such a change of coordinates. 18.2 Transformation Notation A transformation begins with a general description of a point, such as P(x, y) which represents a point P with coordinates x and y. Following this is an arrow [image: Image] and then a description of the point’s image, the place where it ends up after the move. Usually the image of P is called P′, the image of A is called A′, and so on. For example, the transformation P(x, y) [image: Image] P′(x–5, 4–y) means that the point A(2, 1) is moved to A′(2–5, 4–1) = A′(–3, 3), the point B(3, 5) is moved to B′(3–5, 4–5) = B′(–2,–1), and the point C(6, 1) is moved to C′(6–5, 4–1) = C′(1, 3). This transformation flips the triangle ΔABC over and slides it to the left, as shown in Fig. 181. [image: Image] Fig. 181 SOLVED PROBLEMS 18.1 Using transformation notation Name the image of the points A(3, 1), B(3, 4), and C(5, 1) under the following transformations: (a) P(x, y) [image: Image] P′(x + 2, y–1) (b) Q(x, y) [image: Image] Q′(x + 5, y) (c) R(x, y) [image: Image] R′(5x, 5y) (d) S(x, y) [image: Image] S′(–y, x) (e) T(x, y) [image: Image] T′ (y, 5–x) Solutions (a) A′(3 + 2, 1–1) = A′(5, 0), B′(3 + 2, 4–1) = B′(5, 3), and C(5 + 2, 1–1) = C(7, 0) (b) A′(3 + 5, 1) = A′(8, 1), B′(3 + 5, 4) = B′(8, 4), and C′ (5 + 5, 1) = C′ (10, 1) (c) A′(5 · 3, 5 · 1) = A′(15, 5), B′(5 · 3, 5 · 4) = B′(15, 20), and C′(5 · 5, 5 · 1) = C′(25, 5) (d) A′(–1, 3), B′(–4, 3), and C′(–1, 5) (e) A′(1, 5–3) = A′(1, 2), B′(4, 5–3) = B′(4, 2), and C′(1, 5–5) = C′(1, 0) 18.3 Translations A transformation that slides figures without flipping or rotating them is called a translation. The translation that slides everything to the right a units and up b units is P(x, y) [image: Image] P′(x + a, y + b). SOLVED PROBLEMS 18.2 Performing a translation Let rectangle ABCD be formed by A(–1, 4), B(–1, 3), C(3, 3), and D(3, 4). Graph rectangle ABCD and its image under the following translations: (a) P(x, y) [image: Image] P′(x + 4, y + 3) (b) P(x, y) [image: Image] P″(x + 2, y–5) (c) P(x, y) [image: Image] P′″(x–6, y–2) Solutions See Fig. 182. [image: Image] Fig. 182 (a) A′(–1 + 4, 4 + 3) = A′(3, 7), B′(–1 + 4, 3 + 3) = B′(3, 6), C′ (3 + 4, 3 + 3) = C′(7, 6), and D ′ (3 + 4, 4 + 3) = D ′ (7, 7) (b) A″(–1 + 2, 4–5) = A″(1,–1), B″ (–1 + 2, 3–5) = B″ (1,–2), C″ (3 + 2, 3–5) = C″ (5,–2), and D″ (3 + 2, 4–5) = D″ (5,–1) (c) A′″(–1–6, 4–2) = A′″(–7, 2), B′″ (–1–6, 3–2) = B′″ (–7, 1), C′″ (3–6, 3–2) = C”′ (–3, 1), and D′″ (3–6, 4–2) = D′″ (–3, 2) 18.3 Recognizing a translation Name the translation that takes ΔABC to (a) ΔA′ B′ C′, (b) ΔA″ B″ C″, and (c) ΔA′″ B′″ C′″ as illustrated in Fig. 183. [image: Image] Fig. 183 Solutions (a) P(x, y) [image: Image] P″(x–4, y + 2) (b) P(x, y) [image: Image] P″(x + 2, y–5) (c) P(x, y) [image: Image] P′″(x–6, y–3) 18.4 Naming translations Name the translation that moves everything: (a) Up 6 spaces (b) Down 1 space (c) To the right 2 spaces (d) To the left 10 spaces (e) Up 5 spaces and to the right 3 spaces (f) Down 7 spaces and to the right 4 spaces (g) 6 spaces to the left and 4 spaces up Solutions (a) P(x, y) [image: Image] P′(x, y + 6) (b) P(x, y) [image: Image] P′(x, y–1) (c) P(x, y) [image: Image] P′(x + 2, y) (d) P(x, y) [image: Image] P′(x–10, y) (e) P(x, y) [image: Image] P′(x + 3, y + 5) (f) P(x, y) [image: Image] P′(x + 4, y–7) (g) P(x, y) [image: Image] P′(x–6, y + 4) 18.4 Reflections A transformation that flips everything over is called a reflection. This is because the image of an object in a mirror looks flipped over, as illustrated in Fig. 184. [image: Image] Fig. 184 The reflection in Fig. 184 is a reflection across the yaxis because the edge of the mirror is pressed against the yaxis. The line where the mirror meets the plane is called the axis of symmetry. The reflection across the vertical line x = a is given by P(x, y) [image: Image] P′(2a–x, y). The reflection across the horizontal line y = a is given by P(x, y) [image: Image] P′(x, 2a–y). SOLVED PROBLEMS 18.5 Performing reflections Let triangle ABC be formed by A(–1, 1), B(0, 3), and C(3, 1). Graph ΔABC and its image under: (a) Reflection across the x axis (y = 0), P(x, y) [image: Image] P′(x,–y) (b) Reflection across the line x = 4, P(x, y) [image: Image] P″(8–x, y) (c) Reflection across the line y = 5, P(x, y) [image: Image] P′″(x, 10–y) Solutions See Fig. 185. [image: Image] Fig. 185 (a) A′(–1,–1), B′(0,–3), and C′(3,–1) (b) A″(8–(–1), 1) = A″(9, 1), B″(8–0, 3) = B″(8, 3), and C″(8–3, 1) = C″(5, 1) (c) A′″(–1, 10–1) = A′″(–1, 9), B′″(0, 10–3) = B′″(0, 7), and C′″(3, 10–1) = C′″(3, 9) 18.6 Recognizing reflections Name the reflection that takes ΔABC to (a) ΔA′ B′ C′, (b) ΔA″ B″ C″, and (c) ΔA′″ B′″ C′″ as illustrated in Fig. 186. [image: Image] Fig. 186 Solutions (a) Reflection across the y axis, P(x, y) [image: Image] P′(–x, y) (b) Reflection across the line y =–1, P(x, y) [image: Image] P″ (x,–2–y) (c) Reflection across the line x = 4, P(x, y) [image: Image] P′″(8–x, y) 18.7 Naming reflections Name the transformation that (a) Reflects across x = 2 (b) Reflects across y = 6 (c) Reflects across x =–10 (d) Reflects across y = [image: Image] Solutions (a) P(x, y) [image: Image] P′(4–x, y) (b) P(x, y) [image: Image] P′(x, 12–y) (c) P(x, y) [image: Image] P′(–20–x, y) (d) P(x, y) [image: Image] P′(x, 1–y) 18.4A Reflectional Symmetry A figure has reflectional symmetry if it looks the same after being flipped across an axis of symmetry that runs through its center. As illustrated in Fig. 187, a figure can have (a) one, (b) several, or (c) no axes of symmetry. [image: Image] Fig. 187 SOLVED PROBLEMS 18.8 Recognizing reflectional symmetry Which of the figures in Fig. 188 have reflectional symmetry? [image: Image] Fig. 188 Solutions Only (b), (e), and (f) have reflectional symmetry. Note that when (d) is flipped, it will look like Fig. 189, which is different from the original in that the upperlefthand crossing is horizontal instead of vertical. [image: