Review Packet #3 Notes

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1 SCIE 40, Spring 0 Miller Review Pket # Notes Mpping Nottion We use mpping nottion to note how oordinte hnges. For exmple, if the point ( ) trnsformed under mpping nottion of ( x, y) ( x, y), then it eomes ( 6,7). 6,7 is Trnsformtions A trnsformtion is hnge in the position, size, or shpe of figure. The originl shpe is lled the preimge, nd the resulting shpe is lled the imge. There re four types of trnsformtions: Trnsltion (slide) - ll of the oordintes move the sme distne in the sme diretion o The mpping nottion for trnsltion is often done using vetors nd ngle 9, is trnslted long the vetor rkets. For exmple, if the point ( ) x 5, y+ 7, then the point eomes ( 9 5, + 7) or ( ) 4, 5. Refletion (flip) - eh point nd its imge re the sme distne from the line of refletion o Aross the x-xis: the x-oordinte stys the sme, the y-oordinte hnges sign. x, y x, y. o ( ) ( ) Aross the y-xis: the y-oordinte stys the sme, the x-oordinte hnges sign. x, y x, y ( ) ( ) o Aross the line y x : the oordintes trde ples. ( x, y) ( y, x) Rottion (turn) - eh point nd its imge re the sme distne from the point of rottion x, y y, x o 90 out the origin: ( ) ( ) o 80 out the origin: ( x, y) ( x, y) Diltion (size) - if entered out the origin, multiply eh oordinte y the sle ftor Perimeter nd Are Sine perimeter is simply the distne round losed figure, you n lwys just dd up the sides. Beuse of their speil properties, retngles do hve speil formul for perimeter: P l w P l+ w, where l is the length nd w is the width. + or ( ) For irles, perimeter eomes the irumferene. Thousnds of yers go, mthemtiins nd sientists notied tht the irumferene of irle ws lwys out three times s long s its dimeter. This reltionship ws then formlly nmed π, nd the definition of π is tht it is C the rtio of the irumferene to the dimeter: π. This n e rewritten to give us the d

2 formul for the irumferene: Cπ d. Sine the dimeter is twie the rdius, n lternte formul is C π r. Are is the numer of squre units it tkes to over shpe. The most si formul is tht of retngle: A h, where is the se of the retngle nd h is the height. If you look t prllelogrm, you n ut off one orner nd rerrnge it to form retngle: Therefore, the formul for prllelogrm is the sme s retngle: A h, lthough the height is not the sme s one of the sides. For tringles, we n tke two tringles nd rrngle them to form prllelogrm, so the re h formul for one tringle is A h or A. If we tke two trpezoids, we n rrnge them to form prllelogrm: h + or So the re formul for one trpezoid is A h( ) A ( + ) h. For rhomi (or rhomuses, if you prefer) nd kites, we n split them in hlf lengthwise nd rerrnge the piees to form retngle: d d d d So the re formul for kite or rhomus is A dd. Sine squre is oth retngle nd rhomus, it n use either formul. Sine the sides re ongruent s well s the digonls eing ongruent, the formuls eome: A s or A d

3 For regulr polygons, we n ut them into tringulr wedges nd rrnge them into prllelogrm: pothem The segment from the enter of polygon to its side is lled the pothem. The se of the retngle is one-hlf the perimeter of the polygon. Therefore, the re is given y the formul: A P. Arhimedes used similr proedure to estimte the re of irle: P rdius The more wedges the irle is split into, the loser it gets to eoming true prllelogrm (in ft, n infinite numer of wedges would e true prllelogrm). The rdius tkes the ple of the pothem nd the irumferene reples the perimeter. This mkes the re formul for irle: A Cr ( π r) rπ r In summry, re formuls re s follows: irumferene Prllelogrm (inl. retngle) A h Squre (with side length s) A s Tringle h A h or A Trpezoid h + A h( + ) or A Rhomus or Kite d d A dd or A Regulr polygon (with pothem nd A P perimeter P) Cirle Aπ r ( ) Perimeter nd re prolems usully ome down to plugging the vlues you know into the formul nd solving for the unknown vrile.

