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Ò0 Œ„0) Ð Ž„0¥8ð­ÒÅ1ð �„0©ÿ…9h¨h@­Ò  ‘2e1i …1æ2Ðæ3¥2ÿà ð       Å4¥3å5�Þ¥<ð© …<ðÐÆ8ÐÐh©l…Ë©l…Ì  ¢ ½ìðeË…Ë�æÌè½ì‘ËèÐé`h©¨�Ò¢ ŽÒ©y� Ò�*©  \ä­*Ðû�Ò`h©�,¢  \ä`ü&àÀ%8%0ÿà ð       %À€#€À# #### #@#€€"€À!@!`! ! !0!!!!!!!!!!!!! ! !`!@ÿà ð       !€À"€€"@#`#0### #€À$@%0%8%àÀ&ü                                                                ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                      ÿà ð       ÿÿ ë ¹ ä™Àˆ÷©7�Ä©�ÅÈŒÑ !¹ÉEðˆˆˆô`ÈŒÐ©À™©È™`¬Ñ¹[ÈŒÑÀÐH¬Ð© ™©äÈ™h „L`LOAD "D:GEOM.BAS"›RUN›  h Œ@}ÿà ð       
ŒD©R�©�© ��©a�©�Ò Å© �©�©�Ò Åx ¹ ™
ˆ÷X©)�Ò ÅL ŒÓ SäÎÓö`î
­i€��îÎÒÐÚ`àá âA}ÿà ð       ã ë ¹ ä™Àˆ÷©7�Ä©�ÅÈŒÑ !¹ÉEðˆˆˆô`ÈŒÐ©À™©È™`¬Ñ¹[ÈŒÑÀÐH¬Ð© ™©äÈ™h „L`LOAD "D:GEOM.BAS"›RUN›  h Œ ÿà ð       ÿÿPg p©�ð©} ¤ö©@…Ð©œ…Ñ  ©xeÐ…Ð�æÑ©eÐ…Ð�æÑ¹¨kI€‘ÐÈÀÐô  ©(eÐ…Ð�æÑ¹®k‘ÐÈÀÐö  8¥Ðé…Ð°ÆÑ¹¶k‘ÐÈÀÐö  ©(eÐ…Ð�æÑ¹Åk‘ÐÈC}ÿà ð       ÀÐö  ©(eÐ…Ð�æÑ¹Ôk‘ÐÈÀÐö  ©PeÐ…Ð�æÑ¹ãk‘ÐÈÀÐö  ©(eÐ…Ð�æÑ¹òk‘ÐÈÀÐö ©(eÐ…Ð�æÑ¹l‘Ðˆø ©(eÐ…Ð�æÑ¹l‘Ðˆø ©(eÐ…Ð�D}ÿà ð       æÑ¹l‘Ðˆø ©(eÐ…Ð�æÑ¹.l‘Ðˆø©@…Ð©œ…Ñ ©(eÐ…Ð�æÑˆò©eÐ…Ð�æÑ ¹=l‘Ðˆø  ©eÐ…Ð�æÐ¹Il‘Ðˆø ©WeÐ…Ð�æÑ¹jl‘Ðˆø ©$eÐE}ÿà ð       …Ð�æÑ¹|l‘Ðˆø ©3…Ð©�…Ñ©›…Í©ž…Î©M‘Ð©N‘ÍˆõÆÐ¢   ©E…Í©�…Î©B‘Ð©V‘Í©(eÐ…Ð�æÑ©(eÍ…Í�æÎèà
ÐÝ ©Ø…Ð©�…Ñ8¹njé ‘Ðˆõ© �kj�lj�mjªF}ÿà ð       èÐýîkj­kjÐõîlj­ljÉÐë©} ¤ö©ë…Ð©œ…Ñ ¹ˆk8é I€‘Ðˆó æÑ©…Ð¹˜k8é I€‘Ðˆó  ©À…Ð©ž…Ñ¹vj8é ‘ÐÈÀðÐó©°…ÐæÑ !¹fk8é ‘Ðˆõ©¢U�Ž
©R�G}ÿà ð       ©
�©
…$©�©�©…%…& Sä0„ Sä0¥É	°Æ&Æ%Ðí¥&ðÆ$ÐÙ8�©ÿ�Dêêê© �kj�ljªèÐýîkjÐøîlj­ljÉÐî© �ð©} ¤ö©�b© �d©p�e©H}ÿà ð       �j© �k¢  Vä©�b©ú�d©�e© �h©�i¢  Vä0L^jL    GEOMETRY*   "SAT" AND "SCHOLASTIC APTITUDE TEST"ARE REGISTERED TRADEMARI}ÿà ð       KS OF THE COLLEGEENTRANCE    EXAMINATION   BOARD.   THESEMATERIALS  HAVE BEEN DEVELOPED BY  ARROWINSTRUCTIONAL  SYSTEMS, INC.J}ÿà ð         FOR HAYDENSOFTWARE COMPANY,  INC. WHICH BEARS SOLERESPONSIBILITY FOR THEIR CONTENTS.ATARI CONVERSIONBY ANDREW TAYLOR(!9$%.3K}ÿà ð       /&47!2%3#/2%NNNNNNNNNN)-02/6%-%.4NNNN3934%- &/2 4(%NH€€J  HJ  €€€€
€J € HÈÊJ  ÙY  Ê€J  €  €  ÙY   Ê€J €€€€  ÙY  € Ê€ €  €  ÙYL}ÿà ð         Ê€€È €  €  ÙY  $%6%,/0%$ "9!22/7 ).3425#4)/.!, 3934%-3 ).##/092)'(4 #	 "9 (!9$%. 3/&47!2% #/ ).##).#›3830 ;›384M}ÿà ð       0 CHK2 STY TIMER ;timer =1›3850     JSR DSKINV ;do a read...›3860     BMI ERROR  ;stop fo}r any err›3870     LDA TIMER  ;wiN}ÿà ð       thin time limit?›3880     CMP #8+1›3890     BCS CHK3  ; no!›3900     DEC TESTCNT ;yes, dec t}est cnt›3910 CHK3 DEC CURRCNT›O}ÿà ð       3920     BNE CHK2 ;  test more intervals›3930     LDA TESTCNT ;all intervals OK?›3940     CLC  };      assume all is OK›395P}ÿà ð       0     BEQ ACCEPT ;Yes, accept: PROTECTED›3960     DEC RTRYCNT ;else, re-try Op›3970     BNE CHK1›3}980 ERROR SEC  ;    GiveQ}ÿà ð        up!›3990 ACCEPT BCC WAIT›4000  LDA #$FF›4001  STA $0244›4010  JMP $E477›4020 WAIT LDA #0›4030  STA T}IM›4040  STA TIM+1›40R}ÿà ð       50  TAX›4060 WAIT1 INX›4070  BNE WAIT1›4080  INC TIM›4090  BNE WAIT1›4100  INC TIM+1›4110  LDA TIM+1›412}0  CMP #18›4130  BS}ÿà ð       NE WAIT1›4140  LDA #0›4150  STA $02F0›4160  LDA #$7D›4170  JSR $F6A4›4180  LDA #3›4190  STA IOCB+2›4200  D:AUTORUN.BAK210  STT}ÿà ð       A IOCB+4›4220àáPgÐ©œ…Ñ  ©xeÐ…Ð�æÑ©eÐ…Ð�æÑ¹¨kI€‘ÐÈÀÐô  ©(eÐ…Ð�æÑ¹®k‘ÐÈÀÐö  8¥Ðé…Ð°ÆÑ¹¶k‘ÐÈÀÐö  ©(eÐ…Ð�æÑ¹Åk‘ÐÈ ÿà ð          ¿ÀxY#k#MN¤DN¤FX¤A¤B¤BL¤BS¤C¤D¤E¤F¤M¤P¤R¤RE¤SC¤AN¤F1¤F2¤F3¤FL¤IN¤T¤TL¤TX¤DS¤SC¨SK¨TM¨×CP±PÂM±EÒR±WÎR²SËR³R°TØSÁWÇCP²XØYV}ÿà ð       ±Y²LN³X±X²O²O±R´R¸R·VÔRµHÔR¶ØIÖSÈPÓCÅAÑSÃQ¤ÑQÑFÁMÎMÁFØÂÅTÓMTØV°DÃMÔV²DÐV±R×CÌTÔÙ €       €      €      €      €      €  W}ÿà ð           €      €      €      €	      €
      €      €      €      €      €      €      €      €      €      €      €X}ÿà ð             €      €      €      €      @      @      @                                    !       "       #       $    Y}ÿà ð          %       &       '       (       )       *       +       ,       -       .       /       0       1       2       3       4 Z}ÿà ð             5       6       7       8       9       :       ;       <       =       >       ?       @       A      €B       C      [}ÿà ð        D       E       F       G       H       I       J       K       L       M       N       O       P       Q       R       S   \}ÿà ð           T       U       V        GEOMETRY› ,A0   "F:@    , A(   ,
@     A"@    +F:@    ,&A(   ,AB7]}ÿà ð       t  +F:B7t  ,&A(   ,
 ss€;@`    ,�;@    ,‚;@@    ,ƒ;@    ,„;@    ,…;@    ,†;@    ,‡;@    ,ˆ;@  ^}ÿà ð         ,‰;@    , ssŠ;@    ,‹;@    ,Œ;@    ,�;@    ,Ž;@'    ,�;@    ,�;@    ,‘;@    ,’;@    ,“;@ _}ÿà ð          , GG”;@    ,•;@    ,–;@`    ,—;@`    ,˜;@`    ,™;@    , 66š9@    <@    ,›9@E    <@    ,œ9@ `}ÿà ð          , 6�-?:AQ   , )+@    @    )AV   @      Q0@    @	    @    70@          @    Q0@ a}ÿà ð          @	    @     "F:@    , A(   "
@%     A"@    +F:@    ,&A(   ,AB7t  +F:B7t  ,&A(   , 
b}ÿà ð       A"P    A!0   ! T6�-?:ž,6Ÿ- !6�-?:A a   ,+A    J?:A!Q   <C:”,<B:”,,@    T
@0    ( 	6•.�A�   2 c}ÿà ð       '	¡"¢£@%    $6�-?:A d   ,'$< ˜7¢<¤,0~Q
@x    = $6¥-¥%¢6›8¥<§,-¨$6Á-A:˜7¦<¦,,> #›8¥<¦,"§6›8¥<¦,-Ÿd}ÿà ð       #
@u    @ '›8¥<¦," 6�-?:A d   ,'
@u    B "›8¥<¦,ŸAu   "
@u    D -6¨-¨%¢A	    #—7¢<¤,4~S-
@h  e}ÿà ð         G 6¨-¨&¢K 	©§
@€    N @€    
@P    P 6¨-¨%¢A	    R 	©"¤6©-¢$T ˜7¢<¢,4~
A    Z 	f}ÿà ð       ¡"¢$n ,	6–.˜A   "6�-?:A!E   ,,
@€    x —4~RA
AP   Œ —4~RC
A    – ˜7¢<¤,4~Q6�-?:ž,g}ÿà ð       $  —0~SD
Ap   ¢ $6�-?:A d   ,6Ÿ- $6½-      ¤ ¨"›8¥<§,%¢6›8¥<¦,- ¦ 
@€    ª —0~SB
A`   ¬h}ÿà ð        !	6Ÿ-¨¨"›8¥<§,%¢!6›8¥<¦,-¨­ 3¨›8¥<§,%¢6ª-¢6�-?:«,)Ay   3
@€    ® Au   
@€    ¯ /	6ª-§6�-?:«,i}ÿà ð       6Ÿ-›8¥<¦,)Ÿ"¨)Ÿ"¨%¢/6ª-¢± 		6¨-Ÿ³ 6¨-¨%¢A	    ´ "	—4�6�-?:A …   ,"6—.~SP¶ !—4~SPA    !
Ay   j}ÿà ð       ¹ —0~SG
A0   º A   ¼ (¬-­®@    $6�-?:¯<°<¬<±<¬,(	¬¾ (¬-°±@    $6�-?:¯<¬<­<¬<®,(	¬À /6�-?:k}ÿà ð       ¯<°<²<±<²,%6�-?:¯<³<­<³<®,/