Image] Fig. 189 18.5 Rotations If a pin were pushed through the origin on a graph and the paper were to be turned, the result would be a rotation about the origin. A rotation is described by the number of degrees by which the paper is turned. The 90° clockwise rotation (or 270° counterclockwise) about the origin is given by P(x, y) [image: Image] P′(y,–x). The 180° rotation about the origin is given by P(x, y) [image: Image] P′(x,–y). The 270° clockwise (or 90° counterclockwise) rotation about the origin is given by P(x, y) [image: Image] P′(–y, x). In general, the clockwise rotation about the origin of θ° is given by P(x, y) [image: Image] P′(x cos θ + y sin θ, y cos θ–x sin θ) SOLVED PROBLEMS 18.9 Performing rotations Let triangle ABC be given by A(2, 1), B(3, 1), and C(3, 4). Graph the image of ΔABC as rotated about the origin by (a) 90° clockwise, (b) 180°, and (c) 270° clockwise. Solutions See Fig. 1810. [image: Image] Fig. 1810 (a) A′(1,–2), B′(1,–3), and C′(4,–3) (b) A″ (–2,–1), B″(–3,–1), C″ (–3,–4) (c) A′″ (–1, 2), B′″(–1, 3), C′″(–4, 3) 18.10 Recognizing rotations Name the rotation that takes ΔABC to (a) ΔA′ B′ C′, (b) ΔA” B″ C″, and (c) ΔA′″ B′″ C′″ as illustrated in Fig. 1811. [image: Image] Fig. 1811 Solutions (a) 270° clockwise or 90° counterclockwise about the origin, P(x, y) [image: Image] P′(–y, x) (b) 180° about the origin (either clockwise or counterclockwise), P(x, y) [image: Image] P″(–x,–y) (c) 45° clockwise, P(x, y) [image: Image] P′″(xcos 45° + y sin 45°, y cos 45°–x sin 45°) = [image: Image] 18.11 Naming rotations Name the transformation that rotates clockwise about the origin: (a) 20° (b) 30° (c) 60° (d) 75° Solutions (a) P(x, y) [image: Image] P′(xcos 20° + y sin 20°, y cos 20°–x sin 20°)–P′(0.9397x + 0.3420y, 0.9397y–0.3420x) (b) P(x, y) [image: Image] P′(xcos 30° + ysin 30°, ycos 30°–xsin 30°) = [image: Image] = P(0.866x + 0.5y, 0.866y–0.5x) (c) P(x, y) [image: Image] P′(xcos60° + y sin 60°, y cos 60°–x sin 60°) = [image: Image] = P′(0.5x + 0.866y, 0.5y–0.866) (d) P(x, y) [image: Image] P′(xcos75° + y sin 75°, y cos 75°–x sin 75°) = P′(0.2588x + 0.9659y, 0.2588y–0.9659x) 18.5A Rotational Symmetry A figure has rotational symmetry if it can be rotated around its center by fewer than 360° and look the same as it did originally. In Fig. 1812, there is (a) a figure that looks the same under a 72° rotation, (b) a figure that looks the same under a 120° rotation, (c) a figure that looks the same under a 180°, and (d) a figure without rotational symmetry. [image: Image] Fig. 1812 SOLVED PROBLEMS 18.12 Recognizing rotational symmetry For each figure in Fig. 1813, give the smallest angle by which the figure could be rotated around its center and still look the same. [image: Image] Fig. 1813 Solutions (a) 90° (b) 120° (c) 360° (no rotational symmetry) (d) 180° (e) 360° (no rotational symmetry) (f) 90° 18.6 Rigid Motions Any combination of translations, reflections, and rotations is called a rigid motion because figures are moved without changing angles, lengths, or shapes. The image of a figure under a rigid motion will always be congruent to the original. SOLVED PROBLEMS 18.13 Graphing rigid motions Let triangle ABC be formed by A(–4, 2), B(–4, 1), and C(–1, 1). Graph ΔABC and its image under the following combinations of transformations: (a) Reflect across the y axis and then move to the right 4 spaces. (b) Rotate 90° clockwise around the origin then move up 3 spaces. (c) Reflect across y = 2 then move up 2 spaces and to the left 3 spaces. (d) Reflect across the x axis and then reflect across the y axis. (e) Rotate 90° counterclockwise around the origin and then reflect across x =–3. Solutions See Fig. 1814. [image: Image] Fig. 1814 18.14 Combining transformations Name the single transformation that does the same thing as the combination of: (a) P(x, y) [image: Image] P′(x + 7, y–2) and then Q′(x, y) [image: Image] Q″(x + 3, y + 5) (b) P(x, y) [image: Image] P′(–x, y) and then Q′(x, y) [image: Image] Q″(x + 3, y + 2) (c) P(x, y) [image: Image] P′(–x,–y) and then Q′(x, y) [image: Image] Q″(4–x, y) (d) P(x, y) [image: Image] P′(y–x) and then Q′(x, y) [image: Image] Q″(x–5, y + 1) Solutions (a) R(x, y) [image: Image] R″((x + 7) + 3, (y–2) + 5) = R″(x + 10, y + 3) (b) R(x, y) [image: Image] R″(–x + 3, y + 2) (c) R(x, y) [image: Image] R″(4–(–x),–y) = R″(4 + x,–y) (d) R(x, y) [image: Image] R″(y–5,–x + 1) 18.15 Recognizing rigid motions Name the transformation that takes ΔABC to (a) ΔA′ B′ C′, (b) ΔA″ B″ C″, and (c) ΔA′″ B′″ C′″ as illustrated in Fig. 1815. [image: Image] Fig. 1815 Solutions (a) The triangle has been reflected across the y axis and then moved up 1 space, so P(x, y) [image: Image] P′(–x, y + 1). (b) The triangle has been reflected across the line y = 3, then moved up 1 space and to the left 2 spaces, so P(x, y) [image: Image] P′(x–2, 7–y). (c) The triangle has been rotated clockwise around the origin 90° and then moved up 8 spaces and to the right 6 spaces, so P(x, y) [image: Image] P′(y + 6,–x + 8). 18.16 Naming rigid motions Name the transformation that (a) Rotates everything around the origin 180°, then moves everything up 3 spaces (b) Reflects across x = 4, then slides everything down 2 spaces (c) Rotates everything 90° clockwise around the origin, then reflects across the y axis (d) Rotates around the origin 90° counterclockwise, then slides to the left 5 spaces Solutions (a) P(x, y) [image: Image] P′(–x,–y + 3) (b) P(x, y) [image: Image] P′(8–x, y–2) (c) P(x, y) [image: Image] P′(–y,–x) (d) P(x, y) [image: Image] P′(–y–5, x) 18.7 Dihilations A dihilation (also called a scaling or an enlargement) is not a rigid motion because it multiplies all lengths by a single scale factor. The image of a figure under a dihilation will always be similar to the original. The dihilation that enlarges everything by a scale factor of k is P(x, y) [image: Image] P′ (kx, ky). SOLVED PROBLEMS 18.17 Performing dihilations Let triangle