Reteaching - Weebly

Name �     ClassDateReteaching (continued)Midsegments of TrianglesProblemAB is a midsegment of GEF. What is the value of x?E2x2AB GFBA2(2x) 20FG204x 20x 5ExercisesFind the length of the indicated segment.1. AC 302. TU 13BD153. SU 5.3RREUTQ4. MO 4.4V5. GH 30LT10.6SCA26US6. JK 9HMNG8.815IFOPKO4.5JNELAlgebra In each triangle, AB is a midsegment. Find the value of x.7.MAO3x7BJ5A3x 15BTN5x 710.R8.L 112x 172xx 7A3x 11HD15x 6.5 L 1.3K2x 5B11.9.ASME10Q12.BA10x BAFPCopyright by Pearson Education, Inc., or its affiliates. All Rights Reserved. x2713.5BR

Name �     ClassDateReteachingPerpendicular and Angle BisectorsPerpendicular BisectorsThere are two useful theorems to remember about perpendicular bisectors.Perpendicular Bisector TheoremIf a point is on the perpendicularbisector of a segment, then it isequidistant from the endpointsof the segment.X is on the perpendicularbisector, so it isequidistant from theendpoints A and B.AX33BConverse of the PerpendicularBisector TheoremIf a point is equidistant from theendpoints of a segment, then it ison the perpendicular bisector ofthe segment.Because X is equidistantfrom the endpoints Cand D, it is on theperpendicular bisectorof the segment.X33CDProblemAWhat is the value of x?Since A is equidistant from the endpoints of the segment, it is on theperpendicular bisector of EG. So, EF GF and x 4.77xEExercisesG4FFind the value of x.1.8M4.6EGP 42x5.FP 12 x6.R2xR53JGx 2H4SF32x 3MQ2xx 6DL3.55BP2Nx8A2.U10x 3Copyright by Pearson Education, Inc., or its affiliates. All Rights Reserved. 1S5x 2T

Name �     ClassDateReteaching (continued)Perpendicular and Angle BisectorsAngle BisectorsThere are two useful theorems to remember about angle bisectors.Angle Bisector TheoremIf a point is on the bisectorof an angle, then the point isequidistant from the sides ofthe angle.X is on the anglebisector and is thereforeequidistant from thesides of the angle.A4BX4CConverse of the AngleBisector TheoremIf a point in the interior of anangle is equidistant from thesides of an angle, then the pointis on the angle bisector.Because X is in theinterior of the angleand is equidistantfrom the sides, X ison the angle bisector.DXEFProblemWhat is the value of x?BBecause point A is in the interior of the angle and it is equidistantfrom the sides of the angle, it is on the bisector of the angle.C BCA ECA8x40A8Ex 40ExercisesFind the value of x.707.48.109.3x12(4x)3838(2x 20)70xCopyright by Pearson Education, Inc., or its affiliates. All Rights Reserved. 55

Name �     ClassDateReteachingBisectors in TrianglesThe Circumcenter of a TriangleIf you construct the perpendicular bisectors of all three sides ofCircumcentera triangle, the constructed segments will all intersect at one point.This point of concurrency is known as the circumcenter of the triangle.It is important to note that the circumcenter of a triangle can lie inside,on, or outside the triangle.The circumcenter is equidistant from the threevertices. Because of this, you can construct acircle centered on the circumcenter that passesthrough the triangle’s vertices. This is called acircumscribed circle.ProblemFind the circumcenter of right ABC.6First construct perpendicular bisectors of the two legs,AB and AC. These intersect at (2, 2), the circumcenter.4yBD2Notice that for a right triangle, the circumcenter ison the hypotenuse.C2 A22x462K x4ExercisesCoordinate Geometry Find the circumcenter of each right triangle.1.642.y3.yMR4 J422 Q22(3, 2)4S x6xO4Ly23(1, 1)4H4 2O2N(0, 0)Coordinate Geometry Find the circumcenter of ABC.4. A(0, 0) (5, 4)B(0, 8)C(10, 8)5. A(-7, 3) (1, 2)B(9, 3)C(-7, -7)6. A(-5, 2) ( 1, 4)B(3, 2)C(3, 6)Copyright by Pearson Education, Inc., or its affiliates. All Rights Reserved.

