2. P lies on the perpendicular bisector of RS ; Because. 168 ft. 3. P lies on the angle bisector of DEF;

Transcription

1 9. = 9 x 9. = x 95. a. ft b. ft b ft c. 9. a. 0 ft b. ft c. hapter. Start Thinking ft ft The roof lines become steeper; The two top chords will get longer as the king post gets longer, but the two top chords will alwas be congruent to each other. The angles formed b the top chords and the king post are congruent. The angles formed b the top chords and the bottom chord are congruent.. Warm Up umulative Review Warm Up.. STTEMENTS. P is the midpoint of MN and TQ.. Practice RESONS. Given. MP NP. Definition of segment midpoint. PT PQ. Definition of segment midpoint. MPQ NPT. Vertical ngles ongruence (Thm..) 5. MQP NTP 5. SS ongruence (Thm. 5.5) STTEMENTS. D, D RESONS. Given.. Reflexive Propert of Segment ongruence (Thm..). D. SSS ongruence (Thm. 5.). P lies on the perpendicular bisector of RS ; The markings show that TP satisfies the definition of a perpendicular bisector.. P lies on the perpendicular bisector of RS ; ecause T is equidistant from the endpoints R and S, T lies on the perpendicular bisector of RS b the onverse of the Perpendicular isector (Thm..). ecause onl one line can be perpendicular to RS at U, TU must be the perpendicular bisector of RS, and P is on TU.. P lies on the angle bisector of DEF; P is equidistant from sides ED and EF of angle DEF, so P is on the angle bisector of DEF b the onverse of the ngle isector (Thm..).. 0; ecause D is on the perpendicular bisector of, D is equidistant from and. 5. 7; the Perpendicular isector (Thm..), GJ = GH. Solving x + 5 = x + gives x =, so x + = 7.. ; Q is on the angle bisector of PSR, so Q is equidistant from SP and SR ; the onverse of the ngle isector (Thm..), GE bisects DGF. Solving x + = 5x gives x = 0. So, m DGF = = 7.. = x ecause an point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment, ou can draw an isosceles triangle b drawing segments from each endpoint of the segment to the same point on the perpendicular bisector. 0. es; In a right triangle, the bisector of the right angle is also the perpendicular bisector of the hpotenuse when the right angle is isosceles. The figure shows that when right of is bisected b D, D D b either the S ongruence (Thm. 5.0) or the SS ongruence (Thm. 5.5). Then the corresponding sides and must be congruent, so D is isosceles. opright ig Ideas Learning, LL

2 . Practice. neither; ecause PQ is not marked perpendicular to RS, ou cannot be certain that point P is on the perpendicular bisector of RS.. neither; ecause there is no indication that PD = PF, ou cannot be certain that point P is on the angle bisector of DEF.. P lies on the bisector of DEF; Using the HL ongruence (Thm. 5.9), DEQ FEQ, so DQ = FQ and EQ bisects DEF b the onverse of the ngle isector (Thm..). The figure shows that point P is on EQ.. ; Point D lies on the perpendicular bisector of b the onverse of the Perpendicular isector (Thm..). m D = 90 b the Triangle Sum (Thm. 5.), so D is the perpendicular bisector of. Solving x + 5 = x, gives x = 7, so = ; ecause L is units from each side NK and NM, NL bisects KNM. ecause m KNM = 90, m LNM = ( 90 ) = 5.. ; TV bisects UTW because VU = VW. Solving x + = 5x gives x = 9, so m UTW = 9 + =.. Enrichment and Extension. 7. ( x + ) + ( ) = ( x ) + ( ) = x 5 5 ;. x =, =. x =, = 5. STTEMENTS RESONS. PQRST is a regular polgon. SV RV. TP QP TS S QR R. DrawTV and QV.. Given. Definition of regular polgon. Through an points there exists exactl one line.. TSV QRV. SS ongruence (Thm. 5.5) 5. VT VQ 5. orresponding parts of congruent triangles are congruent.. VT = VQ PT = PQ 7. V lies on the perpendicular bisector of TQ. P lies on the perpendicular bisector of TQ.. Definition of congruent segments 7. onverse of the Perpendicular isector (Thm..) 7. = x +. no; es; Point Q does not lie on line m because Q is not the same distance from P and R. Point S lies on line m, the perpendicular bisector of PR, because S is the same distance from P and R. 9. Install the fountain at the point where the bisectors of the angles of the triangle intersect. That point is the same distance from each side of each angle, or equivalentl, each side of the triangle.. PV lies on the perpendicular bisector of TQ.. Perpendicular Postulate (Post..) opright ig Ideas Learning, LL