ABC be formed by A(2, 2), B(4, 2), and C(4, 3). Graph the image of ΔABC under (a) magnification by 2 and (b) scaling by [image: Image]. Solutions See Fig. 1816. [image: Image] Fig. 1816 (a) A′(4, 4), B′(8, 4), and C′(8, 6) (b) A″ (1, 1), B″ (2, 1), and C″ (2, 1.5) 18.18 Naming dihilations Name the transformation that (a) Scales everything 5 times larger (b) Shrinks every length to half size (c) Triples all linear dimensions (d) Depicts everything at [image: Image] scale (e) Dihilates by a scale factor of 12 Solutions [image: Image] SUPPLEMENTARY PROBLEMS 18.1 Name the image of points A(6, 2), B(–1, 4), and C(2, 7) under the transformation (a) P(x, y) [image: Image] P′(x + 3, y) (b) P(x, y) [image: Image] P′(2–x, y) (c) P(x, y) [image: Image] P′(–x + 1,–y + 3) (d) P(x, y) [image: Image] P′(–y, x + 8) (e) P(x, y) [image: Image] P′(2x, 2y) (f) P(x, y) [image: Image] P′(4–3x, 3y) (g) P(x, y) [image: Image] P′(2–y, 5–x) 18.2. Let triangle ABC be defined by A(1,–1), B(2, 2), and C(3,–1). Graph the image of ΔABC under (a) P(x, y) [image: Image] P′(x + 5, y) (b) P(x, y) [image: Image] P′(x, y–4) (c) P(x, y) [image: Image] P′(x–3, y + 2) 18.3. Name the translation that takes ΔABC to (a) ΔA′ B′ C′, (b) ΔA″ B″ C″, and (c) ΔA′″ B′″ C′″ as illustrated in Fig. 1817. [image: Image] Fig. 1817 18.4. Name the translation that moves everything (a) Down 5 spaces (b) To the right 6 spaces (c) Up 3 spaces and 7 spaces to the left (d) Down 2 spaces and 8 spaces to the right (e) Up 4 spaces and to the left 1 space 18.5. Let trapezoid ABCD be formed by A(1, 3), B(5, 3), C(4, 1), and D(2, 1). Graph trapezoid ABCD and its image under (a) reflection across the y axis P(x, y) [image: Image] P′(–x, y), (b) reflection across the line y =–1, P(x, y) [image: Image] P″(x,–2–y), and (c) reflection across the line x = 8, P(x, y) [image: Image] P′″(16–x, y). 18.6. Name the reflection that takes ΔABC to (a) ΔA′ B′ C′, (b) ΔA″ B″ C″, and (c) ΔA′″ B′″ C′″ as illustrated in Fig. 1818. [image: Image] Fig. 1818 18.7. Name the transformation that (a) Reflects across y = 5 (b) Reflects across x =–2 (c) Reflects across y =–1 (d) Reflects across x = [image: Image] 18.8. Which of the figures in Fig. 1819 has reflectional symmetry? [image: Image] Fig. 1819 18.9. Let parallelogram ABCD be defined by A(1, 2), B(4, 2), C(5, 1), and D(2, 1). Graph parallelogram ABCD and its image under (a) a 90° clockwise rotation about the origin, (b) a 180° rotation about the origin, and (c) a 270° clockwise rotation about the origin. 18.10. Name the rotation that takes ΔABC to (a)ΔA′ B′ C′, (b) ΔA″ B″ C″, and (c) ΔA′″ B′″ C′″ as illustrated in Fig. 1820 [image: Image] Fig. 1820 18.11. Name the transformation that rotates clockwise about the origin: (a) 40° (b) 50° (c) 80° 18.12. For each figure in Fig. 1821, give the smallest angle by which the figure could be rotated around its center and still look the same. [image: Image] Fig. 1821 18.13. Let triangle ABC be defined by A(2, 1), B(3, 2), and C(3,–1). Graph ΔABC and its image under the following combinations of transformations: (a) Reflect across the line y = 3 and then move to the right 2 spaces. (b) Rotate about the origin 90° clockwise and then move to the right 1 space and down 3 spaces. (c) Rotate about the origin 270° clockwise and then reflect across the x axis. (d) Reflect across the line x =–1 and then move up 2 spaces. 18.14. Name the single transformation that does the same thing as (a) P(x, y) [image: Image] P′(x + 5, y–3) and then Q′(x, y) [image: Image] Q″(x + 1, y + 2) (b) P(x, y) [image: Image] P′(5–x, y) and then Q′(x, y) [image: Image] Q″(x–4, y + 2) (c) P(x, y) [image: Image] P′(y,–x) and then Q′(x, y) [image: Image] Q″(x + 3, y–6) (d) P(x, y) [image: Image] P′(–y, x) and then Q′(x, y) [image: Image] Q″(x, 4–y) (e) P(x, y) [image: Image] P′(x,–3–y) and then Q′(x, y) [image: Image] Q″ (6–x, y) 18.15. Name the transformation that takes ΔABC to (a) ΔA′ B′ C′, (b) ΔA″ B″ C″, and (c) ΔA′″ B′″ C′″ as illustrated in Fig. 1822. [image: Image] Fig. 1822 18.16. Name the transformation that (a) Reflects across the x axis and then moves everything down 3 spaces (b) Rotates around the origin clockwise 90° and then moves everything to the right 2 spaces (c) Reflects across the line y = 2 and then rotates 180° around the origin (d) Rotates 180° around the origin and then reflects across the line y = 2 (e) Moves everything up 3 spaces and to the left 1 space, then reflects across the line x =–4 18.17. Let rectangle ABCD be formed by A(–1, 2), B(1, 2), C(1, 1), and D(–1, 1). Graph this rectangle and also its image under the transformation P(x, y) [image: Image] P′(3x, 3y). 18.18. Name the transformation that (a) Scales everything to be twice as large (b) Scales everything by scale factor 8 (c) Dihilates everything by a scale factor of [image: Image] CHAPTER 15 Constructions 15.1 Introduction Geometric figures are constructed with straightedge and compass. Since constructions are based on deductive reasoning, measuring instruments such as the ruler and protractor are not permitted. However, a ruler may be used as a straightedge if its markings are disregarded. In constructions, it is advisable to plan ahead by making a sketch of the situation; such a sketch will usually reveal the needed construction steps. Construction lines should be made light to distinguish them from the required figure. The following constructions are detailed in this chapter: 1. To construct a line segment congruent to a given line segment 2. To construct an angle congruent to a given angle 3. To bisect a given angle 4. To construct a line perpendicular to a given line through a given point on the line 5. To bisect a given line segment 6. To construct a line perpendicular to a given line through a given external point 7. To construct a triangle given its three sides 8. To construct an angle of measure 60° 9. To construct a triangle given two sides and the included angle 10. To construct