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1 44. Let be the irumenter of the. Given: = m; so by the properties of -6-9,. So = + = _ 5- = _ () = 4 m. medians and altitudes of Triangles hek it out! 1a. KZ + ZW = KW _ KW + ZW = KW ZW KW 7 KW 1 = KW b. LZ = _ LX = _ (.1) = 5.4. ; 4; possible answer: the -oordinate of the entroid is the average of the -oordinates of the verties of the, and the y-oordinate of the entroid is the average of the y-oordinates of the verties of the.. Possible answer: n equation of altitude to JK is y = - 1_ +. It is true that 4 = - 1_ ( -) +, so (-, 4) is a solution of this equation. Therefore this altitude passes through the orthoenter. think and disuss 1. Possible answer: is isos.. Possible answer: is a rt.. K J L. The ratio of the length of the longer segment to the length of the shorter segment is : entroid Orthoenter efinition Loation (Inside, Outside, or On) The pt. of onurreny of the medians Inside the The pt. of onurreny of the altitudes an be inside, outside, or on the 5. RW = _ RY 14 = _ RY _ (14) = RY RY = WY RW (14) = Understand the Problem nswer will be the oordinates of the entroid of the. Important information is the loation of verties, (, ), (7, 4), and (5. ). Make a Plan The entroid of the is the point of intersetion of the three medians. So write the equations for two medians and find their point of intersetion. Solve Let M be the midpoint of and N be the midpoint of. M = ( _ + 7, _ + 4 ) = (.5, ) N = ( _ 5 + 7, _ + 4 ) = (6, ) N is horizontal. Its equation is y =. Slope of M = _ - = -. Its equation is.5-5 y = -( - 5). t the entroid, y = = -( - 5), so = 5 + (-1) = 4. The oordinates of the entroid are (4, ). 4 Look ak Let L be the midpoint of. Equation for L is y - 4 = _ ( - 7), whih intersets y = at (4, ) y L = 4 K (4, -1) y + = 1_ (- ) 1 M Step Find an equation of the line ontaining the altitude from L to KM. Sine KM is horizontal, the altitude is vertial, so the equation is = 4. Step Find an equation of the line ontaining the altitude from K to LM. Slope of LM = _ = -. Equation is y + ( - ). Step 4 Solve the system to find the oordinates of the orthoenter. = 4 and y + (4 - ) = 1, so y = -1. The oordinates of the orthoenter are (4, -1). eerises guided Pratie 1. entroid. altitude. VW = _ VX 4. WX = _ VX (4) = 16 (4) = 6 9 Holt Mougal Geometry

2 9. - V = -5 U y y = - 1 W (, -) Step Find an equation of the line ontaining the altitude from W to UV. Sine UV is vertial, the altitude is horizontal, so the equation is y = -. Step Find an equation of the line ontaining the altitude from U to VW. Slope of VW = _ = -1. Equation is y + 9 = + 4, or y = - 5. Step 4 Solve the system to find the oordinates of the orthoenter. y = - and - = - 5, so =. The oordinates of the orthoenter are (, -). 1. P y = -5 - (-5, ) R - Q y = +1 4 Step Find an equation of the line ontaining the altitude from P to QR. QR is horizontal, the altitude is vertial, so the equation is = -5. Step Find an equation of the line ontaining the altitude from Q to PR. Slope of PR = _ = -1. Equation is y - 5 = - 4, or y = + 1. Step 4 Solve the system to find the oordinates of the orthoenter. = -5 and y = =. The oordinates of the orthoenter are (-5, ). 11. y 4 = -1 y = E (-1, ) 4 6 Step Find an equation of the line ontaining the altitude from E to. is vertial, the altitude is horizontal, so the equation is y =. Step Find an equation of the line ontaining the altitude from to E. E is horizontal, altitude is vertial, so the equation is = -1. Step 4 y = and = -1. The oordinates of the orthoenter are (-1, ). pratie and problem solving 1. P H 1. H = _ Hp (1.) =.6 = _ (1.) = J = P = (.9) = JP = P = (.9) = Support should be attahed at the entroid. Equation of the median through (4, ) is = 4. The median through (, 1) also passes through ( 4 _ +, _ + 14 ) = (6, 7), and has slope _ 1-7 = - 1_ - 6. Equation of the seond median is y = - 1_ + 1. t intersetion, = 4, so y = - 1_ (4) + 1 =. The oordinates of the entroid are (4, ). 17. Step 1 Find an equation of the line ontaining the altitude through X. YZ is vertial, the altitude is horizontal, so the equation is y = -. Step Find an equation of the line ontaining the altitude through Z. Slope of XY is _ = _. Equation is y + 6 = - _ ( - 6). Step Find the oordinates of the orthoenter. y = -, so = 4 = - _ ( - 6), or = =. The oordinates are (, -). 1. Step 1 Find an equation of the line ontaining the altitude through J. GH is horizontal, the altitude is vertial, so the equation is = 4. Step Find an equation of the line ontaining the altitude through H. Slope of GJ is _ = -1. Equation is y - 5 = - 6. Step Find the oordinates of the orthoenter. = 4, so y - 5 = 4-6, or y = 5 - =. The oordinates are (4, ). 19. Step 1 Find an equation of the line ontaining the altitude through T. RS is horizontal, the altitude is vertial, so the equation is = -. Step Find an equation of the line ontaining the altitude through R. ST is vertial, the altitude is horizontal, so the equation is y = 9. Step Find the oordinates of the orthoenter. = - and y = 9. The oordinates are (-, 9).. Step 1 Find an equation of the line ontaining the altitude through. is vertial, the altitude is horizontal, so the equation is y = -. Step Find an equation of the line ontaining the altitude through. Slope of is _ =. Equation is y + = - 1_ ( - ). Step Find the oordinates of the orthoenter. y = -, so - + = 5 = - 1_ ( - ), or = -1 + = -. The oordinates are (-, -). 99 Holt Mougal Geometry