4 Pythgoren Theorem: + Although Pythgors s nme is on it, the reltionship etween the sides of right tringles hs een known sine Bylonin times. There hve een mny proofs of this theorem (inluding one y Jmes Grfield efore he eme president of the U.S. in 88), ut one of my fvorites is grphil one tht I use with my students: A further pplition of the Pythgoren Theorem is the distne formul. The length of segment on oordinte plne n e determined y using the oordintes of its endpoints to form right tringle, the hypotenuse of whih is the segment: ( ) ( ) d x x + y y Speil Right Tringles If we split squre in hlf long the digonl, we end up with two right tringles whose legs re the sme length. Using the Pythgoren Theorem to find the length of the hypotenuse looks like this: + A + This reltionship holds for ny tringle the length of the hypotenuse is the length of the leg. times Likewise, if we split n equilterl tringle in hlf, we end up with two tringles. In this se, we know tht the hypotenuse is doule the length of the shorter leg, ut we don t know the other leg: 60 ( ) A Sine oth omposites ontin the sme shpes, the res re equl. Therefore, +.

5 Thus, for ny tringle, the long leg is doule the short leg. times the short leg, nd the hypotenuse is Polygons Polygons re nmed y the numer of their sides. In ddition, onve polygon is one in whih nononseutive verties n e onneted y segment tht is outside the polygon (it is ved in ). A onvex polygon is one tht is not onve. Exmples: onve degon onvex otgon One of the si theorems in Geometry is tht the ngles of tringle dd up to 80. We n use tht to investigte the ngles of polygons in generl: 4 Therefore, the sum of the interior ngles of onvex polygon is ( n ) 80, where n is the numer of sides. If the polygon is regulr (i.e. oth equilterl nd equingulr), then eh ngle mesures ( n ) 80. As the numer of sides gets lrger nd lrger, the mesure of n eh interior ngle gets loser nd loser to 80. Solid Geometry Solid geometry (to differentite it from plne geometry) dels with three-dimensionl ojets. One of the hllengs of working with three-dimensionl ojets is urtely representing them in two dimensions. A net is flt representtion of the surfes of the solid suh tht it ould e folded to form tht shpe. One of the most si nets is tht of ue:

6 A polyhedron (pl. polyhedr) is solid omposed of polygons. The Pltoni Solids re solids tht re omposed solely of regulr polygons. There re only five of these: Nme # of fes Polygon Piture Tetrhedron 4 Equilterl Tringles Othedron 8 Equilterl Tringles Ioshedron 0 Equilterl Tringles Hexhedron (ue) 6 Squres Dodehedron Pentgons If solid is omposed of polygons, nd two of the polygons re prllel nd ongruent, nd these polygons re onneted y prllelogrms, then the solid is lled prism. The ongruent polygons re lled ses, nd the prllelogrms re the fes. To lulte the surfe re of the prism, we lulte the re of the fes, lled the lterl re, nd dd the res of the ses. The formul for this is: L Ph where P is the perimeter of the se, nd h is the height of the prism (the distne etween the ses). Adding the ses mkes the formul for the totl surfe re: S L+ B where B is the re of the se. If solid onsists of two ongruent prllel irles onneted y urved surfe, this is lled ylinder. Although it is not polyhedron, it shres mny of the hrteristis of the prism. The surfe tht onnets the two irles is prllelogrm, so its re (lso lled the lterl re) is: L Ch where C is the irumferene of the irle (either πr or πd), nd h is the height of the ylinder. Adding the res of the irle mkes the formul for the totl re: S L+ π r