a triangle given two angles and the included side 11. To construct a triangle given two angles and a side not included 12. To construct a right triangle given its hypotenuse and a leg 13. To construct a line parallel to a given line through a given external point 14. To construct a tangent to a given circle through a given point on the circle 15. To construct a tangent to a given circle through a given point outside the circle 16. To circumscribe a circle about a triangle 17. To locate the center of a given circle 18. To inscribe a circle in a given triangle 19. To inscribe a square in a given circle 20. To inscribe a regular octagon in a given circle 21. To inscribe a regular hexagon in a given circle 22. To inscribe an equilateral triangle in a given circle 23. To construct a triangle similar to a given triangle on a given line segment as base 15.2 Duplicating Segments and Angles CONSTRUCTION 1: To construct a line segment congruent to a given line segment Given: Line segment [image: Image] (Fig. 151) To construct: A line segment congruent to [image: Image] Construction: On a working line w, with any point C as a center and a radius equal to AB, construct an arc intersecting w at D. Then [image: Image] is the required line segment. [image: Image] Fig. 151 CONSTRUCTION 2: To construct an angle congruent to a given angle Given: ∠A (Fig. 152) [image: Image] Fig. 152 To construct: An angle congruent to ∠A Construction: With A as center and a convenient radius, construct an arc (1) intersecting the sides of ∠A at B and C. With A′, a point on a working line w, as center and the same radius, construct arc (2) intersecting w at B′. With B′ as center and a radius equal to BC, construct arc (3) intersecting arc (2) at C’. Draw A’C’. Then ∠A is the required angle. (ΔABC ≐ ΔA′B′C by SSS; hence ∠A≐ ∠A′.) SOLVED PROBLEMS 15.1 Combining line segments Given line segments with lengths a and b (Fig. 153), construct line segments with lengths equal to (a) a + 2b; (b) 2(a + b); (c) b–a. [image: Image] Fig. 153 Solutions Use construction 1. (a) On a working line w, construct a line segment [image: Image] with length a. From B, construct a line segment with length equal to b, to point C; and from C construct a line segment with length b, to point D. Then [image: Image] is the required line segment. (b) Similar to (a). AD = a + b + (a + b). (c) Similar to (a). First construct [image: Image] with length b, then [image: Image] with length a. AC = b–a. 15.2 Combining angles Given ΔABC in Fig. 154, construct angles whose measures are equal to (a) 2A; (b) A + B + C; (c) B–A. [image: Image] Fig. 154 Solutions Use construction 2. (a) Using a working line w as one side, duplicate ∠A. Construct another duplicate of ∠A adjacent to ∠A, as shown. The exterior sides of the copied angles form the required angle. (b) Using a working line w as one side, duplicate ∠A. Construct ∠B adjacent to ∠A. Then construct ∠C adjacent to ∠B. The exterior sides of the copied angles A and C form the required angle. Note that the angle is a straight angle. (c) Using a working line w as one side, duplicate ∠B. Then duplicate ∠A from the new side of ∠B as shown. The difference is the required angle. 15.3 Constructing Bisectors and Perpendiculars CONSTRUCTION 3: To bisect a given angle Given: ∠A (Fig. 155) [image: Image] Fig. 155 To construct: The bisector of ∠A Construction: With A as center and a convenient radius, construct an arc intersecting the sides of ∠A at B and C. With B and C as centers and equal radii, construct arcs intersecting in D. Draw [image: Image]. Then [image: Image] is the required bisector. (ΔABD ≐ ΔACD by SSS; hence, ∠1 ≐ ∠2.) CONSTRUCTION 4: To construct a line perpendicular to a given line through a given point on the line Given: Line w and point P on w (Fig. 156) [image: Image] Fig. 156 To construct: A perpendicular to w at P Construction: Using construction 3, bisect the straight angle at P. Then [image: Image] is the required perpendicular; [image: Image] is the required line. CONSTRUCTION 5: To bisect a given line segment (to construct the perpendicular bisector of a given line segment) Given: Line segment [image: Image] (Fig. 157) [image: Image] Fig. 157 To construct: The perpendicular bisector of [image: Image] Construction: With A as center and a radius of more than half [image: Image], construct arc (1). With B as center and the same radius, construct arc (2) intersecting arc (1) at C and D. Draw [image: Image]. [image: Image] is the required perpendicular bisector of [image: Image]. (Two points each equidistant from the ends of a segment determine the perpendicular bisector of the segment.) CONSTRUCTION 6: To construct a line perpendicular to a given line through a given external point Given: Line w and point P outside of w (Fig. 158) [image: Image] Fig. 158 To construct: A perpendicular to w through P Construction: With P as center and a sufficiently long radius, construct an arc intersecting w at B and C. With B and C as centers and equal radii of more than half [image: Image], construct arcs intersecting at A. Draw [image: Image]. Then [image: Image] is the required perpendicular. (Points P and A are each equidistant from B and C.) SOLVED PROBLEMS 15.3 Constructing special lines in a triangle In scalene ΔABC [Fig. 159(a)], construct (a) a perpendicular bisector of [image: Image] and (b) a median to [image: Image]. In ΔDEF [Fig. 159(b)], D is an obtuse angle; construct (c) the altitude to [image: Image] and (d) the bisector of ∠E. [image: Image] Fig. 159 Solutions (a) Use construction 5 to obtain [image: Image] the perpendicular bisector of [image: Image]. (b) Point M is the midpoint of [image: Image]. Draw [image: Image], the median to [image: Image]. (c) Use construction 6 to obtain [image: Image], the altitude to [image: Image] (extended). (d) Use construction 3 to bisect ∠E. [image: Image] is the required bisector. 