3 1. GL = _ GP = _ () = 1. PL GP () = 4. HL = LJ = 5 4. GL is the perpendiular bisetor of HJ, so GJ GH. GJ = GH = GK = (6.5) = 1 5. P = GJ + GH + HJ = GH + LJ = (1) + (5) = 6 units 6. ( HJ)(GL) (1)(1) = 6 square units 7. G = ( 1_ ( ), 1_ ( ) ) = (1, -). G = ( 1_ ( + + 5), 1_ ( ) ) = (5, 1) 9. PZ = ZX = (7) = Step 1 Find n. n + 17 = 54 n = n = 1.5 Step Find QZ. QZ = 4n - 6 = 4(1.5) - 6 = 4. YZ ( QZ) (4) = 4. Possible answer: the perpendiular bisetor of base; the bisetor of verte ; the median to the base; the altitude to the base 4. sometimes 5. always 6. never 7. always. PX = ZX = (7) = 1. Statements Reasons 1. PS and RT are medians of PQR. 1. Given PS RT. PS = RT. ef. of segs.. _ PS = _ RT. Mult. Prop. of = 4. PZ = _ PS, RZ = _ RT 4. entroid Thm. 5. PZ = RZ 5. Subst. 6. PZ RZ 6. ef. of segs. 7. SPR TRP 7. Isos. Thm.. PR PR. Refle. Prop. of 9. PTR RSP 9. SS 1. QPR QRP 1. PT 11. PQ RQ 11. on. of Isos. Theorem 1. PQR is an isos. 1. ef. of isos. 9. Possible answer: The entroid of a is also alled its enter of gravity beause the weight of the shape is evenly distributed in every diretion from this point. This means the shape will rest in a horizontal position when supported at this point. 4a. G = ( 1_ ( + + ), 1_ ( + + ) ) = ( _, _ ) b. G = ÇÇÇÇÇ ( _ ) + ( _ ) = _ Ç. mi. Perpendiular from G rosses EF at H(4, 4), distane = ÇÇÇÇÇ ( 4_ ) + ( 4_ ) = 4_ Ç 1.9 mi test prep G; I, III true sine inenter, entroid always inside II false sine obtuse 4. hallenge and etend 44a. Possible answer: is equil., and l is the perpendiular bisetor of. Sine is equil., by definition. So = by the definition of segs. Therefore by the onverse of the Perpendiular isetor Theorem, is on line l. Similarly, is on the perpendiular bisetor of, and is on the perpendiular bisetor of. l 1 Holt Mougal Geometry

4 b. Possible answer: y the definition of the perpendiular bisetor,. So is the midpoint of by definition, and is a median of by the definition of median. Therefore l ontains the median of through. lso by the definition of the perpendiular bisetor,. So is the altitude of by the definition. Therefore l ontains the altitude of through. gain by the definition of the perpendiular bisetor,. by the definition of equil., and by the Refle. Prop. of. So by SSS. Then by PT, and is the bisetor of by the definition of bisetor. Therefore l ontains the bisetor of through. The same reason an be applied to the other two bisetors.. Possible answer: The perpendiular bisetors of a are onurrent at the irumenter, and the bisetors are onurrent at the inenter. The medians of a are onurrent at the entroid, and the altitudes of a are onurrent at the orthoenter. ut in an equil., the perpendiular bisetor through a given verte also ontains the bisetor, the median, and the altitude through that verte. So the points of onurreny must all be the same point That is, the irumenter, the inenter, the entroid, and the orthoenter in an equil. are the same point. 45a. slope of RS = _ ; slope of ST = _ b b - a ; slope of RT =. b. Sine l b_ RS, slope of l = -. Sine m ST, slope of m = - _ b - a = _ a - b. Sine n RT, n is a vertial line, and its slope is undefined.. n equation b_ of l is n equation of m is y - = - ( - a) y - = _ a - b ( - ) b_ y = - + ab_ y = _ a - b n equation of n is = b. _ d. (b, ab - b ) e. Sine the equation of line n is = b and the -oordinate of P is b, P lies on n. f. Lines l, m, and n are onurrent at P. 5 The Triangle midsegment theorem hek it out! 1. The midpoints are M(1, 1), N(, 4); slope of MN = _ ; slope of RS = 6_ 4 = _ ; sine the slopes are =, MN ǁ RS. MN = ÇÇÇÇ + = ÇÇ 1 ; RS = ÇÇÇÇ = ÇÇ 5 = 1 ÇÇ ; and MN RS. a. JL = PN = (6) = 7. m MLK = m JMP = 1. HF E (155) = 775 b. PM KL (97) = 4.5 The distane she measures between H and F is 775 m. think and disuss 1. The endpoints of XY are not the midpoints of the sides of the.. efinition: a seg. joining the mdpts. of sides of a Eample: Midsegment eerises guided Pratie 1. midpoints Triangle Midsegment E Properties: to the third side; half the length of the third side Noneample:. The midpoints are S(-1, 4), T(4, 6); slope of ST = _ ; slope of PR = 4_ 5 1 = _ 5 ; sine the slopes are =, ST ǁ PR. ST = ÇÇÇÇ + 5 = ÇÇ 9 ; PR = ÇÇÇÇ = ÇÇ 116 = ÇÇ 9 ; and ST PR.. NM XY (1.) = NZ XZ (11.) = m LMN = m MNZ = 9 E 4. XZ = LM = (5.6) = Holt Mougal Geometry