7 For oth prisms nd ylinders, the volume is the re of the se multiplied y the height of the prism: V Bh A solid omposed of polygonl se nd tringulr fes tht meet t ommon vertex is lled pyrmid. Like the prism, it hs lterl nd totl surfe re. If the se is regulr polygon, the lterl re is hlf the perimeter of the se multiplied y the slnt height (whih is the height of the tringulr fes): P L Pl or L l The totl surfe re dds the one se to the lterl re: S L+ B A solid omposed of irle onneted to urved surfe with vertex is lled one. Like the ylinder, it is not polyhedron, ut it n e treted in similr fshion. The formul for its lterl re is: L Cl ( π r) lπrl nd the totl surfe re is the sme s the pyrmid: S L+ B or Sπ rl +πr For oth pyrmids nd ones, the volume is one-third the re of the se multiplied y the height of the solid (the distne etween the se nd the vertex): V Bh A sphere is the set of ll points in spe tht re given distne from enter point. Like the ylinder nd the one, sphere is not polyhedron. Around 0 B.C., Arhimedes derived the formuls for the surfe re nd volume of sphere, whih he onsidered one of his gretest hievements. For the surfe re, Arhimedes showed tht the surfe re of the sphere ws the sme s the lterl re of ylinder whose dimeter nd height were the sme s the dimeter of the sphere. (He onsidered it one of his gretest hievements.) Therefore, the re formul eomes: A Ch π r r 4π r ( )( ) To find the volume formul, we first need to look t hemisphere, or hlf-sphere, nd ompre it to ylinder with one-shped hole. Cvlieri s Priniple sttes tht if two threedimensionl ojets hve the sme height nd the sme ross-setionl res t every level, then they hve the sme volumes. Let s tke look t the ross-setions of eh solid.

8 r h h For given height, h, the re of the rosssetion of the hemisphere n e found using the Pythgoren Theorem: hemi ( ) ( ) A π r h π r h For the ylinder, the rdius of the inner ring is the sme s the height of the ross-setion, so the re of the ross-setion would e yl ( ) A πr π h π r h Sine the ross-setions hve the sme re, the volume of the hemisphere is the sme s the volume of the ylinder. The formul for the volume of the ylinder would then e (rememer, the height is the sme s the rdius): Vhemi Vylinder Vone πr ( r) π r ( r) Vhemi πr π r π r For the sphere, we simply doule this formul: 4 V π r Prllel Lines If two prllel lines re onneted y third line (lled the trnsversl), the resulting ngles re either ongruent or supplementry. Angle pirs re nmed s follows: Angle Pir Angles Reltionship Corresponding & 5, Congruent & 6, & 7, 4& 8 Alternte & 6, Congruent Interior 4& 5 Alternte & 8, Congruent Exterior & 7 Sme-side & 5, Supplementry Interior 4& 6 Sme-side Exterior & 7, & 8 Supplementry The most importnt ngle pir to e le to reognize is orresponding ngle pirs; ll of the others n e derived from tht.

9 Proofs The proof in mthemtis is relly ll out reking prolem down into smller piees nd explining the resons ehind the proedures used to solve tht prolem. At its most forml level, proof should ontin every step used to prove the onlusion s well s the justifitions for eh step. A proof n lso onsist of more informl outline of the proess used to solve the prolem. The si two-olumn proof lwys strts with the informtion tht hs een given, nd the lst sttement should e whtever is eing proven. The resons used n e lgeri properties, geometri definitions, postultes, or theorems. Here is n exmple of n lgeri proof: Given: x x 8 Prove: x Sttements Resons. x x 8. Given. x 8. Sutrtion prop. of. x 6. Add. prop. of 4. x 4. Div. prop. of You should e fmilir with these properties s well s the multiplition, sustitution, trnsitive, distriutive, ssoitive, nd reflexive properties. Geometri proofs re more diffiult to set up, prtly euse they require more speifi geometry knowledge, ut for this ourse (nd the ompeteny test, s fr s I n tell), you relly only need to e fmilir with some of the generl theorems. These re: Definitions o Bionditionl: if nd only if mens tht the sttement nd onverse re true (it goes oth wys). Exmple: Tody is Mondy if nd only if yesterdy ws Sundy, reks down into If tody is Mondy, then yesterdy ws Sundy; nd if yesterdy ws Sundy, then tody is Mondy. o Congruene: Two figures re ongruent if nd only if their mesures re equl. (If two figures re ongruent then their mesures re equl, nd if their mesures re equl then the figures re ongruent.) Severl of the lgeri properties hve orresponding ongruene properties. These re the trnsitive, sustitution, nd reflexive properties of ongruene. o Supplementry: Two ngles re supplementry if nd only if the sum of their mesures is 80. o Complementry: Two ngles re omplementry if nd only if the sum of their mesures is 90. o Right ngle: An ngle is right ngle if nd only if its mesure is 90 Midpoint nd isetor theorems o A point is the midpoint of segment if nd only if it divides the segment into two segments of equl length.