3 . STTEMENTS RESONS. LQ NQ. Given.. ase ngles (Thm. 5.).. Given. + = MLR. ngles ddition + = MNR Postulate (Post..) 5. + = MLR 5. Substitution. MLR MNR. Substitution 7. MN MN 7. onverse of the ase ngles (Thm. 5.7). MQ is the perpendicular bisector of LN.. onverse of Perpendicular isector (Thm..) 9. LP NP 9. onverse of Perpendicular isector (Thm..)... umulative Review Warm Up. x D D = x. Puzzle Time TOOK HIS MEDIINE ND FORGOT TO SHKE IT.. Start Thinking = The intersect at one point that appears to be the center of the circle. The also appear to be both angle bisectors and perpendicular bisectors.. Warm Up.. x X Y. Practice. 9; ircumcenter (Thm..5). 0; ircumcenter (Thm..5). ; Incenter (Thm..). (, ) 5. (, ) opright ig Ideas Learning, LL

4 Sample answer: 9. onstruct the circumcenter of the triangle formed b the locations of the three buildings. The tower is located at the circumcenter. 0. no; The circumcenter of an obtuse triangle lies outside the triangle, and the circumcenter of a right triangle lies on the hpotenuse.. Practice. ; ircumcenter (Thm..5). 5; ircumcenter (Thm..5). ; Incenter (Thm..). 9, 5. MG = 0, NG = 0. GJ =, GE = 7 7. a. Find the incenter b finding the intersection of two angle bisectors. b. Find the circumcenter b finding the intersection of the perpendicular bisectors of two sides. c. the circumcenter; The circumcenter is centered between the three vertices, whereas the incenter is toward the bottom of the triangular lawn in the figure. 5.. egin b drawing a line segment from point to point F, as shown. You are given F is the perpendicular bisector of and FE is the perpendicular bisector of D. the Perpendicular isector (Thm..) ou know that F = F and FD = F. Using the Transitive Propert of Equalit, ou can conclude that F = FD. the definition of congruent segments, ou know that F FD. STTEMENTS. GJ is the bisector of HK. RESONS. Given. HJ JK. Definition of segment bisector. MH MK. Perpendicular isector (Thm..). GH GK. Perpendicular isector (Thm..) 5. GM GM 5. Reflexive Propert of Segment ongruence (Thm..). GHM GKM 7. GHM GKM. SSS ongruence (Thm. 5.) 7. orresponding parts of congruent triangles are congruent.. The circumcenter lies outside the triangle when the triangle is obtuse. F 9.. Enrichment and Extension. (, ) Q. units 5 5. ( 5.,. ). (.,.0 ). Puzzle Time HD YTE E D opright ig Ideas Learning, LL