Ay   Ò 56³-+A:˜7´<´%¤,,,$µ'¶56²-A:˜7@    <@    %¤,,Ó G%6°-+A:˜7@    <@    %¤,,,$µl}ÿà ð       '¶G6±-+A:˜7@    <@    %¤,,,$µ'¶Ô ;6­-A:˜7@    <@    %¤,,;6®-A:˜7@    <@    %¤,,Õ $Ü A   Þ 6±-±$¶m}ÿà ð       'µ6­-­$µ'¶à A%6¬-+A:˜7@"    <@"    %¤,,,$µ'¶A6�-A:˜7@%    <@%    %¤,,ä 6�-?:¯<³<²<°<±<­<®<¬<�,$æ —0~SH
n}ÿà ð       A@   è )6�-?:·<A:˜7´<¸,,,)6�-?:¹<A:˜7º<¶,,,ê .$6�-?:A!	   <C:˜7µ,,<¼<B:˜7µ,,,.
Ay   ð —0~SF
Ay   ô 26�-?o}ÿà ð       :ž,6�-?:A!9   ,&6�-?:¹<¢,26½-@    ö 	ª"§6¨-›8¥<§,ø $—4~ET6¡-¢
AP   @
A0   ^½      p}ÿà ð       6�-?:A!9   ,c.6�-?:·<@!    ,6–.Ž$A€   .A0   h	•4…
A    |•4†*+©§)¾"¤,
A    †!•4Q)•4�)•q}ÿà ð       4‹)•4†!
A   �A   
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A    '6�-?:A d   ,6Ÿ- ©"¢$6©-§'$A    56�-?:ž,¥-§@E    "»-§¦-6›8¥<»,-§1	»5	¥,Á-§~}ÿà ð       @    »-§¤$6š8Á<»,-§(	»,	Á0	6Á-§6¨-§6¡-§6¥-§!6©-§'6¾-§-6Ÿ- 0$ •4‹)—4~ET6¡-¢*	¡"¢6�-?:ž,}ÿà ð       $46�-?:ž,
@€    >Ad   AU   R"F:Ad   ,"AU   "
AP   \6•.>:?:A!   ,,$„;'?:A!   <¨<@$  €}ÿà ð         <C:˜,,@    -6¡-¢86˜.~   ;$†B:˜,"§6˜.      ‡B:˜,"¢6˜.      ˆ$B:˜,!¢˜7¢<¤,4~Q$6˜.~Q0‰�}ÿà ð       B:˜, @    6—.   $Ž6—.˜7¢<¦,˜—4~ET6¡-¢¢$¬16�-?:ž,6Ã-§Ä-¢¥%›8Ä<¤,"§+6Ã-¢16Ä-¥®	Ä‚}ÿà ð       Ã"§$°6�-?:A p   ,²-A0   •0U*•0…#A   -
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A@   ì	•4‹
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A@   �}ÿà ð       l6�-?:A y   ,$Š6�-?:A ‚   ,¼,6�-?:¹<¢,6�-?:¿,,6»-@!    &+B:–,'¤,½36�-?:¹<»,6¼-A(   $A   06¼-�}ÿà ð             3$ÆM6�-?:¹<¢,6–.] A   16�-?:¹<@9    ,;A   J6�-?:A!E   ,M$ÚV6�-?:¹<¢,6�-?:·<@#    ,.6�-?:Ì‘}ÿà ð       <¤<Ð<È,46–.‚>A   G6�-?:¿,V6�-?:A!E   ,ÛZ6�-?:·<@$    ,#6�-?:Ì<¤<Ð<Å,)6–.‚3A   <6�-?:¿,K6�-?:A!E   ,Z6’}ÿà ð       �-?:A!E   ,ä$îG6�-?:·<º,)6–.NUMBER CORRECT    = =6–7B:–,,.=:š8§<¢,,GA€   ïM6�-?:·<@    ,/6–.NUMBER IN“}ÿà ð       CORRECT  = C6–7B:–,,.=:š8§<¤,,MA€   ðM6�-?:·<@    ,/6–.NUMBER UNANSWERED = C6–7B:–,,.=:š8§<§,,MA€   ñ6�”}ÿà ð       -?:A R   ,$R"6™.G"6™7B:™,%@    ,.=:Î,\2	Ñ"Î(?:A!Q   <C:™,<B:™,,"@    2
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D:GEOM.BAS¤TX¤DS¤SC¨SK¨TM¨×CP±PÂM±EÒR±WÎR²SËR³R°TØSÁWÇCP²XØY Kÿà ð       ›Welcome to the MATH MODULE of the›HAYDEN SCORE IMPROVEMENT SYSTEM FOR›THE SAT& one of three modules designed›to help you rai�}ÿà ð       se your SAT scores.›This Geometry Section is an effective›tool to begin your preparation for the›Mathematical section of the ž}ÿà ð       Scholastic›Aptitude Tests.The system is easy to›operate so that you can concentrate on›its content. All of the information›yoŸ}ÿà ð       u need to answer questions appears›on the screen& as do instructions for›moving from one part of the program to›another. More }ÿà ð        detailed information›follows in the User's Guide.›~RA›This GEOMETRY SECTION provides›instruction and practice in solving›the¡}ÿà ð        entire range of geometry problems›of the types found on the SAT. All›figures needed to solve the problems›are illustrated on¢}ÿà ð        the screen.››~RA›This section includes eleven parts in›addition to the User's Guide~›`  Complementary and Supplementary›`  A£}ÿà ð       ngles  and Angle Sums›`  Parallel Lines›`  Angles of a Circle›`  Equalities of Angles and Sides›`  Pythagorean Theorem›`  Spe¤}ÿà ð       cial Triangles›`  Perimeters›`  Areas›`  Similar Figures›`  Volumes›`  All Examples Without Tutormode››~RA›Other areas in the¥}ÿà ð        Mathematical›section of the SAT are Algebra& and›Quantitative Comparisons and Word›Problems; two separate sections within›th¦}ÿà ð       is Math Module provide review›material in these areas.›This section contains a two-sided disk›which has a Program Disk side a§}ÿà ð       nd a›Review Disk side.  Always load the›Program Disk side first.  Instructions›on the screen will prompt you to›insert the ot¨}ÿà ð       her side at the›appropriate time.›~RA›MENUS››The MAIN MENU lets you move easily›from one section of the program to›another. S©}ÿà ð       imply press the key›corresponding to the letter next to›the section you wish to see.›~RA›The first ten topics on the MAIN MENª}ÿà ð       U›combine two key aspects of the Hayden›System~ DEFINITIONS& ANALYSIS AND›STRATEGIES and EXAMPLES WITH›TUTORMODE. First& DEFI«}ÿà ð       NITIONS provides›the background information you will›need to tackle the problem type on the›SAT and demonstrates problem-solv¬}ÿà ð       ing›methods& including valuable tricks and›shortcuts.›~RA›Second& }TUTORMODE} gives you a›detailed& step-by-step explanation ­}ÿà ð       of›how to arrive at the correct answer.›By reviewing and practicing& you›develop more efficient problem-solving›techniques.›~®}ÿà ð       RA›The ALL EXAMPLES WITHOUT TUTORMODE›option provides quick drill and›practice in all the problem types›listed on the MAIN ME¯}ÿà ð       NU so that you›can improve speed and accuracy. If you›answer incorrectly& you are shown the›correct answer& but no detailed›e°}ÿà ð       xplanation is provided. At the end of›the section& the computer tallies the›number of questions answered correctly›and incorr±}ÿà ð       ectly& providing an›indication of how well you have›mastered the material.›~RA›When you make a selection from the›MAIN MENU& ²}ÿà ð       the System starts›presenting the material or asks for›the other side of the disk to be›loaded into the drive.›~RA›FUNCTION KE³}ÿà ð       YS››A function key is a key which has a›specific effect on the program's›operation each time it is pressed.››Whenever a menu ´}ÿà ð       is on your screen the›following function keys are›operational~››`    M (Main Menu)›`    Q (Quit)››Pressing }M} always brings µ}ÿà ð       you back to›the MAIN MENU. Pressing }Q} causes the›computer to ask if you really want to›quit. If you answer }Y}& you end the¶}ÿà ð       ›program. If you answer }N}& you›continue where you left off.