15.4 Constructing bisectors and perpendiculars to obtain required angles (a) Construct angles measuring 90°, 45°, and 135°. (b) Given an angle with measure A (Fig. 1510), construct an angle whose measure is 90°+ A. [image: Image] Fig. 1510 Solutions (a) In Fig. 1510(a), m∠DAB = 90°, m∠BAE = 45°, m∠BAE = 135° (b) In Fig. 1510(b), m∠GHJ = 90°+ A. 15.4 Constructing a Triangle 15.4A Determining a Triangle A triangle is determined when a set of given data fix its size and shape. Since the parts needed to prove congruent triangles fix the size and shape of the triangles, a triangle is determined when the given data consist of three sides, or two sides and the angle included by those sides, or two angles and a side included by those angles, or two angles and a side not included by those angles, or the hypotenuse and either leg of a right triangle. 15.4B Sketching Triangles to be Constructed Before doing the actual construction, it is very helpful to make a preliminary sketch of the required triangle. In this sketch: 1. Show the position of each of the given parts of the triangle. 2. Draw the given parts heavy, the remaining parts light. 3. Approximate the sizes of the given parts. 4. Use small letters for sides to agree with the capital letters for the angles opposite them. [image: Image] Fig. 1511 As an example, you might make a sketch like that in Fig. 1511 before constructing a triangle given two angles and an included side. 15.4C Triangle Constructions CONSTRUCTION 7: To construct a triangle given its three sides Given: Sides of lengths a, b, and c (Fig. 1512) [image: Image] Fig. 1512 To construct: ΔABC Construction: On a working line w, construct [image: Image] such that AC = b. With A as center and c as radius, construct arc (1). Then with C as center and a as radius, construct arc (2) intersecting arc (1) at B. Draw [image: Image] and [image: Image]. ΔABC is the required triangle. CONSTRUCTION 8: To construct an angle of measure 60° Given: Line w (Fig. 15–13) [image: Image] Fig. 1513 To construct: An angle of measure 60° Construction: Using a convenient length as a side, construct an equilateral triangle using construction 7. Then any angle of the equilateral triangle is the required angle. CONSTRUCTION 9: To construct a triangle given two sides and the included angle Given: ∠A, segments of lengths b and c (Fig. 1514) To construct: ΔABC Construction: On a working line w, construct [image: Image] such that AC = b. At A, construct ∠A with one side [image: Image]. On the other side of ∠A, construct [image: Image] such that AB = c. Draw [image: Image]. Then the required triangle is ΔABC. [image: Image] Fig. 1514 CONSTRUCTION 10: To construct a triangle given two angles and the included side Given: ∠A, ∠C, and a segment of length b (Fig. 1515) [image: Image] Fig. 1515 To construct: ΔABC Construction: On a working line w, construct [image: Image] such that AC = b. At A, construct ∠A with one side on [image: Image], and at C, construct ∠C with one side on [image: Image]. Extend the new sides of the angles until they meet at B. CONSTRUCTION 11: To construct a triangle given two angles and a side not included Given: ∠A, ∠B, and a segment of length b (Fig. 1516) [image: Image] Fig. 1516 To construct: ΔABC Construction: On a working line w, construct [image: Image] such that AC = b. At C, construct an angle with measure equal to m∠A + m∠B so that the extension of [image: Image] will be one side of the angle. The remainder of the straight angle at C will be ∠C. At A, construct ∠A with one side on [image: Image]. The intersection of the new sides of the angles is B. CONSTRUCTION 12: To construct a right triangle given its hypotenuse and a leg Given: Hypotenuse with length c and leg with length b of right triangle ABC (Fig. 1517) [image: Image] Fig. 1517 To construct: Right triangle ABC Construction: On a working line w, construct [image: Image] such that AC = b. At C construct a perpendicular to [image: Image]. With A as center and a radius of c, construct an arc intersecting the perpendicular at B. SOLVED PROBLEMS 15.5 Constructing a triangle Construct an isosceles triangle, given the lengths of the base and an arm (Fig. 1518). [image: Image] Fig. 1518 Solution Use construction 7, since all three sides of the triangle are known. 15.6 Constructing angles based on the construction of the 60° angle Construct an angle of measure (a) 120°; (b) 30°; (c) 150°; (d) 105°; (e) 75°. Solutions (a) Use construction 8 [Fig. 15–19(a)] to construct 120° as 180°–60°. (b) Use constructions 8 and 3 to construct 30° as ¿(60°) [Fig. 15–19(b)]. (c) Use (b) to construct 150° as 180°–30° [Fig. 15–19(b)]. (d) Use constructions 3, 4, and 8 to construct 105° as 60°+ [image: Image](90°) [Fig. 15–19(c)]. (e) Use (d) to construct 75° as 180°–105° [Fig. 15–19(c)]. [image: Image] Fig. 1519 15.5 Constructing Parallel Lines CONSTRUCTION 13: To construct a line parallel to a given line through a given external point Given: [image: Image] and external point P (Fig. 15–20) [image: Image] Fig. 1520 To construct: A line through P parallel to [image: Image] Construction: Draw a line [image: Image] through P intersecting [image: Image] in Q. Construct ∠SPD ≐ ∠PQB. Then [image: Image] is the required parallel. (If two corresponding angles are congruent, the lines cut by the transversal are parallel.) SOLVED PROBLEM 15.7 Constructing a parallelogram Construct a parallelogram given the lengths of two adjacent sides a and b and of a diagonal d (Fig. 1521). [image: Image] Fig. 1521 Solution Three vertices of the parallelogram are obtained by constructing ΔABD by construction 7. The fourth vertex, C, is obtained by constructing ΔBCD upon diagonal [image: Image] by construction 7. Vertex C may also be obtained by constructing [image: Image][image: Image] and [image: Image][image: Image]. 