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Answer Key 1 8.1 Circles and Similarity Chapter 8 – Circles Answer Key 8.1 Circles and Similarity Answers 3 1. Translate circle A one unit left and 11 units down. The, dilate about its center by a scale factor of . 4 2. Translate circle A 11 units to the left and 2 units up. Then, dilate about its center by a scale factor of 5 . 3 3. Translate circle A two units to the right and 6 units down. Then, dilate about its center by a scale 7 factor of . 2 4. Translate circle A two units to the left and 5 units down. Then, dilate about its center by a scale 4 factor of . 3 5. Translate circle A 1 unit to the right and 14 units down. Then, dilate about its center by a scale factor of 2. 5 6. Dilate circle A about its center by a scale factor of . 4 7. Translate circle A 5 units to the right and 10 units up. Then, dilate about its center by a scale factor of 4. 8. Translate circle A 6 units to left and one unit down. Then, dilate about its center by a scale factor of 8 . 5 6 9. Dilate circle A about its center by a scale factor of . 5 10. Translate circle A 8 units to the right and 10 units down. 2 11. 3 6 12. √ 1 25 13. 81 2 3 14. √ √5 15. Any reflection or rotation on a circle could more simply be a translation. Therefore, reflections and rotations are not necessary when looking to prove that two circles are similar. CK-12 Geometry Honors Concepts 1 Chapter 8 – Circles Answer Key 8.2 Area and Circumference of Circles Answers 360° 180 1. = 2푛 푛 180 2. 2 sin 푛 180 3. cos 푛 180 180 4. sin ⋅ cos 푛 푛 180 180 5. 푛 ⋅ sin ⋅ cos 푛 푛 180 6. 2푛 sin 푛 7. 퐴 = 3.141592; 푃 = 6.2831852 8. 퐴 = 3.1415926; 푃 = 6.2831853. It makes sense that the area of a circle with radius 1 unit is 휋 and the circumference of a circle with radius 1 unit is 2휋. 9. The polygon gets closer and closer to the circle so its area gets closer and closer to the area of the circle. 10. The polygon gets closer and closer to the circle so its perimeter gets closer and closer to the circumference of the circle. 11. The scale factor for a circle with radius 1 and a circle with radius r is 푘 = 푟. Therefore, the ratio of their circumferences is 푟: 1. Since the circumference of the circle with radius 1 is 2휋, the circumference of the circle with radius 푟 is 2휋푟. 12. The ratio of the areas is 9: 25. The ratio of the circumferences is 3: 5. 13. 5 units. 14. 6 units. 15. ≈ 14.31휋 CK-12 Geometry Honors Concepts 2 Chapter 8 – Circles Answer Key 8.3 Central Angles and Chords Answers 1. Answers vary. 2. They have the same measure. 3. Answers vary. 4. Diameter 5. 푚퐹퐸̂ = 30°. 6. 푚퐶퐷̂ = 130°. 7. 푚∠퐸퐴퐺 = 75° 8. 푚퐺퐵̂ = 75° 9. 퐴퐺̅̅̅̅ is the perpendicular bisector of ̅퐸퐵̅̅̅. 10. ̅퐸퐹̅̅̅ ≅ 퐹퐷̅̅̅̅ by assumption, ̅퐸퐴̅̅̅ ≅ 퐴퐷̅̅̅̅ because they are both radii of the circle, and 퐴퐹̅̅̅̅ ≅ 퐴퐹̅̅̅̅ by the reflexive property. Therefore, Δ퐴퐹퐸 ≅ Δ퐴퐹퐷 by 푆푆푆 ≅. ∠퐴퐹퐸 and ∠퐴퐹퐷 are both congruent (corresponding parts of congruent triangles) and supplementary, so they must both be right angles. Therefore, 푚∠퐴퐹퐷 = 90°. 