5 . Start Thinking. Practice. QL =, L =. JQ =, Q =. Q 5, K 5 = =. (, ) 5. inside;,. outside; (, 9) Two of the altitudes coincide with the two legs of the right triangle, and the three altitudes intersect at the vertex of the right angle.. Warm Up. P. m. umulative Review Warm Up = +. x = 5. ( x ). 9 ( x ) + =. + 7 = ( x ) 5. 5 ( x ) + =. Q =. Practice. P =, PL =. PL =, L =. P =, PL = 9. PL = 5, L = 5 5. (, ). (, ) 7. on; ( 0, 0 ). outside; (, 5) 9. orthocenter; When the strings are pulled tight, right angles are formed on opposite sides of each vertex. Three altitudes are formed, which are concurrent at the orthocenter. 7. ; Sample answer: The two vertices give one side of the triangle. The centroid and each given vertex can be used to find the exact location of the midpoint of another side. The two midpoints and the two given vertices can be used to find the third vertex of the triangle.. ; Sample answer: The two vertices give one side of the triangle. Each vertex and the orthocenter can be used to draw a line containing the altitude from that vertex. The line perpendicular to that line and passing through the other vertex contains another side of the triangle. The third vertex is the intersection of the new sides. 9. ( 0, ) 0. (, ). no; Sample answer: congruent pair of adjacent sides in a triangle have a perpendicular bisector, angle bisector, median, and altitude from the shared vertex that are all the same segment. n equilateral triangle has three pairs of congruent adjacent sides, so the special segments will all be the same, as will their points of concurrenc.. es; Sample answer: The median is the same as the altitude onl when the sides that share the vertex are congruent, so the triangles formed b this median will be congruent b the HL ongruence (Thm. 5.9).. no; The circumcenter of an obtuse triangle is outside the triangle and the incenter of an triangle is inside the triangle.. Enrichment and Extension. x + x + x, + + N, z + z + z a b c. N,, 0. no; The orthocenter and the centroid are the same point in an isosceles triangles. opright ig Ideas Learning, LL 5

6 a b. T,,0 ; The distance of FN = a b c + + = a + b + c and the distance of NT = a b c + + = a + b + c, so FN = NT.. 5. or 7. or square units. about.7 square units 9. M = c ac = x, b a b a N = c = x, a + b P = c ac = x ; b a b a ( a + b,c). Puzzle Time OOKWORM. Start Thinking. umulative Review Warm Up. 79.,., 50,. Practice. The slope of ED = =, and the slope of = =. So, ED. 5 ED = + = 0, = 5 + = 0, so ED =.. E(, ), F (, ). The slope of EF = =, and the slope of = =. So, EF. 5 EF = + =, = 5 + =, so EF =.. D(, ), F (, ) 5. The slope of DF = = 0, and the slope of = = 0. So, DF. ( ), = =, so DF = = and DF = M N The measure of MN is one-half the measure of.. Warm Up.. (, 0), es; ecause each midsegment is half as long as the corresponding side, the sum of the lengths of the midsegments (the perimeter of the midsegment triangle) will be half the sum of the lengths of the corresponding sides (the perimeter of the original triangle).. ft opright ig Ideas Learning, LL

7 . Practice. D(,, ) E(,, ) F(, ).. The slope of slope of The slope of slope of The slope of slope of FD = =, and the 5 = =. So, FD. FE = =, and the = =. So, FE. 5 DE = =, and the 5 = =. So, DE. 5. ( [ ]) ( [ ]) FD = + =, ( [ ]) ( [ ]) = + 5 =, so FD =. ( [ ]) ( [ ]) FE = + = 0, ( [ ]) = 5 + = 0, so FE =. ( [ ]) DE = + = 7, ( [ ]) = = 7, so DE F D =. E x Sample answer: The two pairs of equal measures determine the midpoints of two sides of a triangle. The midsegment of the triangle is represented b the surface of the bottom step. So, the bottom step is parallel to the floor, which represents the bottom side of the triangle. 0. no; Sample answer: The midsegments of a triangle with side lengths of a, b, and c divide the triangle into triangles with side lengths of a, b, and c. So, the smaller triangle has onl one-fourth the area of the larger triangle.. Enrichment and Extension. a. b. c. d. D H F Stage n 0 5 Midsegment length n = = w. G(,, ) H( 0,, ) J (,0) J G. perimeter = 7. units, n E Midsegments of Triangles 0 Midsegment length 0 5 x Stage area =.5 units opright ig Ideas Learning, LL 7