›~RA›While text is on the screen pressing›}R} restarts the secti·}ÿà ð       on (erasing any›previous answers that you may have›entered)& pressing }M} takes you to›the main menu and pressing }Q} enables¸}ÿà ð       ›you to quit.›~RA›The left-arrow key lets you page›backwards through the text one screen›at a time until the first screen of›¹}ÿà ð       the section is reached. When a›question appears on the screen& your›previous answer& if any& is shown. You›can replace that aº}ÿà ð       nswer by entering›another one& or you can leave your›answer undisturbed by pressing the›left-arrow again.›~RA›Pressing  the l»}ÿà ð       etter }O} leaves the›current question temporarily›unanswered and displays the next›question. At the end of the section›you ha¼}ÿà ð       ve a chance to review all the›unanswered questions.›~ET›~ET››YSTEM FOR›THE SAT& one of three modules designed›to help you rai =ÿà ð       The HAYDEN SCORE IMPROVEMENT SYSTEM›FOR THE SAT is organized into three›modules. It includes both simulated›SAT exams and com¾}ÿà ð       plete reviews of the›areas typically covered by the Verbal›and Mathematical sections of the SAT.›In addition to this Math Mod¿}ÿà ð       ule, the›following modules are available:›~RA›The PRACTICE TESTS MODULE contains an›Analysis of the SAT, a Pre-Test, and›two À}ÿà ð       Practice Tests.››The ANALYSIS OF THE SAT gives you›insight into the workings of the›actual exam -- its organization and›scoriÁ}ÿà ð       ng& plus test-taking strategies›and tips for raising your scores.›~RA›The PRE-TEST is a›diagnostic/prescriptive tool for›deteÂ}ÿà ð       rmining your strengths and›weaknesses in the areas typically›covered by the SAT. It is a two-hour›test consisting of a mix ofÃ}ÿà ð        Math and›Verbal questions similar to that on an›actual SAT. After you complete the›test your computer will provide scores›inÄ}ÿà ð        each of sixteen subjects which›contribute to your Math and Verbal›scores. This profile of your›performance indicates which aÅ}ÿà ð       dditional›modules in the Hayden System will be›useful in your preparation.›~RA›The PRACTICE TESTS are two-hour›simulated examÆ}ÿà ð       s with complete›Mathematical and Verbal sections timed›and formatted to be representative of›the latest SATs and scored on thÇ}ÿà ð       e SAT›scale. After reviewing your weak›areas& take these Practice Tests and›see how your performance would measure›up on the È}ÿà ð       actual exam.›~RA›The VERBAL MODULE provides tutorials,›drill and analysis in the Verbal areas›normally covered on the SAT.››TÉ}ÿà ð       he VOCABULARY SECTION provides a›thorough review of antonyms& analogies›and sentence completions& as well as›an on-screen dicÊ}ÿà ð       tionary with 1000›words.›~RA›The READING COMPREHENSION SECTION›offers strategies and practice in›responding to questions abouË}ÿà ð       t the›material just read. Working with›passages drawn from the most›up-to-date sources in a variety of›fields will help you iÌ}ÿà ð       mprove your›ability to determine main ideas& to›recognize logical implications and to›extract factual information from what›yÍ}ÿà ð       ou read.›~RA›Each topic in a given section can be›approached in three ways~››`acquiring background with›`DEFINITIONS& ANALYSIÎ}ÿà ð       S AND STRATEGIES››`gaining practice and instruction with›`EXAMPLES WITH TUTORMODE››`drilling with›`EXAMPLES WITHOUT TUTORMODEÏ}ÿà ð       ››*NOTE~  This Geometry Section combines›Definitions& Analysis and Strategies›and Examples With Tutormode.›~ET›~ET››s and com tÿà ð       ›~SB›~SH1127A  D  y    B›~SH0431C›~SH0537E›~SH0932k›~SP189089210089210032210032›~SP258040210089254089254089›~SF›B. ComplementÑ}ÿà ð       ary and Supplementary›`      Angles and Angle Sums.››Make sure you know these›theorems& definitions and›assumptions.›››1. TwoÒ}ÿà ð        angles whose sum›is a right angle are›complementary.››`    <k + <y = 90*››2. Two angles whose sum is›180* are supplementary.Ó}ÿà ð       ››`  <ADE + <y = 180*›~RA›~SB›~SH0828A   a  b B›~SH0536C›~SH1230D›~SH1030d  c›~SP196064252064252064252064›~SP2470402240642100Ô}ÿà ð       88210088›~SF›3. The sum of the angles›about a point is 360*.››<a + <b + <c + <d = 360*››4. The sum of the angles›about a poinÕ}ÿà ð       t on one side›of a straight line is›180 degrees.››`   <a + <b = 180*&››`   <c + <d = 180*›~RA›~SB›~SH0927A           C›~SH033Ö}ÿà ð       3B›~SP189064224028259064189064›~SF›5. The sum of the angles›of a triangle is 180*.››<A + <B + <C = 180*›~RA›~SB›~SH0827A›~SH0×}ÿà ð       735B›~SH0939C›~SH1432D›~SP189064244056266072217104›~SP217104189064189064189064›~SF›6. The sum of the angles of a›quadrilateraØ}ÿà ð       l is 360*.››<A + <B + <C + <D = 360*›~RA›~SB›~SP196024210016245032252056›~SP252056217064196048196024›~SH0328A›~SH0231B›~SH043Ù}ÿà ð       6C›~SH0737D›~SH0932E›~SH0628F›~SF›7. The sum of the angles›of a polygon of n sides›is 180 (n - 2) degrees.››Note~ In the diagÚ}ÿà ð       ram›(hexagon ABCDEF)& the›sum of the angles is~››180(6 - 2) degrees = 720*›~RA›~SB›~SH0829A   E   B›~SH0431C›~SH0636D›~SP2030Û}ÿà ð       55252055252055252055›~SP217032224055245048245048›~SF›~Q›1. Given~ CE[ED;›<AEC = 70*. Find~ <BED››(a) 20*  (b) 70*  (c) 90*››(Ü}ÿà ð       d) 110*  (e) 180*›~RCA›1. (a) 20* Ans.››The sum of the angles›about a point on one side›of a straight line is 180*›therefore~Ý}ÿà ð       ››`<BED + <DEC + <AEC = 180*››Since   CE[ED& <DEC =  90*››`<BED +  90* +  70* = 180*››`              <BED =  20* Ans.›~RA›~Q›Þ}ÿà ð       ~SB›~SH0828A         B›~SH0338C›~SH1336D›~SH0930E›~SP198060257060257060257060›~SP254016210060245096245096›~SF›2. Given~ <CED ß}ÿà ð       is a right angle. <CEB›is 50* Find~ <AED››(a) 40*  (b) 90*  (c) 130*››(d) 140*  (e) It cannot›be determined from the›informatà}ÿà ð       ion given.›~RCD›2. (d) 140* Ans.