11. Δ퐴퐹퐸 and Δ퐴퐹퐷 are both right triangles by assumption, ̅퐸퐴̅̅̅ ≅ 퐴퐷̅̅̅̅ because they are both radii of the circle, and 퐴퐹̅̅̅̅ ≅ 퐴퐹̅̅̅̅ by the reflexive property. Therefore, Δ퐴퐹퐸 ≅ Δ퐴퐹퐷 by 퐻퐿 ≅. ̅퐸퐹̅̅̅ ≅ 퐹퐷̅̅̅̅ because they are corresponding parts. 12. 퐷퐹 = 8 13. 퐴퐶 = 12 14. 퐴퐹 ≈ 8.94 15. 퐶퐹 ≈ 3.06 CK-12 Geometry Honors Concepts 3 Chapter 8 – Circles Answer Key 8.4 Inscribed Angles Answers 1. If an inscribed angle and a central angle intercept the same arc, the measure of the inscribed angle will be half the measure of the central angle. 2. 120° 3. 30° 4. 60° 5. 30° 6. Equilateral 7. 푥 = 7.2 8. 푥 = 50° 9. 푥 = 3 10. 푥 = 42.5° 11. 푥 = 29° 12. 푥 = 29° 13. 푥 = 6√2 14 . 푥 = 43° 15. 퐵퐷̅̅̅̅ ∥ 퐸퐶̅̅̅̅, so alternate interior angles are congruent. This means that ∠퐵퐸퐶 ≅ ∠퐷퐶퐸 and thus 푚∠퐵퐸퐶 = 푚∠퐷퐶퐸. 푚퐵퐶̂ = 2푚∠퐵퐸퐶 and 푚퐷퐸̂ = 2푚∠퐷퐶퐸. By substitution, 푚퐵퐶̂ = 푚퐷퐸̂ and 퐵퐶̂ ≅ 퐸퐶̂ . CK-12 Geometry Honors Concepts 4 Chapter 8 – Circles Answer Key 8.5 Inscribed and Circumscribed Circles of Triangles Answers 1-3: Answers vary. See Examples A-C for help. 4. The third angle bisector will intersect in the same point of intersection as the first two angle bisectors. The third angle bisector does not provide any new information. 5. The distance between a point and a line is the length of the segment perpendicular to the line that passes through the point. The segments perpendicular to each of the sides of the triangle that pass through the incenter are radii of the inscribed circle. Therefore, the incenter is equidistant from each of the sides of the triangle. 6-7: Answers vary. See Guided Practice for help. 8. The circumcenter is the center of the circumscribed circle and each of the three vertices are on the circle. From the circumcenter to each of the vertices is a radius, so the distance from the circumcenter to each of the vertices is the same. 9. circumcenter 10. Construct the circumcenter by drawing a triangle and finding the point of intersection of the perpendicular bisectors. 11. Fold the map so the playground overlaps with the basketball court and make a crease. This should be the perpendicular bisector of the line segment connecting those two locations. Fold the map again so the playground overlaps with the parking lot and make a crease. The point where the creases intersect is the circumcenter. 12. incenter CK-12 Geometry Honors Concepts 5 Chapter 8 – Circles Answer Key 13. Construct the incenter by following the steps from the Examples. 14. Fold the map so Main St overlaps with Redwood Rd and make a crease. This should be the angle bisector of the angle formed by those two roads. Fold the map again so Main St overlaps with Springfield Ave and make a crease. The point where the creases intersect is in incenter. 