5 9. The midpoint of is (1, ); slope of = _ -1 = -1, so the slope of the 1 perpendiular bisetor is 1; the equation of the perpendiular bisetor is y = The midpoint of XY is (4, 6); slope of XY = _ = 4, so the slope of the perpendiular bisetor is -.5; the equation of the perpendiular bisetor is y - 6 = -.5( - 4). 11. No; to apply the onverse of the ngle isetor Theorem, you need to know that P and P. 1. Yes; sine P, P, and P P, P is on the bisetor of by the onverse of the ngle isetor Theorem. 5- isetors of Triangles 1. GY = HY = GP = JP = GJ = GX 16. PH = JP = 46 = (.) = distane from to UV = distane from to UW = 1 1. m WVU + m VUW + m UWV = 1 m WV + ()+ 66 = 1 m WV = 74 m WV = MO is vertial, so the equation of the horizontal perpendiular bisetor is y = ; NO is horizontal, so the equation of the vertial perpendiular bisetor is = 4. The irumenter is at (4, ).. OR is vertial, so the equation of the horizontal perpendiular bisetor is y = -.5; 5- OS is horizontal, so the equation of the vertial perpendiular bisetor is = -6. The irumenter is at (-6, -.5). Medians and ltitudes of Triangles 1. Z = _ = _. EZ = Z 11.6 = Z Z = 5. (4.6) = = Z 4.6 = Z Z =. 4. E = Z = (5.) = JK is vertial, so the equation of the altitude from L is y = ; KL is horizontal, so the equation of the altitude from J is = -6. The orthoenter is at (-6, ). 6. is horizontal, so the equation of the altitude from is = 1; is vertial, so the equation of the altitude from is y =. The orthoenter is at (1, ). 7. RT is horizontal, so the equation of the altitude from S is = 7; RS has slope 5_ = 1, so the equation of the altitude 5 from T is y - = -( - ). t the orthoenter, = 7 and y - = -(7 - ) = 1 y = 4, so the orthoenter is at (7, 4).. XY is horizontal, so the equation of the altitude from Z is = ; XZ has slope 6_ = -1, so the equation of the altitude -6 from Y is y - = - 5 or y = -. t the orthoenter, = and y = - =, so the orthoenter is at (, ). 9. G = ( 1_ ( + + 6), 1_ (4 + + ) ) = (, 4) The Triangle Midsegment Theorem 5. XY (7.) = 5.1. X XZ = =.4 4. m X = 1 - m = 1-4 = 1 1. XZ = = (.4) = 64.. m Z = m = 4 5. m YXZ = m Z = 4 6. V = (-1, -1); W = (6, 1); slope of VW = _ 7 ; slope of GJ = 4_ 14 = _ ; sine the slopes are the 7 same, VW ǁ GJ. VW = ÇÇÇÇ + 7 = 5 ÇÇ ; GJ = ÇÇÇÇ = 5 ÇÇ, so VW GJ. 5-5 indiret Proof and Inequalities in One Triangle 7. is the smallest, so is the shortest side; is the largest, so is the longest side; From shortest to longest, the order is,,.. GH is the shortest side, so F is the smallest ; FH is the longest side, so G is the largest ; From smallest to largest, the order is F, H, G > > > 9 1 > Range of the values: > 9 m and < 1 m > 14. Yes; possible answer: the sum of eah pair of lengths is greater than the third length. 11 Holt Mougal Geometry

95 Holt McDougal Geometry

95 Holt McDougal Geometry 1. It is given that KN is the perpendicular bisector of J and N is the perpendicular bisector of K. B the Perpendicular Bisector Theorem, JK = K and K =. Thus JK = b the Trans. Prop. of =. B the definition