››` <CEB + <BED =  90*››`  50* + <BED =  90*››`        <BED =  40*››Since the sum of the›anglá}ÿà ð       es about a point›on one side of a›straight line is 180*&››` <BED + <AED = 180*››`  40* + <AED = 180*››`        <AED = 140* Anâ}ÿà ð       s.›~RA›~Q›3. <CED is a right angle& and <AEC =›<AED. Find the number›of degrees in <CEB.››(a) 45  (b) 90  (c) 135››(d) 180  (ã}ÿà ð       e) 20›~RCA›3. (a) 45 Ans.››The sum of the angles›around a point is 360*.›Therefore~››<CED + <AEC + <AED = 360*››`90* + <AEC +ä}ÿà ð        <AED = 360*››`      <AEC + <AED = 270*››`                    270*›`      <AEC = <AED = ----›`                     2››`      å}ÿà ð                   = 135*›~RA›`       <AEC + <CEB = 180*››`       135* + <CEB = 180*››`              <CEB =  45* Ans.›~RA›~SD›~Q›4. æ}ÿà ð       In {ABC& <A~<B~<C = 1~2~3. Find the›number of degrees in <B.››(a) 15  (b) 30  (c) 45  (d) 60  (e) 90›~RCD›4. (d) 60 Ans.››`  ç}ÿà ð       <A + <B + <C = 180*››`   y + 2y + 3y = 180››`            6y = 180››`             y = 30››`       <B = 2y = 60*  Ans.›~RA›~Q›5è}ÿà ð       . In {ABC& <A = 3<B& and <B = 2<C.›Find the number of degrees in <C.››(a) 10  (b) 20  (c) 40  (d) 60››(e) 120›~RCB›5. (b) 20 é}ÿà ð       Ans.››`      <A + <B + <C = 180*››`       6y + 2y + y = 180››`                9y = 180››`                 y = 20››`          ê}ÿà ð         <C = y = 20*  Ans.›~RA›~Q›~SB›~SP196055266055217032210055›~SP266055279055279055279055›~SH0828D  A      B E›~SH0432C›~SF›6. ë}ÿà ð       If <DAC = 120* and <EBC = 150*& then›{ABC is~››(a) equiangular›(b) equilateral›(c) isosceles›(d) right›(e) It cannot be›`   dì}ÿà ð       etermined from the›`   information given›~RCD›6. (d) right Ans.››Since the sum of angles›about a point on one side›of a straií}ÿà ð       ght line is 180*››`   <DAC + <CAB = 180*››`   120* + <CAB = 180*››`          <CAB =  60*›and››`   <EBC + <ABC = 180*››`   150î}ÿà ð       * + <ABC = 180*››`          <ABC =  30*›~RA›The sum of the angles of›a triangle is 180*.›Therefore~››<CAB + <ABC + <ACB = 180ï}ÿà ð       *››`60* +  30* + <ACB = 180*››`             <ACB =  90*&››making {ABC a right›triangle. Ans.›~RA›~Q›~SB›~SH0330A›~SH1127D  C ð}ÿà ð              B›~SH1238E›~SP203079203024273096273096›~SP182079256079256079256079›~SF›7. If <DCA = 2<A& and›<CBE = 3<A& then the›numbñ}ÿà ð       er of degrees in›<A is~››(a) 15  (b) 30  (c) 45››(d) 74  (e) 75›~RCC›7. (c) 45* Ans.››The sum of the angles›of a triangle is ò}ÿà ð       180*››`<A + <B + <C = 180*››Since the sum of the›angles about a point›on one side of a›straight line is 180*~››<C = (180* - 2ó}ÿà ð       A)››<B = (180* - 3A)››<A + (180* - 3A) + (180* - 2A) = 180*››<A = 45* Ans.›~RA›~Q›~SB›~SP189064244056266072217104›~SP21710418ô}ÿà ð       9064189064189064›~SH0827A›~SH0735B›~SH0939C›~SH1432D›~SF›8. In quadrilateral ABCD& <A = y*& <B =›2y*& <C = y - 30*& and <D = õ}ÿà ð       y + 30*›Find the number of degrees in <A.››(a) 36  (b) 43  (c) 72››(d) 102  (e) 144›~RCC›8. (c) 72 Ans.››<A + <B + <C + <D = ö}ÿà ð       360*››y + 2y + y - 30 + y + 30››`                 = 360››`              5y = 360››`               y = 72*››`     <A = y = 72*÷}ÿà ð         Ans.›~RA›~SD›~Q›9. Each angle of an equiangular›pentagon (a polygon of 5 sides) is~››(a) 72*  (b) 90*  (c) 108*  (d) 180*››ø}ÿà ð       (e) 540*›~RCC›9. (c) 108* Ans.››Note~ The sum of the angles of a›polygon = (N - 2)180››`         (5 - 2)180››`           = (3ù}ÿà ð       )180››`           = 540* in all››`     540/5 = 108* each  Ans.›~RA›~Q›~SB›~SH0528A        B›~SH0928C        D›~SH0739E›~SP196ú}ÿà ð       036245036266052245068›~SP245068196068196068196068›~SF›10. Given~ AB ^ CD; <B = 140*;›<D = 150*. Find~ <E.››(a) 10*  (b) 30*  û}ÿà ð       (c) 40*››(d) 50*  (e) 70*›~RCE›10. (e) 70* Ans.››NOTE~ Draw a line›perpindicular to lines AB›and CD.››The sum of the angles iü}ÿà ð       n›polygon ABEDC =››(5 - 2) x 180* = 540*››Both <A and <C = 90*››90* + 140* + <E + 150* + 90* = 540*››`                       ý}ÿà ð        <E = 70* Ans.›~ET›~ET›B›~SH0431C›~SH0537E›~SH0932k›~SP189089210089210032210032›~SP258040210089254089254089›~SF›B. Complement ÿà ð       ›~SB›~SC226039210056233056233056›~SH0534C›~SH0735B›~SH0830O›~SF›D. Angles of a Circle.››Make sure you know these theorems&›de ÿ}ÿà ð       finitions and assumptions.››1. A central angle is›measured by its›intercepted arc. Stated›otherwise& a central›angle is equal! }ÿà ð        in degrees›to its intercepted arc.››<COB = degrees in BC›~RA›~SB›~SC210032186056226038226038›~SH0726A›~SH0534B›~SH0430C›~SF›!}ÿà ð       2. An inscribed angle is equal in›degrees to one-half its intercepted›arc.››<CAB = 1/2  degrees in BC››3. In a circle& a›diam!}ÿà ð       eter is twice the›radius and divides the›circle into two›semicircles.›~RA›~SB›~SC186056235056225038186056›~SH0726A›~SH0735B›~!}ÿà ð       SH0534C›~SF›4. An angle inscribed in a semicircle›is a right angle.››<ACB = 90*›~RA›~SB›~SC210056192071226071226071›~SH0731O›!}ÿà ð       ~SH1027A›~SH1034B›~SF›Example~››Given~ Circle 0; AB = 60*&›OA = 6} Find~ Length of AB.››1. Draw radius OB.›2. Since AB = 60*&!}ÿà ð       ›`  <AOB = 60*.›3. Since OA = OB&›`  <A = <B.›4. But <A + <B›`  = 180* - <O›`  = 180* -60* = 120*›5. Then& <A = <B = 60* = <O!}ÿà ð       ›6. Since { AOB is equiangular& it is›equilateral; OA = OB = AB = 6} Ans.›~RA›~Q›~SB›~SC186055236055210055214033›~SH0726A›~SH!}ÿà ð       0735B›~SH0431C›~SH0831O›~SH0730y›~SF›1. Given~ Circle O&›diameter AB& radius OC&›BC = 80*.››Find~  Number of›degrees in <y.››!}ÿà ð       (a) 40  (b) 80  (c) 100››(d) 140  (e) 0›~RCC›1. (c) 100 Ans.››The central angle of a›circle is equal to its›intercepted arc.›!	}ÿà ð       Therefore~››`  <COB = 80*››Since y + <COB are on›the same side of a›straight line about a›central point& they›total 180*››`y !