15. Any three non-collinear points define a triangle. All triangles have exactly one circumcenter and therefore all triangles have exactly one circumscribed circle. The circumscribed circle will be the circle that passes through the three points. CK-12 Geometry Honors Concepts 6 Chapter 8 – Circles Answer Key 8.6 Quadrilaterals Inscribed in Circles Answers 1. A cyclic quadrilateral is a quadrilateral that can be inscribed in a circle. 2. supplementary 3. 127° 4. 92° 5. 53° 6. 60° 7. 60° 8. 60° 9. 120° 10. 푥 = 15 11. 푦 = 109 12. 푥 = 90 13. 푥 = 90 14. No, but two opposite vertices will create a diameter of the circle, and it will have two right angles. 15. First note that 푚퐶퐷퐸̂ + 푚퐶퐵퐸̂ = 360° because these two arcs make a full circle. 2푚∠퐵 = 푚퐶퐷퐸̂ and 2푚∠퐷 = 푚퐶퐵퐸̂ because the measure of an inscribed angle is half the measure of its intercepted arc. By substitution, 2푚∠퐵 + 2푚∠퐷 = 360°. Divide by 2 and you have 푚∠퐵 + 푚∠퐷 = 180°. Therefore, ∠퐵 and ∠퐷 are supplementary. CK-12 Geometry Honors Concepts 7 Chapter 8 – Circles Answer Key 8.7 Tangent Lines to Circles Answers 1. A tangent line is a line that intersects a circle exactly once. 2. 퐴푃 = 5 3. 퐴퐶 ≈ 10.3 4. 푚∠퐶퐴푄 = 61° 5. 푄퐶 = 9.62 6. 퐴푄 = 5.33 7. 푃퐶 = 9.62 8. 푚푃푄̂ = 140° 9. 푚푃퐸푄̂ = 220° 10. 223.2° 11. 43.2° 12. Δ퐴퐵퐼~Δ퐻퐺퐼 13. ∠퐴퐵퐼 and ∠퐻퐺퐼 are right angles, so ∠퐴퐵퐼 ≅ ∠퐻퐺퐼. ∠퐴퐼퐵 and ∠퐻퐼퐺 are vertical angles, so ∠퐴퐼퐵 ≅ ∠퐻퐼퐺. Δ퐴퐵퐼~Δ퐻퐺퐼 by 퐴퐴~. CK-12 Geometry Honors Concepts 8 Chapter 8 – Circles Answer Key 14. Construct the perpendicular bisector of 퐴퐶̅̅̅̅ in order to find its midpoint. CK-12 Geometry Honors Concepts 9 Chapter 8 – Circles Answer Key Then construct a circle centered at point M that passes through point C. The circle should also pass through point A. Find the points of intersection and connect them with point C. 15. 퐴퐶̅̅̅̅ is a diameter of circle M, so it divides circle M into two semicircles. ∠퐴푃퐶 and ∠퐴푄퐶 are inscribed angles of these semicircles, so they must be right angles. 푃퐶̅̅̅̅ meets radius 퐴푃̅̅̅̅ at a right angle, so 푃퐶̅̅̅̅ is tangent to circle A. Similarly, 푄퐶̅̅̅̅ meets radius 퐴푄̅̅̅̅ at a right angle, so 푄퐶̅̅̅̅ is tangent to circle A. CK-12 Geometry Honors Concepts 10 Chapter 8 – Circles Answer Key 8.8 Secant Lines to Circles Answers 1. A secant intersects a circle in two points while a tangent intersects a circle in one point. 2. 휃 = 94° 3. 푥 = 9.6 4. 휃 = 50° 5. 푥 ≈ 7.3 6. 휃 = 47.5° 7. 푥 ≈ 2.3 푚퐸퐵̂ 8. Both angles are equal to . 2 9. ∠퐹퐸퐵 ≅ ∠퐸퐻퐵 and both triangles share ∠퐸퐹퐵, so Δ퐸퐻퐹~Δ퐵퐸퐹 by 퐴퐴~. 퐹퐵 퐹퐸 10. Because Δ퐸퐻퐹~Δ퐵퐸퐹, corresponding sides are proportional. This means = , so 퐹퐵 ⋅ 퐹퐻 = 퐹퐸 퐹퐻 2 퐹퐸 . 푚퐻퐺퐸̂ 푚퐵퐸̂ 11. 푚∠퐸퐵퐻 = and 푚∠퐹퐸퐵 = because they are inscribed angles. 푚∠퐸퐵퐻 = ∠퐹퐸퐵 + 2 2 ∠퐵퐹퐸 because the measure of an exterior angle of a triangle is equal to the sum of the measures of 푚퐻퐺퐸̂ 푚퐵퐸̂ the remote interior angles.
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