Name �     ClassDateReteaching (continued)Bisectors in TrianglesThe Incenter of a TriangleIf you construct angle bisectors at the three vertices of a triangle, thesegments will intersect at one point. This point of concurrency wherethe angle bisectors intersect is known as the incenter of the triangle.IncenterIt is important to note that the incenter of a triangle will always lieinside the triangle.The incenter is equidistant from the sidesof the triangle. You can draw a circle centeredon the incenter that just touches the threesides of the triangle. This is called aninscribed circle.ProblemBFind the value of x.The angle bisectors intersect at P. The incenter P is equidistantfrom the sides, so SP PT . Therefore, x 9.SxA9Note that PV , the continuation of the angle bisector, is not thecorrect segment to use for the shortest distance from P to AC.TP10VExercisesCFind the value of x.147.28.129.5x 6xx 62x142x 610.3.511.1612.4x2x 74 x2x 12Copyright by Pearson Education, Inc., or its affiliates. All Rights Reserved. 4x 885x

Name �     ClassDateReteachingMedians and AltitudesA median of a triangle is a segment that runs from one vertex of the triangleto the midpoint of the opposite side. The point of concurrency of themedians is called the centroid.AThe medians of ABC are AM, CX , and BL.XLThe centroid is point D.CMDBAn altitude of a triangle is a segment that runs from one vertex perpendicularto the line that contains the opposite side. The orthocenter is the point ofconcurrency for the altitudes. An altitude may be inside or outside the triangle,or a side of the triangle.QThe altitudes of QRS are QT , RU , and SN .The orthocenter is point V.NVURS TDetermine whether AB is a median, an altitude, or neither.A altitude1. OA median2.3B3.2TXA3.altitudeZBneither4. AEBCBD5. Name the centroid. ZS6. Name the orthocenter. PQWUFZBTSTPRCopyright by Pearson Education, Inc., or its affiliates. All Rights Reserved.

Name �     ClassDateReteaching (continued)Medians and AltitudesThe medians of a triangle intersect at a point two-thirds of the distance from avertex to the opposite side. This is the Concurrency of Medians Theorem.CJ and AH are medians of ABCand point F is the centroid.ACF 23 CJFCJHProblemBPoint F is the centroid of ABC. If CF 30, what is CJ?CF 23 CJConcurrency of Medians Theorem30 23 * CJFill in known information.32 * 30 CJ45 CJMultiply each side by 32 .Solve for CJ.ExercisesIn VYX , the centroid is Z. Use the diagram to solve the problems.7. If XR 24, find XZ and ZR. 16; 8V8. If XZ 44, find XR and ZR. 66; 229. If VZ 14, find VP and ZP. 21; 7ROZXPY10. If VP 51, find VZ and ZP. 34; 1711. If ZO 10, find YZ and YO. 20; 3012. If YO 18, find YZ and ZO. 12; 6In Exercises 13–16, name each segment.D13. a median in DEF DL14. an altitude in DEF FK15. a median in EHFHLKG HJIFL16. an altitude in HEK HK or KECopyright by Pearson Education, Inc., or its affiliates. All Rights Reserved. E