}ÿà ð       + 80* = 180*››`      y = 100* Ans.›~RA›~Q›~SB›~SC193038226038214078214033›~SP214033192038193038193038›~SH0527A›~SH1131B›~SH04!}ÿà ð       31C›~SH0534D›~SF›2. Given~ Circle O& <A = 30*. Find~›Number of degrees in <B.››(a) 15  (b) 30  (c) 60››(d) 90  (e) 10›~RCB›2.!}ÿà ð        (b) 30 Ans.››Angle A and Angle B›intercept the same arc›CD and are therefore›equal.››`   <B = <A = 30* Ans.›~RA›~Q›~SB›~SC18!}ÿà ð       6056236056210056206033›~SP186056206033206033206033›~SH0726A›~SH0735B›~SH0430C›~SH0831O›~SF›3. Given~ Circle O& diameter AB&›r!}ÿà ð       adius OC& <COB = 100*. Find~ Number›of degrees in <A.››(a) 50  (b) 90  (c) 100››(d) 200  (e) 150›~RCA›3. (a) 50 Ans.››An insc!}ÿà ð       ribed angle is›equal in degrees to one›half its intercepted›arc. Since <A intercepts›the same arc as <COB&››`<A = 1/2 <COB››`!}ÿà ð          = 1/2 (100*)››`   = 50* Ans.›~ET›~ET›››~SH0735B›~SH0830O›~SF›D. Angles of a Circle.››Make sure you know these theorems&›de  )ÿà ð       ›F. The Pythagorean Theorem››Make sure you know these theorems&›definitions and assumptions.›~RA›~SB›~SP205024251070205070205%}ÿà ð       024›~SH1029B   b  A›~SH0329C›~SH0729a›~SH0534c›~SF›The Pythagorean Theorem›states~›››In any right triangle&›the square of the%}ÿà ð       ›hypotenuse equals the›sum of the squares of›the other two sides.›››In symbols this theory is expressed~››`    2    2    2›` %}ÿà ð         c  = a  + b››where c is always the hypotenuse.›~RA›Example~››The perimeter of right›triangle ABC is 3 ft. The›sides forming%}ÿà ð        the right›angle are in a 3~4 ratio.›Find the length of each of›the three sides.››Let p = number of feet in›perimeter of tria%}ÿà ð       ngle ABC.››Let c = number of feet in›hypotenuse AB.››`    Let 3y = side a.›`    Let 4y = side b.›~RA›In right triangle ABC~››%}ÿà ð       `2    2    2›c  = a  + b››`2       2      2›c  = (3y) + (4y)››`2     2      2›c  = 9y  + 16y››`2      2›c  = 25y››c  = 5y››p %}ÿà ð        = a + b + c››3  = 3y + 4y + 5y›~RA›~SB›~SP205024251070205070205024›~SH1029B   b  A›~SH0329C›~SH0729a›~SH0534c›~SF›`   3  = 1%}ÿà ð       2y››`   y  = 1/4››`  Therefore~››side a = 3y = 3/4ft.››side b = 4y››`      = 4/4››`      = 1ft.››side c = 5y››`      = 5/4››`%}ÿà ð             = 1 1/4ft.›~RA›~SD›Note the following multiple›combinations of the sides of a few›typical right triangles(integers›only%}ÿà ð       ).››-----------------------------------›`          LEG   LEG   HYPOTENUSE›-----------------------------------›`           3  %}ÿà ð          4        5›`   A       6     8       10›`           9    12       15›-----------------------------------›`           5    %}ÿà ð       12       13›`   B      10    24       26›-----------------------------------›`           8    15       17›`   C      16    30%}ÿà ð              34›-----------------------------------›~RA›~Q›1. A ladder 20 ft. long is placed›against a building so that it just›rea%}ÿà ð       ches a window 16 ft. high. How far›from the wall is the foot of the›ladder?››(a) 4ft. (b) 12 ft. (c) 18 ft.›`      ___›(d) 2\% }ÿà ð       /139 ft.››(e) It cannot be determined from the›information given.›~RCB›1. (b) 12 ft. Ans.››`     2    2    2›`    a  + b  = c%!}ÿà ð       ››`     2    2     2›`   16  + b  = 20››`          2›`         b  = 400 - 256››`          2›`         b  = 144››`          b %"}ÿà ð       = 12 ft. Ans.›~RA›~Q›2. How much distance is saved in›walking diagonally across a›rectangular vacant lot 30 ft. by 40›ft& ins%#}ÿà ð       tead of walking around its›length and width?››(a) 10 ft. (b) 20 ft. (c) 40 ft.››(d) 70 ft. (e) 0 ft.›~RCB›2. (b) 20 ft. Ans.›%$}ÿà ð       ›`       40 + 30 = 70  walking around››`        2    2    2›`       a  + b  = c››`       2     2    2›`     30  + 40  = c››` %%}ÿà ð                        2›`    900 + 1600 = c››`                  2›`          2500 = c››`            50 = c   walking across››70 ft%&}ÿà ð       . - 50 ft. = 20 ft. saved Ans.›~RA›~Q›3. Find the altitude of an isosceles›triangle if its base is 10 inches long›and each of%'}ÿà ð        its equal sides is 13›inches long.››(a) 8 in. (b) 12 in. (c) 24 in.›`     ___›(d) \/269 in. (e) 0 in.›~RCB›3. (b) 12 in. Ans%(}ÿà ð       .››`      2    2    2›`     a  + b  = c››`      2    2     2›`     5  + h  = 13››`           2›`          h  = 169 -25››`    %)}ÿà ð              2›`          h  = 144››`           h = 12 in.  Ans.›~RA›~Q›4. The hypotenuse of a right triangle›is 25 in. The other t%*}ÿà ð       wo sides are in›the ratio of 3~4. Find the length of›each side in inches.››(a) 3&4  (b) 6&8  (c) 9&12  (d) 12&16››(e) 15&20›~%+}ÿà ð       RCE›4. (e) 15&20 Ans.››`        2    2›`       a  + b  = c››`     2       2       2›` (3y)  + (4y)  = (25)››`      2      2›`%,}ÿà ð           9y  + 16y  = 625››`             2›`          25y  = 625››`             2›`            y  = 25››`             y = 5››There%-}ÿà ð       fore~ 3y = 15 in.   4y = 20 in.›~RA›~Q›5. A certain right triangle has a›perimeter of 4 ft. The sides forming›the right angle%.}ÿà ð        are in a 3~4 ratio.›Find the length of each of the three›sides in inches.››(a) 3&4&5  (b) 1&4/3&5/3››(c) 36&48&60  (d) 12&16%/}ÿà ð       &20››(e) 15&20&13›~RCD›5. (d) 12&16&20 Ans.››`    3y + 4y + c = 4 ft.››`      2       2    2›`  (3y)  + (4y)  = c››`       2 %0}ÿà ð            2    2›`     9y  + 16y  = c››`              2    2›`           25y  = c››`             5y = c›~RA›`          4 ft. = 48 %1}ÿà ð       in.››`   3y + 4y + 5y = 48››`            12y = 48››`              y = 4››Therefore~    3y = 12 in.››`             4y = 16 in.%2}ÿà ð       ››`             5y = 20 in. Ans.›~ET›~ET›› know these theorems&›definitions and assumptions.›~RA›~SB›~SP205024251070205070205$ *ÿà ð       ›C. Parallel Lines›`    (AB^CD)››Make sure you know these theorems&›definitions and assumptions.›~RA›~SB›~SH11311   3›~SH0933)4}ÿà ð       2›~SH13334›~SP196052259116259116259116›~SP259052196116196116196116›~SF›1. If two straight lines intersect&›the pairs of oppos)5}ÿà ð       ite angles››Vertical Angles~››`    angle <1 = <3›`          <2 = <4››Supplementary Angles~››`     <1 + <2 = 180*›`     <2 + <)6}ÿà ð       3 = 180*›`     <3 + <4 = 180*›`     <4 + <1 = 180*›~RA›~SB›~SH0927l›~SH1327m›~SH08361  2›~SH10334   3›~SH12315  6›~SH14298  7)7}ÿà ð       ›~SP193068266068266068266068›~SP193100266100266100266100›~SP265050196120196120196120›~SF›2. If two straight lines are paralle)8}ÿà ð       l›(never meeting) and are cut by a›transversal (a line that touches each›of the parallel lines)~››a. the pairs of alternate›i)9}ÿà ð       nterior angles are equal;››b. the pairs of›corresponding angles›are equal;››c. the pairs of interior›angles on the same side›):}ÿà ð       of the transversal are›supplementary.