10 o A segment, ry, or line intersets segment t its midpoint if nd only if it is segment isetor. o A segment, ry, or line divides n ngle into two ongruent ngles if nd only if it is n ngle isetor. Angle nd segment ddition postultes o If segment is divided into prts, then the mesure of the segment is equl to the sum of the mesures of its prts. o If n ngle is divided into two or more ngles, then the mesure of the ngle is equl to the sum of the mesures of the ngles. Angles o Two djent ngles form stright line if nd only if they re liner pir. o If two ngles form liner pir, then they re supplementry. o If two ngles re vertil ngles, then they re ongruent. o If two lines re perpendiulr, then they form right ngle. Angles formed y prllel lines o Two prllel lines re ut y trnsversl if nd only if orresponding ngles re ongruent, lternte interior/exterior ngles re ongruent, or sme-side interior/exterior ngles re supplementry Congruent tringles o Two tringles re ongruent if nd only if ll sides of one tringle re ongruent to ll sides of the seond tringle (SSS), two sides nd the inluded ngle of one tringle re ongruent to two sides nd the inluded ngle of the seond tringle (SAS), two ngles nd the inluded side of one tringle re ongruent to two ngles nd the inluded side of the seond tringle (ASA), two ngles nd the non-inluded side of one tringle re ongruent to two ngles nd the non-inluded side of the seond tringle (AAS), or the hypotenuse nd leg of one right tringle re ongruent to the hypotenuse nd leg of the seond right tringle (HL) o Two tringles re ongruent if nd only if their orresponding prts re ongruent (CPCTC Corresponding Prts of Congruent Tringles re Congruent) o Isoseles Tringle Theorem: Two ngles of tringle re ongruent if nd only if the sides opposite those ngles re ongruent. o Equilterl Tringle Theorem: A tringle is equilterl if nd only if it is equingulr. Similr Tringles o Two tringles re similr if nd only if the sides of one tringle re proportionl to the sides of the seond tringle (SSS~), two ngles of one tringle re ongruent to two ngles of the seond tringle (AA~), or

11 two sides of one tringle re proportionl to two sides of the seond tringle nd the inluded ngles re ongruent (SAS~) Here is n exmple of geometri proof: Given: MA TH ; AMT HTM Prove: A H M A Sttements Resons. MA TH ; AMT HTM. Given. MT TM. Reflexive prop. of ongruene. AMT HTM. SAS 4. A H 4. CPCTC H I ve tried to simplify ll of this s muh s I n, ut looking t oth the smple test nd the study mteril, it s hrd for me to e sure wht I n sfely leve out. From wht I n tell, however, they don t seem to require you to tully set up proof from srth. As you n see on the worksheet (nd s I will do on the quiz), I hve given you the si struture of the proof, nd you will need to fill in the missing piees. T Cirles: Ar Length nd Setor Are A entrl ngle is n ngle whose vertex is on the enter of irle, nd whose sides interset the irle. An r is portion of irle. A setor is region ounded y entrl ngle. The R r mesure is the mesure of the entrl ngle tht interepts the r; the r length is its proportion of the irumferene of the irle. The re of setor is its proportion of the re of the irle. A To find the r length or the setor re, we n set up proportion: G r length r mesure setor re r mesure irumferene 60 irle re 60 L m S m π r 60 π r 60 Solving for the length or the re gives us m L r m π Sπr You n work the prolems using either the proportion or the formul, whihever mkes more sense to you (I prefer the formul, ut one of my fellow tehers prefers proportions). Tringle trigonometry is now onsidered prt of Geometry, ut I m going to hold off on it until fter Alger II.