Name �     ClassDateReteachingIndirect ProofIn an indirect proof, you prove a statement or conclusion to be true by proving theopposite of the statement to be false.There are three steps to writing an indirect proof.Step 1: S tate as a temporary assumption the opposite (negation) of what youwant to prove.Step 2: Show that this temporary assumption leads to a contradiction.Step 3: C onclude that the temporary assumption is false and that what you wantto prove must be true.ProblemGiven: There are 13 dogs in a show; some are long-haired and the rest areshort-haired. There are more long-haired than short-haired dogs.Prove: There are at least seven long-haired dogs in the show.Step 1: Assume that fewer than seven long-haired dogs are in the show.Step 2: Let / be the number of long-haired dogs and s be the number of shorthaired dogs. Because / s 13, s 13 - /. If / is less than 7, s isgreater than or equal to 7. Therefore, s is greater than /. This contradictsthe statement that there are more long-haired than short-haired dogs.Step 3: Therefore, there are at least seven long-haired dogs.ExercisesWrite the temporary assumption you would make as a first step in writing anindirect proof.1. Given: an integer q; Prove: q is a factor of 34. Assume q is not a factor of 34.2. Given: XYZ; Prove: XY XZ 7 YZ.Assume XY XZ " YZ .3. Given: rectangle GHIJ; Prove: m G 90 Assume mjG 90.4. Given: XY and XM; Prove: XY XMAssume XY XM.Write a statement that contradicts the given statement.5. Whitney lives in an apartment. Whitney does not live in an apartment.6. Marc does not have three sisters. Marc has three sisters.7. 1 is a right angle. j1 is an acute angle.8. Lines m and h intersect. Lines m and h do not intersect.Copyright by Pearson Education, Inc., or its affiliates. All Rights Reserved.

Name �     ClassDateReteaching (continued)Indirect ProofProblemGiven: A and B are not complementary.Prove: C is not a right angle.AStep 1: Assume that C is a right angle.CBStep 2: If C is a right angle, then by the Triangle Angle-Sum Theorem,m A m B 90 180. So m A m B 90. Therefore, Aand B are complementary. But A and B are not complementary.Step 3: Therefore, C is not a right angle.ExercisesComplete the proofs.9. Arrange the statements given at the right to complete the steps ofYthe indirect proof.Given: XY R YZProve: 1 R 43 4Z1 2XStep 1:?BA. But XY R YZ.Step 2:Step 3:Step 4:?DB. Assume 1 4.FC. Therefore, 1 R 4.ED. 1 and 2 are supplementary, andStep 5:?AStep 6:?C 3 and 4 are supplementary.E. According to the Converse of the IsoscelesTriangle Theorem, XY YZ or XY YZ.F. If 1 4, then by the CongruentSupplements Theorem, 2 3.10. Complete the steps below to write a convincing argument using indirectEreasoning.Given: DEF with D R FProve: EF R DEStep 1:?Assume EF DE .Step 2:?If EF DE , then by the Isosceles Triangle Theorem, jD jF.Step 3:?But jD R jF.Step 4:?Therefore, EF R DE .Copyright by Pearson Education, Inc., or its affiliates. All Rights Reserved. DF

Name �     ClassDateReteachingInequalities in One TriangleFor any triangle, if two sides are not congruent, then the larger angle is oppositethe longer side (Theorem 33). Conversely, if two angles are not congruent, thenthe longer side is opposite the larger angle (Theorem 34).ProblemCUse the triangle inequality theorems to answer the questions.a. Which is the largest angle of ABC?86AB is the longest side of ABC. C lies opposite AB.A C is the largest angle of ABC.B9b. What is m E? Which is the shortest side of DEF?m D m E m F 180Triangle Angle-Sum Theorem30 m E 90 180FSubstitution120 m E 180Additionm E 60Subtraction Property of Equality30DE D is the smallest angle of DEF. Because FE lies opposite D,FE is the shortest side of DEF.Exercises1. Draw three triangles, one obtuse, one acute, and one right. Label the vertices.Exchange your triangles with a partner.a. Identify the longest and shortest sides of each triangle.b. Identify the largest and smallest angles of each triangle.c. Describe the relationship between the longest and shortest sides andthe largest and smallest angles for each of your partner’s triangles.Check students’ work. The longest side will be opposite the largest angle.The shortest side will be opposite the smallest angle.Which are the largest and smallest angles of each triangle?F2.7D64largest: jDEF;smallest: jDFE8PE3QCWhich are the longest and shortest sides of each triangle?5.longest: DF; shortest: FEF15D6.longest: PQ; shortest: RQR7P70QBlongest: SV ; shortest: ST7. S35130Elargest: jACB;smallest: jCBA10R largest: jPQR; 4. Asmallest: jPRQ653.T100Copyright by Pearson Education, Inc., or its affiliates. All Rights Reserved. 35V