8 ecause. You are given DEF. corresponding parts of congruent triangles are congruent, ou can conclude DE, EF, and FD. the definition of congruent segments, ou can also conclude = DE, = EF, and = FD. You are also given that T, U, and V are the midpoints of and X, Y, and Z are midpoints of DEF. So, TV = and XZ = EF. the Substitution Propert of Equalit, ou have TV = EF and TV = XZ. You know UV = and YZ = DE. So, b the Substitution Propert of Equalit, ou have UV = DE and UV = YZ. Finall, ou know TU = and XY = FD. So, b the Substitution Propert of Equalit, ou have TU = FD and TU = XY. the definition of congruent segments, ou can conclude TV XZ, UV YZ and TU XY. the SSS ongruence (Thm. 5.), ou can conclude TUV XYZ.. Puzzle Time STIK WITH ME ND WE WILL GO PLES.5 Start Thinking. is the longest side and is the largest angle.. YZ is the longest side and X is the largest angle.. NP is the longest side and M is the largest angle. The largest angle is alwas opposite the longest side..5 Warm Up. If there is no right angle in a triangle, then it is not a right triangle.. If two lines do not have the same slope, then the are not parallel.. Sample answer: If a quadrilateral does not have four right angles, then the quadrilateral is not a rectangle. 5. If the sum of the measures of the interior angles of a polgon is not 0, then the polgon is not a triangle.. Sample answer: If a triangle does not contain three congruent angles, then it is not equiangular..5 umulative Review Warm Up Practice. N, L, M. F, D, E.,,. QP, PR, RQ 5. no; The sum of the first two sides is = 5, which is not greater than 5.. es; The sum of the lengths of an two sides of the triangle is greater than the length of the third side. 7. ssume that a triangle has more than one obtuse angle. n obtuse angle is an angle that is greater than 90. This makes the sum of the angles in the triangle greater than 0. However, the sum of the angles of a triangle must be equal to, not greater than 0. This is a contradiction, so the assumption that a triangle has more than one obtuse angle must be false, which proves that a triangle has, at most, one obtuse angle.. x > 9. D, D, ; Using the properties of exterior angles, ou can solve for x =, and then obtain all of the angle measures. 0. Sample answer: In terms of the Triangle Inequalit (Thm..), this can be thought of as a direct route between two points b traveling along one side, or an indirect route between two points b traveling along the other two sides. ecause the sum of the lengths of an two sides of a triangle is greater than the length of the third side, the direct route along a single side is the shortest distance between two points..5 Practice. L, M, N. V, U, W. QS, RQ, SR., D, D. Sample answer: If no two angles of a triangle are congruent, then the triangle is a scalene triangle. opright ig Ideas Learning, LL

9 5. ssume temporaril that a right triangle has three acute angles. So the measure of each angle is less than 90, but b the definition of a right triangle, one angle must have a measure of 90. So, the assumption that a right triangle has three acute angles must be false. Next assume temporaril that a right triangle has exactl one acute angle, which means the triangle has one right angle and one obtuse angle. The measure of an obtuse angle plus 90 is greater than 0. the Triangle Sum (Thm. 5.), m + m + m = 0. So, the assumption that a triangle has exactl one acute angle must be false, which proves that a right triangle has exactl two acute angles.. es; Substituting the given value of x into the expressions for the measures of the sides gives 0, 07, and. ecause the sum of an two of these lengths is greater than the third, it is possible to construct such a triangle. Use the Triangle Inequalit 7. DEF, FG; (Thm..). For DEF, DE + DF > EF, DE + EF > DF, and DF + EF > DE. For FG, F + G > FG, G + FG > F, and FG + F > G.. es; If ou know all three angles measures, ou can use a protractor and a straightedge to construct a triangle with the given angles that obes the Triangle Inequalit (Thm..)..5 Enrichment and Extension. D,, D,, D. DE, E, D,, D,, D. Sample answer: STTEMENTS. and median M. Extend M to point D such that M DM. Draw D. RESONS. Given. onstruction. M M. Definition of median. M DM. Vertical ngles ongruence (Thm. 5.5) 5. M DM 5. SS ongruence (Thm. 5.5). D. orresponding parts of congruent triangles are congruent. 7. = D, 7. Definition of M = DM congruent segments. M + MD = D. Segment ddition Postulate (Post..) 9. M + M = D 9. Substitution Propert of Equalit 0. M = D 0. Simplif.. D < + D. Triangle Inequalit (Thm..). M < +. Substitution. M ( ) < +. Division Propert of Equalit. ( + ) < ( + + ) 5. M ( + + ) <. Properties of real numbers 5. Transitive Propert of Inequalit opright ig Ideas Learning, LL 9