›~RA›If line l is parallel to line m~››a. Alternate interior angles››`    <3 = <5›`    <);}ÿà ð       4 = <6››b. Corresponding angles››`    <1 = <5›`    <4 = <8›`    <2 = <6›`    <3 = <7››c. Interior angles on›`  the same side )<}ÿà ð       of›`  the transversal››`  <4 + <5 = 180*›`  <3 + <6 = 180*›~RA›~SB›~SH0927l›~SH1327m›~SH08361  2›~SH10334   3›~SH12315  6›~SH)=}ÿà ð       14298  8›~SP193068266068266068266068›~SP193100266100266100266100›~SP265050196120196120196120›~SF›Notice that there are many p)>}ÿà ð       airs of›vertical angles and supplementary›angles.››Vertical Angles››`  <1 = <3›`  <2 = <4›`  <5 = <7›`  <6 = <8››Supplementar)?}ÿà ð       y Angles››<1+<2 = 180*  <5+<6 = 180*›<2+<3 = 180*  <6+<7 = 180*›<3+<4 = 180*  <7+<8 = 180*›<4+<1 = 180*  <8+<5 = 180*›~RA›~Q›)@}ÿà ð       ~SB›~SP198028264028264028264028›~SP198070264070264070264070›~SP244012217082217082217082›~SH0428A›~SH0439B›~SH0928C›~SH0939D›~)A}ÿà ð       SH0236E›~SH1133F›~SH0533a›~SH0834b›~SH0336c›~SH0535d›~SH1031e›~SF›1. If <d = 150* and›AB^CD& find the number›of degrees in <e)B}ÿà ð       .››(a) 30  (b) 60  (c) 150››(d) 180 (e) It cannot be›determined from the›information given.›~RCA›1. (a) 30 Ans.››Because <a a)C}ÿà ð       nd <d›are supplementary››<a = 180* - <d››<a =  30*››<a and <e are›corresponding angles›and therefore equal››<e = <a = 30* Ans)D}ÿà ð       .›~RA›~Q›2. Given~ <a = 3y + 5*&›<b = 2y + 11*. Find the›number of degrees in <a.››(a) 6  (b) 16  (c) 23››(d) 43  (e) 0›~RCC›)E}ÿà ð       2. (c) 23 Ans.››Since <a and <b are›alternate interior›angles&››`    <a = <b››`3y + 5 = 2y + 11››`     y = 6*››`     a = 3 (6)F}ÿà ð       *) + 5*››`       = 23* Ans.›~RA›~Q›~SB›~SP198028264028264028264028›~SP198070264070264070264070›~SP244012217082217082217082›~S)G}ÿà ð       H0428A›~SH0439B›~SH0928C›~SH0939D›~SH0236E›~SH1133F›~SH0533a›~SH0834b›~SH0336c›~SH0535d›~SH1031e›~SF›3. Angle b = 4m* and›ang)H}ÿà ð       le d = 5m*. Find the›value of m.››(a) 10  (b) 20  (c) 40››(d) 50  (e) 80›~RCB›3. (b) 20 Ans.››Angle b and <d are›interior ang)I}ÿà ð       les on›the same side of the›transversal and›therefore total 180*››` 4m* + 5m* = 180*››`        m  =  20* Ans.›~RA›~Q›~SB›~SH0)J}ÿà ð       528A           B›~SH0828C           D›~SH0434y›~SH0932140*›~SP198036275036275036275036›~SP198060275060275060275060›~SP2520242)K}ÿà ð       03072203072203072›~SF›4. AB ^ CD. Find the number of degrees›in angle y.››(a) 40  (b) 50  (c) 90››(d) 140  (e) 180›~RCD›4. (d)L}ÿà ð       ) 140 Ans.››y is a corresponding›angle to the vertical›angle equal to 140*›and is therefore also›equal to 140* Ans.›~RA›~Q›~S)M}ÿà ð       B›~SH0427A      x    B›~SH0627C           D›~SH0827E    G›~SH0234F    H›~SH0734y›~SP189032266032266032266032›~SP1890472660472)N}ÿà ð       66047266047›~SP231016189056189056189056›~SP266016224056224056224056›~SF›5. AB ^ CD& EF ^ GH& and <x = 30*.›Find the number of)O}ÿà ð       ›degrees in angle y.››(a) 30  (b) 60  (c) 90››(d) 150  (e) 180›~RCD›5. (d) 150 Ans.››y is a corresponding›angle to the›supple)P}ÿà ð       mentary angle›equal to 30* therefore~››`  y + 30* = 180*››`        y = 150* Ans.›~ET›~ET››tions.›~RA›~SB›~SH11311   3›~SH0933( Zÿà ð       ›~SB›~SH0432a b›~SP208008245048245048245048›~SP245008208048208048208048›~SF›E. Equalities of Angles and Sides.››Make sure you-R}ÿà ð        know these›theorems& definitions and›assumptions.››1. Vertical angles are equal.››`    <a = <b›~RA›~SB›~SH1828A        B›~SH-S}ÿà ð       0933C›~SP196136252136224072193136›~SF›2. An isosceles triangle has two sides›equal in length. The angle between the›equal sid-T}ÿà ð       es is called the vertex›angle. The other two angles are called›base angles and are equal.››If AC = BC& then <A = <B.››3. If t-U}ÿà ð       wo angles of a›triangle are equal& the›sides opposite are equal.››If <A = <B& then AC = BC.›~RA›~SB›~SH1531A       B›~SH0735C-V}ÿà ð       ›~SH14311›~SP196112266112243062217112›~SF›4. An equilateral triangle (ABC) is›equiangular& and an equiangular›triangle is equ-W}ÿà ð       ilateral.››If AB = BC = AC&›then <A = <B = <C.››5. An exterior angle›of a triangle is the›sum of the non-adjacent›interior an-X}ÿà ð       gles.››<1 = <B + <C.›~RA›~Q›~SB›~SH0627A      B    D›~SH0332C›~SH0739E›~SP266039189039220024266048›~SF›1. Given~ Isosceles {A-Y}ÿà ð       BC›with AC = BC& and <DBE =›20*. Find <C.››(a) 20*  (b) 70*  (c) 140*››(d) 160*  (e) 150*›~RCC›1. (c) 140* Ans.››Since <DBE a-Z}ÿà ð       nd <B of {ABC›are vertical angles& they›are equal.››`  <B = <DBE = 20*››Since AC = BC the angles›opposite them are equal.››` -[}ÿà ð                <A = <B››`<A + <B + <C = 180*››`20*+ 20*+ <C = 180*››`          <C = 140* Ans.›~RA›~Q›~SB›~SH1328A      B›~SH0532C›~-\}ÿà ð       SH0230D›~SP217046196096238096206016›~SF›2. Given~ Isosceles {ABC›with <ACD = 130*. Find~ <B.››(a) 50*  (b)  60*  (c) 65*››(d)-]}ÿà ð        130*  (e) 75*›~RCC›2. (c) 65* Ans.››<ACD and <ACB are›supplementary.››`<ACD + <ACB = 180*››`130* + <ACB = 180*››`       <ACB-^}ÿà ð        =  50*›~RA›The angles of triangle›total 180*››`  A + <B + <C = 180*››`<A + <B + 50* = 180*››`      <A + <B = 130*››In the is-_}ÿà ð       osceles›triangles& <A = <B››`<B + <B = 130*››`     <B =  65* Ans.›~RA›~Q›~SB›~SH0435C›~SH0928 y›~SH1028D   A     B›~SH0729E›~-`}ÿà ð       SH1236F›~SP189071259071241038224071›~SP196056252088252088252088›~SF›3. Given~ {ABC is equilateral and <CAF›is a right angle. -a}ÿà ð       Find~ <y.››(a) 30*  (b) 60*  (c) 90*››(d) 150*  (e) 45*›~RCA›3. (a) 30* Ans.››Since ABC is equilateral&››<CAB = 60*››<BAF = <-b}ÿà ð       CAF - <CAB››<BAF = 90* - 60*››`    = 30*››<y is a vertical angle to <BAF››` <y = <BAF = 30* Ans.›~RA›~Q›~SB›~SH1032A     B›~S-c}ÿà ð       H0531D   C   E›~SP224072259072241038224072›~SP215039266039266039266039›~SF›4. Given~ {ABC with AB = 10in.; <B =›60*; DCE ^ AB-d}ÿà ð       ; <DCA = 60*. Find~›Length of BC.››(a) 4}  (b) 10}  (c) 20}››(d) 50} (e) 8}›~RCB›4. (b) 10} Ans.››<A = <DCA because they›are -e}ÿà ð       alternate interior›angles››`            <A =  60*››`  <A + <B + <C = 180*››` 60* + 60*+ <C = 180*››`            <C =  60*››Th-f}ÿà ð       erefore {ABC is equilateral and BC =›AB = 10}. Ans.›~RA›~Q›~SB›~SP224071259071241038224071›~SP279032259071279071279071›~SH103-g}ÿà ð       2A     B D›~SH0435C    E›~SF›5. BE^AC& and BE›bisects <CBD. Then&››(a) AB = AC  (b) AB = BC››(c) AC = BC  (d) AC = BE››(e) AB-q}ÿà ð       Ãc                                                                                    ÿÿÿÿÿÿÿÿÿÿÿÿ                            ÿà ð       b7  PRNTS   SYSB > G0         B ? AUTORUN BAKB B AUTORUN SYSBG U GEOM    BASB! œ AA         B ½ AB         B. Ð BBB        ÿà ð       B þ DBB        B" FBB        B 3CBB        B QEBB        B' uDOS     SYSB œG1         B4 ¤GBB        B ØJBB        ÿà ð       B îKBB        B) 