15.6 Circle constructions CONSTRUCTION 14: To construct a tangent to a given circle through a given point on the circle Given: Circle O and point P on the circle (Fig. 1522) [image: Image] Fig. 1522 To construct: A tangent to circle O at P Construction: Draw radius [image: Image] and extend it outside the circle. Construct [image: Image] ⊥ [image: Image] at P. [image: Image] is the required tangent. (A line perpendicular to a radius at its outer extremity is a tangent to the circle.) CONSTRUCTION 15: To construct a tangent to a given circle through a given point outside the circle Given: Circle O and point P outside the circle (Fig. 1523) [image: Image] Fig. 1523 To construct: A tangent to circle O from P Construction: Draw [image: Image], and make [image: Image] the diameter of a new circle Q. Connect P to A and B, the intersections of circles O and Q. Then [image: Image] and [image: Image] are tangents. (∠OAP and ∠OBP are right angles, since angles inscribed in semicircles are right angles.) CONSTRUCTION 16: To circumscribe a circle about a triangle Given: ΔABC (Fig. 1524) [image: Image] Fig. 1524 To construct: The circumscribed circle of ΔABC Construction: Construct the perpendicular bisectors of two sides of the triangle. Their intersection is the center of the required circle, and the distance to any vertex is the radius. (Any point on the perpendicular bisector of a segment is equidistant from the ends of the segment.) CONSTRUCTION 17: To locate the center of a given circle Given: A circle (Fig. 1525) [image: Image] Fig. 1525 To construct: The center of the given circle Construction: Select any three points A, B, and C on the circle. Construct the perpendicular bisectors of line segments [image: Image] and [image: Image]. The intersection of these perpendicular bisectors is the center of the circle. CONSTRUCTION 18: To inscribe a circle in a given triangle Given: ΔABC (Fig. 1526) [image: Image] Fig. 1526 To construct: The circle inscribed in ΔABC Construction: Construct the bisectors of two of the angles of ΔABC. Their intersection is the center of the required circle, and the distance (perpendicular) to any side is the radius. (Any point on the bisector of an angle is equidistant from the sides of the angle.) SOLVED PROBLEMS 15.8 Constructing tangents A secant from a point P outside circle O in Fig. 1527 meets the circle in B and A. Construct a triangle circumscribed about the circle so that two of its sides meet in P and the third side is tangent to the circle at A. [image: Image] Fig. 1527 Solution Use constructions 14 and 15: At A construct a tangent to circle O. From P construct tangents to circle O intersecting the first tangent in C and D. The required triangle is ΔPCD. 15.9 Constructing circles Construct the circumscribed and inscribed circles of isosceles triangle DEF in Fig. 1528. [image: Image] Fig. 1528 Solution Use constructions 16 and 18. In doing so, note that the bisector of ∠E is also the perpendicular bisector of [image: Image]. Then the center of each circle is on [image: Image]. I, the center of the inscribed circle, is found by constructing the bisector of ∠D or ∠F. C, the center of the circumscribed circle, is found by constructing the perpendicular bisector of [image: Image] or [image: Image]. 15.7 Inscribing and Circumscribing Regular Polygons CONSTRUCTION 19: To inscribe a square in a given circle Given: Circle O (Fig. 1529) [image: Image] Fig. 1529 To construct: A square inscribed in circle O Construction: Draw a diameter, and construct another diameter perpendicular to it. Join the end points of the diameters to form the required square. CONSTRUCTION 20: To inscribe a regular octagon in a given circle Given: Circle O (Fig. 1530) [image: Image] Fig. 1530 To construct: A regular octagon inscribed in circle O Construction: As in construction 19, construct perpendicular diameters. Then bisect the angles formed by these diameters, dividing the circle into eight congruent arcs. The chords of these arcs are the sides of the required regular octagon. CONSTRUCTION 21: To inscribe a regular hexagon in a given circle Given: Circle O (Fig. 1531) [image: Image] Fig. 1531 To construct: A regular hexagon inscribed in circle O Construction: Draw diameter [image: Image] and, using A and D as centers, construct four arcs having the same radius as circle O and intersecting the circle. Construct the required regular hexagon by joining consecutive points in which these arcs intersect the circle. CONSTRUCTION 22: To inscribe an equilateral triangle in a given circle Given: Circle O (Fig. 1532) [image: Image] Fig. 1532 To construct: An equilateral triangle inscribed in circle O Construction: Inscribed equilateral triangles are obtained by joining alternately the six points of division obtained in construction 21. 15.8 Constructing Similar Triangles CONSTRUCTION 23: To construct a triangle similar to a given triangle on a given line segment as base Given: ΔABC and line segment [image: Image] (Fig. 1533) [image: Image] Fig. 1533 To construct: ΔA′B′C′ ~ ΔABC on [image: Image] as base Construction: On [image: Image] construct ∠A′ ≐ ∠A and ∠C‘ ≐ ∠C using construction 2. Extend the other sides until they meet, at B. (If two angles of one triangle are congruent to two angles of another triangle, the triangles are similar.) SOLVED PROBLEM 15.10 Constructing similar triangles Construct a triangle similar to triangle ABC in Fig. 1534, with a base twice as long as the base of the given triangle. [image: Image] Fig. 1534 Solution Construct [image: Image] twice as long as [image: Image], and then use construction 23. Alternative method (Fig. 1535): Extend two sides of ΔABC to twice their lengths and join the endpoints. [image: Image] Fig. 1535 SUPPLEMENTARY PROBLEMS 15.1. Given line segments with lengths a and b as follows: [image: Image] Construct a line segment whose length equals (a) a + b; (b) a–b; (c) 2a + b; (d) a + 3b; (e) 2(a + b); (f) 2(3b–a). (15.1) 15.2. Given line segments with lengths a, b, and c: [image: Image] Construct a line segment whose length equals (a) a + b + c; (b) a+ c–b; (c) a+ 2(b + c); (d) b + 2(a–c); (e) 3(b + c–a). (15.1) 15.3. Given angles with measures A and B (Fig. 1536). Construct an angle with measure (a) A + B; (b) A–B; (c) 2B–A; (d) 2A–B; (e) 2(A–B). (15.2) [image: Image] Fig. 1536 15.4. Given angles with measures A, B, and C (Fig. 1537). Construct an angle with measure (a) A + C; (b) B + C–A; (c) 2C; (d) B–C; (e) 2(A–B). (15.2) [image: Image] Fig. 1537 15.5. In a right triangle, construct (a) the bisector of the right angle; (b) the perpendicular bisector of the hypotenuse; (c) the median to the hypotenuse. (15.3) 15.6. For each kind of triangle (acute, right, and obtuse), show that the following sets of rays and segments are concurrent, that is, they intersect in one point: (a) the angle bisectors; (b) the medians; (c) the altitudes; (d) the perpendicular bisectors. (15.3) 15.7. Given ΔABC in Fig. 1538, construct (a) the supplement of ∠A; (b) the complement of ∠B; (c) the complement of [image: Image]∠c. (15.4) [image: Image] Fig. 15.38 15.8. Construct an angle with measure equal to (a) [image: Image]; (b) [image: Image]° ; (c) [image: Image] (15.4) 15.9. Given an acute angle, construct (a) its supplement; (b) its complement; (c) half its supplement; (d) half its complement. (15.4) 15.10. By actual construction, illustrate that the difference between the measures of the supplement and complement of an acute angle equals 90°. (15.4) 15.11. Construct a right triangle given its (a) legs; (b) hypotenuse and a leg; (c) leg and an acute angle adjacent to the leg; (d) leg and an acute angle opposite the leg; (e) hypotenuse and an acute angle. (15.5) 15.12. Construct an isosceles triangle given (a) an arm and a vertex angle; (b) an arm and a base angle; (c) an arm and the altitude to the base; (d) the base and the altitude to the base. (15.5) 15.13. Construct an isosceles right triangle given (a) a leg; (b) the hypotenuse; (c) the altitude to the hypotenuse. (15.5) 15.14. Construct a triangle given (a) two sides and the median to one of them; (b) two sides and the altitude to one of them; (c) an angle, the angle bisector of the given angle, and a side adjacent to the given angle. (15.5) 15.15. Construct angles of measure 15° and 165°. (15.6) 15.16. Given an angle with measure A, construct angles with measure (a) A + 60°; (b) A + 30°; (c) A + 120°. (15.6) (15.6) 15.17. Construct a parallelogram, given (a) two adjacent sides and an angle; (b) the diagonals and the acute angle at their intersection; (c) the diagonals and a side; (d) two adjacent sides and the altitude to one of them; (e) a side, an angle, and the altitude to the given side. (15.7) 15.18. Circumscribe a triangle about a given circle, if the points of tangency are given. (15.8) 15.19. Secant [image: Image] passes through the center of circle O in Fig. 1539. Circumscribe a quadrilateral about the circle so that A and B are opposite vertices. (15.8) [image: Image] Fig. 1539 15.20. Circumscribe and inscribe circles about (a) an acute triangle; (b) an obtuse triangle. (15.9 15.21. Circumscribe a circle about (a) a right triangle; (b) a rectangle; (c) a square. (15.9) 15.22. Construct the inscribed and circumscribed circles of an equilateral triangle. (15.9) 15.23. Locate the center of a circle drawn around the outside of a halfdollar piece. (15.9) 15.24. In a given circle, inscribe (a) a square; (b) a regular octagon; (c) a regular 16gon; (d) a regular hexagon; (e) an equilateral triangle; (f) a regular dodecagon. 15.25. Construct a triangle similar to a given triangle with a base (a) three times as long; (b) half as long; (c) one and onehalf times as long. (15.10) CHAPTER 10 Regular Polygons and the Circle 10.1 Regular Polygons A regular polygon is an equilateral and equiangular polygon. The center of a regular polygon is the common center of its inscribed and circumscribed circles. A radius of a regular polygon is a segment joining its center to any vertex. A radius of a regular polygon is also a radius of the circumscribed circle. (Here, as for circles, we may use the word radius to mean the number that is “the length of the radius.”) A central angle of a regular polygon is an angle included between two radii drawn to successive vertices. An apothem of a regular polygon is a segment from its center perpendicular to one of its sides. An apothem is also a radius of the inscribed circle. Thus for the regular pentagon shown in Fig. 101, AB = BC = CD = DE = EA and m∠A = m∠B = m∠C = m∠D = m∠E. Also, its center is O, [image: Images] and [image: Images] are its radii; ∂AOB is a central angle; and [image: Images] and [image: Images] are apothems. [image: Image] Fig. 101 10.1A RegularPolygon Principles PRINCIPLE 1: If a regular polygon of n sides has a side of length s, the perimeter is p = ns. PRINCIPLE 2: A circle may be circumscribed about any regular polygon. PRINCIPLE 3: A circle may be inscribed in any regular polygon. PRINCIPLE 4: The center of the circumscribed circle of a regular polygon is also the center of its inscribed circle. PRINCIPLE 5: An equilateral polygon inscribed in a circle is a regular polygon. PRINCIPLE 6: Radii of a regular polygon are congruent. PRINCIPLE 7: A radius of a regular polygon bisects the angle to which it is drawn. Thus in Fig. 101, [image: Images] bisects ∠ABC. PRINCIPLE 8: Apothems of a regular polygon are congruent. PRINCIPLE 9: An apothem of a regular polygon bisects the side to which it is drawn. Thus in Fig. 101, [image: Images] bisects [image: Images], and [image: Images] bisects [image: Images]. PRINCIPLE 10: For a regular polygon of n sides: 1. Each central angle c measures [image: Images] 2. Each interior angle i measures [image: Images] 3. Each exterior angle e measures [image: Images] Thus for the regular pentagon ABCDE of Fig. 102, [image: Image] [image: Image] Fig. 102 and SOLVED PROBLEMS 10.1 Finding measures of lines and angles in a regular polygon (a) Find the length of a side s of a regular pentagon if the perimeter