10 . Sample answer: In, let X, Y,andZ be medians. the entroid (Thm..7) and the Triangle Inequalit (Thm..), X + Y >, X + Z >, and Y + Z >. dding the left sides and right sides of the three inequalities gives X + Y + Z > + +. Multipling each side b gives X + Y + Z > + +. Finall, > and the Transitive Propert of Inequalit gives X + Y + Z > ( + + ), as desired < x < 0; 0 < x < 0. a. < x < b. x > 7. If a line segment is perpendicular to a plane, then it is perpendicular to ever line segment in the plane. So, P D, and PD is a right triangle. The largest angle in a right triangle is the right angle, so m PD > m PD. Finall, ou can conclude that PD > P because if one angle of a triangle is larger than another angle, then the side opposite the larger angle is longer than the side opposite the smaller angle..5 Puzzle Time JUMP ROPE. Start Thinking The length increases; The angle must be less than 0 degrees;. Practice. > DF; the Hinge (Thm..), because is the third side of the triangle with the larger included angle, it is longer than DF.. m HGI = m IGJ; The triangles are congruent b the SSS ongruence (Thm. 5.). So, because corresponding parts of congruent triangles are congruent, m HGI = m IGJ.. m < m ; the onverse of the Hinge (Thm..), because is the included angle in the triangle with the shorter third side, its measure is less than that of.. KL < MN; the Hinge (Thm..), because KL is the third side of the triangle with the smaller included angle, it is shorter than MN. 5. x + 7 < x, x > 0 x + > x 0, x <. 7. STTEMENTS RESONS. TV UW. Given. UV UV. Reflexive Propert of Segment ongruence (Thm..). TU > VW. Given. m TVU > m WUV. onverse of the Hinge (Thm..). Warm Up. es; E and DE are congruent b the S ongruence (Thm. 5.).. no. no. es; D and E are congruent b the SS ongruence (Thm. 5.5).. umulative Review Warm Up. rotation. dilation. translation 70 opright ig Ideas Learning, LL

11 . STTEMENTS RESONS. D E. Given.. Definition of segment midpoint. m > m. Given. D > E. Hinge (Thm..) 5. DF EF 5. Given. DF = EF. Definition of congruent segments 7. D + DF > E + DF 7. ddition Propert of Inequalit. D + DF > E + EF. Substitution 9. D + DF = F, E + EF = F 9. Segment ddition Postulate (Post.) 0. F > F 0. Substitution 9. no; You cannot appl either the Hinge (Thm..) or the onverse of the Hinge (Thm..) in this situation; These theorems require that two sides of one triangle are congruent to two sides of the other triangle. In this case, the ladders are different heights, so ou onl have one pair of congruent sides.. Practice. > DE; the Hinge (Thm..), because is the third side of the triangle with the larger included angle, it is longer than DE.. JI > GH; the Hinge (Thm..), because JI is the third side of the triangle with the larger included angle, it is longer than GH.. m > m ; the onverse of the Hinge (Thm..), because is the included angle in the triangle with the longer third side, its measure is greater than that of.. m U < m R; the onverse of the Hinge (Thm..), because U is the included angle in the triangle with the shorter third side, its measure is less than that of R. x > x +, x > 5. x + > x, x <. 7. STTEMENTS RESONS. PQ SR. Given. PS + SR > PR. Triangle Inequalit (Thm..). PR = PQ +. Segment ddition QR Postulate (Post..). PS + SR > PQ + QR. Substitution 5. PQ = SR 5. Definition of congruent segments. PS + SR >. Substitution SR + QR 7. PS > QR 7. Subtraction Propert of Inequalit. m PQS >. onverse of the m RSQ Hinge (Thm..). Sailboat ; ecause 5 > 9, the distance Sailboat traveled is a greater distance than the distance Sailboat traveled b the Hinge (Thm..). 9. Each theorem refers to the included angles of two triangles when two sides of one triangle are congruent to two sides of the other triangle, The Hinge (Thm..) refers to the case when the included angle of the first is larger than the included angle of the second. The SS ongruence (Thm. 5.5) refers to the case when the included angles are congruent.. Enrichment and Extension. a. never b. never c. alwas d. never e. never f. sometimes opright ig Ideas Learning, LL 7