LCC        B 3HBB        B" NIBB                                                                        ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð        = AC = BC›~RCB›5. (b) AB = BC Ans.››Since <ACB and <CBE are›alternate interior angles&›they are equal. The sum of›the angles-r}ÿà ð        around a point›on one side of a straight›line is 180*.››<CBA + <CBE + <EBD = 180*››`<CBE = <EBD››`<CBA = 180* - 2 <CBE››`<CB-s}ÿà ð       A = 180* - 2 <ACB›~RA›The sum of the angles of›a triangle is 180*››<CAB + <ACB + <CBA = 180*››<CAB = 180* - <ACB - <CBA››<CAB-t}ÿà ð        = 180* - <ACB›`      - (180* - 2<ACB)››<CAB = <ACB››Since sides opposite equal angles are›equal& AB = BC Ans.›~ET›~ET››e you, xÿà ð       ª¾
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Special Triangles››Make sure you know these theorems&›definitions and assumptions.›~RA›~SB›~SP266095224095224031266095›~S9¥}ÿà ð       H1335s›~SH0736h›~SF›`A. The 30*& 60*& 90* Triangle.››1. The side(s) opposite›the 30* angle equals›one-half the hypotenuse (h)9¦}ÿà ð       .›`    s = h/2›››››2. The side(s) opposite›the 60* angle equals›one-half the hypotenuse (h)›`       _›times \/3.››`       h  9§}ÿà ð         _›`   s = -  \/3›`       2›~RA›~SB›~SH1433s›~SH0729s      s›~SH0934a›~SP182095266095224031182095›~SP224031224095224095224099¨}ÿà ð       5›~SF›3. The altitude (a) of›an equilateral triangle›equals one-half the›`               _›side(s) times \/3.››`       s    _9©}ÿà ð       ›`   a = -  \/3›`       2›~RA›~SB›~SH0733h    s›~SH1134s›~SP210079258079258031210079›~SF›`B. The 45*& 45*& 90* Triangle››1. T9ª}ÿà ð       he side(s) opposite the›45* angle equals one-half›`                          _›the hypotenuse (h) times \/2.››`       h    _›9«}ÿà ð       `   s = -  \/2›`       2››2. The hypotenuse (h)       _›equals the side (s) times \/2.›`           _›`    h = s\/2›~RA›~SB›~S9¬}ÿà ð       H1134s›~SH0730s  d    s›~SH0434s›~SP210079258079258031210079›~SP258031210031210079210079›~SF›3. The side of a square(s)›equal9­}ÿà ð       s half the diagonal (d)›`       _›times \/2.››`        d   _›`    s = - \/2›`        2›››4. The diagonal (d) of a›square equa9®}ÿà ð       ls the side(s)›`       _›times \/2.›`            _›`    d = s \/2›~RA›~SD›~Q›1. Find the altitude of an equilateral›triangle 9¯}ÿà ð       whose side is 10.›`             _         _›(a) 5  (b) 5\/2  (c) 5\/3›`       _          _›(d) 10\/2  (e) 10\/3›~RCC›`       9°}ÿà ð         _›1. (c) 5\/3 Ans.››Note~ Draw an altitude (a) to create a›30*& 60*& 90* triangle.››`          1    _›`      a = -10\/3›`  9±}ÿà ð               2›`             _›`      a = 5\/3  Ans.›~RA›~Q›~SB›~SH0729s      s›~SH09336›~SH1326A      s     B›~SH0333C›~SP18209529²}ÿà ð       66095224024182095›~SP224024224096224095224095›~SF›2. Find the side s of›equilateral triangle ABC›whose altitude is 6.›`      9³}ÿà ð              _         _›(a) 3  (b) 2\/3  (c) 3\/3›`      _         _›(d) 4\/3  (e) 6\/3›~RCD›`         _›2. (d) 4\/3 Ans.››Altitud9´}ÿà ð       e of an equilateral›`               S  _›triangle equals -\/3›`               2››`  S  _›`  -\/3 = 6›`  2›`     _›` S \/3 = 19µ}ÿà ð       2›`             _›`    3S = 12\/3›`            _›`     S = 4\/3 Ans.›~RA›~Q›~SB›~SH0330A›~SH1329B›~SH1029D     C›~SP2030242039¶}ÿà ð       096238078203024›~SP203078238078238078238020›~SF›3. In {ABC& AC = 8} and <A = 30*. Find›the length of altitude CD.›`          9·}ÿà ð           _          _›(a) 4}  (b) 4\/2}  (c) 4\/3}›`      _          _›(d) 8\/2}  (e) 8\/3}›~RCA›3. (a) 4} Ans.››The side opposite9¸}ÿà ð        the 30*›angle of a 30*& 60*& 90*›triangle equals h/2››Altitude CD = h/2››`             8}›`           = --›`             2››9¹}ÿà ð       `           = 4}›~RA›~Q›~SB›~SH0733D›~SH1130A       C›~SH0438B›~SP210080259080249031210080›~SP234055259080259080259080›~SF›4.9º}ÿà ð        Find the length of›altitude CD to the hypotenuse›of isosceles right triangle›ABC each of whose equal›sides is 6 inches long.9»}ÿà ð       ›`              _›(a) 6}  (b) 6\/2}  (c) 3}›`      _          _›(d) 3\/2}  (e) 3\/3}›~RCD›`         _›4. (d) 3\/2} Ans.››The 9¼}ÿà ð       original hypotenuse›`     _                 _›is 6\/2& each part = 3\/2.›~RA›~SD›~Q›5. Find the diagonal of a square whose›si9½}ÿà ð       de is 10.›`             _         _›(a) 5  (b) 5\/2  (c) 5\/3›`       _          _›(d) 10\/2  (e) 10\/3›~RCD›`          _›5. 9¾}ÿà ð       (d) 10\/2 Ans.››Note~ The diagonal creates a 45*& 45*&›90* triangle. According to the›`                            _›formula 9¿}ÿà ð       the hypotenuse is s\/2. Since›`                         _›s = 10 the diagonal = 10\/2.  Ans.›~RA›~Q›~SB›~SH1130A       B›~SH09À}ÿà ð       430D       C›~SP210080259080259031210080›~SP259031210031210080210080›~SF›6. Find to the nearest foot›the distance saved by›wa9Á}ÿà ð       lking along diagonal AC of›square lot ABCD& instead of›along its sides AB and BC&›each of which is 100 feet›`           _›lon9Â}ÿà ð       g (use \/2 = 1.414.)››(a) 58 (b) 58.6  (c) 59››(d) 117  (e) 117.2›~RCC›6. (c) 59 Ans.››Note~ The diagonal›creates a 45*& 45*&9Ã}ÿà ð       ›90* triangle.›According to the›formula the hypotenuse›`     _›is s\/2. Therefore›`          _›AC = 100 \/2››= 141.4››AB + BC9Ä}ÿà ð        = 200 along sides››200 - 141.4 = 58.6››To nearest foot = 59 ft. Ans.›~RA›~Q›~SB›~SP252036210036189079273079›~SP27307925203629Å}ÿà ð       52079252079›~SH1125  A      F  E  D›~SH0430B       C›~SH0836y›~SF›7. Given~ BC^AD& BC = 4}& AD = 8}& <A =›<D = 60*& CE[AD. Fi9Æ}ÿà ð       nd y in inches.››Note~ Draw CF^AB&›thus completing›parallelogram ABCF.›`             _›(a) 2  (b) 2\/2›`      _›(c) 2\/3  (d)9Ç}ÿà ð        4›`      _›(e) 4\/3›~RCC›`         _›7. (c) 2\/3  Ans.››Drawing CF^AB creates›{CFD which is›equilateral since <D›= 60*. In p9È}ÿà ð       arallelogram›ABCF&››`    AF = BC = 4››`    FD = AD - AF››`       = 8} - 4}››`       = 4}›~RA›In right triangle CED y is oppos9É}ÿà ð       ite the›60* angle››`         h  _›`     y = -\/3›`         2››The hypotenuse›CD = FD = 4}››`         4  _›`     y = -\/3›`   9Ê}ÿà ð             2›`            _›`     y = 2\/3  Ans.›~RA›~Q›~SB›~SH0733D›~SH1130A       C›~SH0438B›~SP210080259080259031210080›~SP23409Ë}ÿà ð       55259080259080259080›~SF›8. Given~ Right isosceles triangle ABC&›with <C = 90* and›AC = BC; CD bisects <C;›AB = 20}. Find the9Ì}ÿà ð        length›of CD.›`                _›(a) 10}  (b) 10\/2}›`       _›(c) 10\/3}  (d) 5}›`      _›(e) 5\/2}›~RCA›8. (a) 10} Ans.››S9Í}ÿà ð       ince AC = BC& <A = <B››`    <A + <B + <C = 180*››`   <A + <B + 90* = 180*››`         <A + <B = 90*››`            2 <A = 90*››9Î}ÿà ð       `              <A = 45*››`              <B = 45*›~RA›Since CD bisets <C& <ACD›= <BCD = 45* DC = AD and›DC = BD because they a9Ï}ÿà ð       re›sides opposite equal angles.