p is 35. (b) Find the length of the apothem a of a regular pentagon if the radius of the inscribed circle is 21. (c) In a regular polygon of five sides, find the measures of the central angle c, the exterior angle e, and the interior angle i. (d) If an interior angle of a regular polygon measures 108°, find the measures of the exterior angle and the central angle and the number of sides. Solutions (a) p = 35. Since p = 5s, we have 35 = 5s and s = 7. (b) Since an apothem r is a radius of the inscribed circle, it has length 21. (c) n = 5. Then m∠c = [image: Images] = 72°; m∠e = [image: Images] = 72° ; m∠i = 180°  m∠e = 108°. (d) m∠i = 108°. Then m∠c = 180°  m∠i = 72°. Since m∠c =, n = 5. (See Fig. 103.) [image: Image] Fig. 103 10.2 Proving a regularpolygon problem stated in words Prove that a vertex angle of a regular pentagon is trisected by diagonals drawn from that vertex. Solution [image: Image] PROOF: [image: Image] 10.2 Relationships of Segments in Regular Polygons of 3, 4, and 6 Sides In the regular hexagon, square, and equilateral triangle, special right triangles are formed when the apothem r and a radius R terminating in the same side are drawn. In the case of the square we obtain a 45°45°90° triangle, while in the other two cases we obtain a 30°60°90° triangle. The formulas in Fig. 104 relate the lengths of the sides and radii of these regular polygons. [image: Image] Fig. 104 SOLVED PROBLEMS 10.3 Applying line relationships in a regular hexagon In a regular hexagon, (a) find the lengths of the side and apothem if the radius is 12; (b) find the radius and length of the apothem if the side has length 8. Solutions (a) Since R = 12, s = R = 12 and r = [image: Images] R [image: Images] = 6 [image: Images]. (b) Since s = 8, R = s = 8 and r = [image: Images] R [image: Images] = 4 [image: Images]. 10.4 Applying line relationships in a square In a square, (a) find the lengths of the side and apothem if the radius is 16; (b) find the radius and the length of the apothem if a side has length 10. Solutions (a) Since R = 16, s = R [image: Images] = 16 [image: Images] and r = [image: Images] = 8 [image: Images]. (b) Since s = 10, r = [image: Images] = 5 and [image: Images] [image: Images] = 5 [image: Images]. 10.5 Applying line relationships in an equilateral triangle In an equilateral triangle, (a) find the lengths of the radius, apothem, and side if the altitude has length 6; (b) find the lengths of the side, apothem, and altitude if the radius is 9. Solutions (a) Since h = 6, we have r = [image: Images] h =2; R = [image: Images]n = 4; and s = R[image: Images] = 4 [image: Images]. (b) Since R = 9, s = R[image: Images] = 9[image: Images]; r =[image: Images] R = 4[image: Images]; and h =[image: Images] R = 13[image: Images]. 10.3 Area of a Regular Polygon The area of a regular polygon equals onehalf the product of its perimeter and the length of its apothem. As shown in Fig. 105, by drawing radii we can divide a regular polygon of n sides and perimeter p = ns into n triangles, each of area [image: Images]. Hence, the area of the regular polygon is [image: Images] [image: Image] Fig. 105 SOLVED PROBLEMS 10.6 Finding the area of a regular polygon (a) Find the area of a regular hexagon if the length of the apothem is 5 [image: Images]. (b) Find the area of a regular pentagon to the nearest integer if the length of the apothem is 20. Solutions (a) In a regular hexagon, [image: Images] Since r = 5[image: Images], s = 10 and p = 6(10) = 60. Then [image: Images] (b) In Fig. 106, m∠AOE = 360°/5 = 72° and m∠AOF = [image: Images]m ∠AOE = 36°. Then tan 36° = [image: Images]s/20 = s/40 or s = 40 tan 36°. Now [image: Images] (40 tan 36°) (20) = 1453. [image: Image] Fig. 106 10.4 Ratios of Segments and Areas of Regular Polygons PRINCIPLE 1: Regular polygons having the same number of sides are similar. PRINCIPLE 2: Corresponding segments of regular polygons having the same number of sides are in proportion. “Segments” here includes sides, perimeters, radii or circumferences of circumscribed or inscribed circles, and such. PRINCIPLE 3: Areas of regular polygons having the same number of sides are to each other as the squares of the lengths of any two corresponding segments. SOLVED PROBLEMS 10.7 Ratios of lines and areas of regular polygons (a) In two regular polygons having the same number of sides, find the ratio of the lengths of the apothems if the perimeters are in the ratio 5:3. (b) In two regular polygons having the same number of sides, find the length of a side of the smaller if the lengths of the apothems are 20 and 50 and a side of the larger has length 32.5. (c) In two regular polygons having the same number of sides, find the ratio of the areas if the lengths of the sides are in the ratio 1:5. (d) In two regular polygons having the same number of sides, find the area of the smaller if the sides have lengths 4 and 12 and the area of the larger is 10,260. Solutions (a) By Principle 2, r: r′ = p: p′ = 5:3. (b) By Principle 2, s: s′= r: r′; thus, s : 32.5 = 20:50 and s = 13. (c) By Principle 3, [image: Images] (d) By Principle 3, [image: Images] 10.5 Circumference and Area of a Circle π (pi) is the ratio of the circumference C of any circle to its diameter d; that is, π = C/d. Hence, [image: Image] Approximate values for p are 3.1416 or 3.14 or [image: Images]. Unless you are told otherwise, we shall use 3.14 for π in solving problems. A circle may be regarded as a regular polygon having an infinite number of sides. If a square is inscribed in a circle, and the number of sides is continually doubled (to form an octagon, a 16gon, and so on), the perimeters of the resulting polygons will get closer and closer to the circumference of the circle (Fig. 107). [image: Image] Fig. 107 Thus to find the area of a circle, the formula A = [image: Images]pr can be used with C substituted for p; doing so, we get [image: Image] All circles are similar figures, since the