12 . D and E because the are radii of the circles and all radii are congruent. the Hinge (Thm..), because m > m DE, then > DE...5 < < Puzzle Time DUK STTEMENTS. D and D are supplementar.. m D + m D = 0. m D = 0 m D umulative Review RESONS. m D = 00. Given 5. m D = Linear Pair Postulate (Post..). Definition of supplementar angles. Subtraction Propert of Equalit 5. Substitution. m D = 0. Subtraction Propert of Equalit 7. m D > 7. Definition of m D congruent angles. D is the midpoint of.. Given 9. D D 9. Definition of midpoint 0. D D 0. Reflexive Propert of Segment ongruence (Thm..). >. Hinge (Thm..). m > m. Triangle Longer Side (Thm..9). c > 9 or c <. < s <. 0 < m 7. p < 5 5. j. < v < 7. 5 r <. b 7 or b < 9. m = 0. m =. m is undefined m =. m = m is undefined m = 0. 5 m =. 9 m = 5 m = 5 m = 9 m = 7 m = 7 m = 5 m =. = x 0 5. = x = x = x. x = 9. x = = x. a. =.5t b. $.5 c. $.75 7 = x +. = x 5. =. = = x = x 7. = x 7 7 opright ig Ideas Learning, LL

13 9. 5 = x 0. 5 = x +. x = 5, =. x = 77, = 0. x =, = 77. x =, = 5. x = 5, =. x =, = a. 7 b.. a. b (, ) 50. (, ) x 5. ( 5, ) 5. D( 0, ) 5. ( 5, ) 5. (, 9) 55. ( 5, ) 5. D(, ) x x opright ig Ideas Learning, LL 7

14 0.. x x.. 0 D F F D 0 0 x E x E 0 7 opright ig Ideas Learning, LL

15 . 7. Z Z Y M N P W W Y x P x X X. N M 5. V 0 0 x T V 0 0 x S T 0 S U 0 U L G M L 0 M 0 0 x J J 0 H E F E F x H G opright ig Ideas Learning, LL 75

16 70. equilateral triangle 7. acute isosceles triangle 7. right scalene triangle 7. obtuse scalene triangle ; obtuse scalene triangle ; acute isosceles triangle 7. ; right scalene triangle ; obtuse scalene triangle hapter 7 7. Start Thinking Warm Up umulative Review Warm Up. x =. = 5. + = ( x 5). = ( x + ) 7. Practice gon. interior:, exterior: m X = m Y = m X = m Y = people 7. Practice. 0.. m X = m Y =. m X = m Y = interior: 5, exterior: 5 9. ; The sum of the interior angle measures of the polgon is ( 90 ) + ( 0 ) + 0 = 0. So, the polgon has 0 + = sides a. 70 b Enrichment and Extension regular decagon 0. sides. You know that a + b + c = 0, d + e + f = 0, and g + h + i = 0 because the sum of the interior angles of a triangle equals 0. You can add those three equations to obtain a + b + c + d + e + f + g + h + i = 50. m YZV = f + i, m ZVW = h, m VWX = g + d + a, m WXY = b, and m XYZ = c + e, so m YZV + m ZVW + m VWX + m WXY + m XYZ = Puzzle Time THEY DIDN T WNT TO WIT FORTY YERS FOR TRIN 7. Start Thinking es; Sample answer: The scout could use the Pthagorean to determine the distance that should be between opposite corner posts, the length of the hpotenuse. It 7 should be approximatel 5 feet 7 inches. Or, the scout could make sure that the distances between the two pairs of opposite corners are the same and not be concerned about the exact measure. This method uses the SSS ongruence (Thm. 5.). 7. Warm Up. STTEMENTS RESONS. MN PO, NO MP. Given. NP NP. Reflexive Propert of Segment ongruence (Thm..). PMN NOP. SSS ongruence (Thm. 5.). 0 7 opright ig Ideas Learning, LL