›Therefore&››`        AD = BD››`   AD + BD = AB››`   AD + BD = 20}››`      2 AD = 20}››`      9Ð}ÿà ð         AD = 10} Ans.›~RA›~Q›~SB›~SH1117A   E               B›~SH1917C                   D›~SH1723F›~SP112088252088119144259144›~SP9Ñ}ÿà ð       119144140088154128154128›~SF›9. Given~ AB^CD& EF[BC& <AEC = 75*&›<BCD = 30*& EB = 2}. Find~ Length of›EC.››(a) 1}  (b)  2}  (9Ò}ÿà ð       c)  3}››(d) 4}  (e)  6}›~RCB›9. (b) 2} Ans.››Since the angles of a triangle total›180*››<FEB + <EBF + <BFE = 180*›<FEB +  30*9Ó}ÿà ð        +  90* = 180*›`             <FEB =  60*››{FEB is therefore a 30*& 60*& 90*{›~RA›`               h›`          EF = -›`       9Ô}ÿà ð               2››`               2}›`          EF = --›`               2››`             = 1}›~RA›The sum of the angles about a poin9Õ}ÿà ð       t on›the same side of a straight line is›180*››`<AEC + <CEF + <FEB = 180*››`  75* + <CEF + 60* = 180*››`              <CEF = 9Ö}ÿà ð        45*›~RA›Therefore& {CEF is an isosceles right›triangle in which the sides are 1}. The›hypotenuse  EC = 2s››`                9×}ÿà ð         = 2 (1})››`                  = 2} Ans.›~ET›~ET›››ngle in which the sides are 1}. The›hypotenuse  EC = 2s››`                8 3ÿà ð       ›~SB›~SP182063224063196016182063›~SP196016196063196063196063›~SP238063259063245040238063›~SP245040245063245063245063›~SH0927A=Ù}ÿà ð       'D' B'A D B›~SH0229C'›~SH0629h'›~SH0536C›~SH0836h›~SF›J.Similar Figures.››Make sure you know these›theorems& definitions and›=Ú}ÿà ð       assumptions.››1. Corresponding lines&›sides& and perimeters (p)›of similar polygons are›in proportion.››`      h    s    p›th=Û}ÿà ð       en&  -- = -- = --›`      h'   s'   p'›~RA›~SD›2. Areas (A) of similar polygons are to›each other as the squares of›correspond=Ü}ÿà ð       ing lines& sides& and›perimeters.››`   {ABC is similar to {A'B'C'››`   then~››`        2      2     2›`   A   h      s     p›=Ý}ÿà ð       `   - = --- = --- = ---›`         2     2     2›`   A'  h'    s'    p'›~RA›3. In any two circles~››`    C    r    d›`   -- = =Þ}ÿà ð       -- = --›`   C'   r'   d'››`         2     2     2›`   A    r     d     C›`   - = --- = --- = ---›`         2     2     2›`   =ß}ÿà ð       A'  r'    d'    C'›~RA›Example~››Given~ Circle O is 4 times as large as›circle O'. Radius OP = 10}.›Find~ Length of radius O'=à}ÿà ð       P'.››Let O' P' = y››a) In circles O and O'&››`              2›`       A    OP›`       - = -----›`               2›`       A' =á}ÿà ð        O'P'›~RA›`              2›`        4   10›b) then~ - = ---›`              2›`        1    y››`        2›`      4y  = 100››` =â}ÿà ð              2›`       y  = 25››`       y  = 5} Ans.›~RA›~Q›1. The sides of two similar polygons›are in the ratio 4~1. Their›corres=ã}ÿà ð       ponding perimeters are in the›ratio~››(a) 4~1  (b) 8~1  (c) 12~1  (d) 16~1››(e) 2~1›~RCA›1. (a) 4~1 Ans.››`  p     s›`  -- = =ä}ÿà ð       --›`   1    1›`  p    s›››`   p   4›`  -- = - Ans.›`   1›`  p    1›~RA›~Q›2. The areas of two similar triangles›are in the ra=å}ÿà ð       tio 4~9. Their›corresponding sides are in the ratio~››(a) 16~81  (b) 4~9  (c) 2~9/2  (d) 2~3››(e) 3~2›~RCD›2. (d) 2~3 Ans.››`=æ}ÿà ð          2›`  s     A›` ---- = --›`    2›`   1     1›` (s )   A›››`   2›`  s     4›` ---- = -›`    2›`   1›` (s )   9›~RA›`        =ç}ÿà ð        _›`   s   \/4›`  -- = ---›`   1     _›`  s    \/9›››`   s   2›`  -- = -  Ans.›`   1›`  s    3›~RA›~Q›3. The circumferences o=è}ÿà ð       f two circles›are 4#} and 9#}& respectively. Their›areas are in the ratio~››(a) 4~9  (b) 2~3  (c) 8~18  (d) 16~81››(e) 16#~81=é}ÿà ð       ›~RCD›3. (d) 16~81 Ans.››`            2›`           c    a›`       -----  = --›`           2›`       (c')     a'››`          =ê}ÿà ð         2›`       (4#r)    a›`       ------ = --›`            2›`       (9#r)    a'››`          2 2›`       16# r    a›`       ----=ë}ÿà ð       -- = --  Ans.›`          2 2›`       81# r    a'›~RA›~Q›~SB›~SP217016217095252095217016›~SP238064238095238095238095›~SH0831y›=ì}ÿà ð       ~SH13333 2›~SH10344›~SF›4. Find y in the diagram~››(a) 10  (b) 6  (c) 2.5››(d) 1.5  (e) 5›~RCA›4. (a) 10 Ans.››`  y     4›`--=í}ÿà ð       --- = -›`(3+2)   2›››`    y›`    - = 2›`    5››`    y = 10 Ans.›~ET›~ET››››(d) 1.5  (e) 5›~RCA›4. (a) 10 Ans.››`  y     4›`--< Iÿà ð       ›K. Volumes.››Make sure you know these theorems&›definitions and assumptions.››Volume is the measure in cubic units of›the spAï}ÿà ð       ace contained in or occupied by›an object.››1. Rectangular Solid.››`      Volume = Bh = lwh››2. Cube.›`                3›`   Að}ÿà ð          Volume = e›~RA›3. Right Circular Cylinder.››`                      2›`      Volume = Bh = #r h›~RA›4. Sphere›`            Añ}ÿà ð               3        3›`      Volume = 4/3#r  = 1/6#d›~RA›Example~››A cylindrical bar of metal having a›circular base of radius 5Aò}ÿà ð        inches and a›height of 3 inches weighs 1000 grams.›What would a cylindrical bar of the›same metal weigh if its circular baseAó}ÿà ð       ›has a radius of 3 inches and its height›is 5 inches?›~RA›`            2                  2›`      V = #r h          V' = #r Aô}ÿà ð       h››`           2                  2›`      V = 5  x 3#       V' = 3  x 5#››`      V = 75#           V' = 45#››`          DireAõ}ÿà ð       ct proportion››`            V    W›`            -- = --›`            V'   W'››`           75#   1000›`           --- = ----›`Aö}ÿà ð                  45#     y››`           75y = 45000›`             y = 600 grams  Ans.›~RA›~Q›1. A tank 8' x 6' x 5' is filled with A÷}ÿà ð       a›liquid weighing 120 lbs. What would be›the weight of a quantity of this liquid›which fills a tank 4' x 3' x 1'?››(a) 20 lbsAø}ÿà ð       . (b) 12 lbs. (c) 10 lbs.››(d) 6 lbs.  (e) 4 lbs.›~RCD›1. (d) 6 lbs. Ans.››`    V     W›`    -- = --›`     1    1›`    V    WAù}ÿà ð       ››` 8x6x5   120 lbs.›` ----- = --------›`             1›` 4x3x1      W››`   240   120›`   --- = ---›`           1›`    12    Aú}ÿà ð       W››`     1›`    W  = 6 Ans.›~RA›~Q›2. A cylinder with base diameter of 10}›and height of 6} weighs 15 lbs. What›would a cylinAû}ÿà ð       der of base diameter 6}›and height of 10} weigh if it is made›of the same material?››(a) 90# lbs. (b) 360# lbs.  (c) 25 lbs.›Aü}ÿà ð       ›(d) 9 lbs.  (e) 3# lbs.›~RCD›2. (d) 9 lbs. Ans.››`           V    W›`          -- = --›`           1    1›`          V    W›Aý}ÿà ð       ›`     2›` #(5)   (6)    15›` -----  ---- = --›`     2          1›` #(3)   (10)   W››`        150#   15 lbs.›`        ---- = Aþ}ÿà ð       -------›`                  1›`         90#     W››`           1›`          W  =  9 lbs. Ans.›~RA›~Q›3. Find the edge of a cubAÿ}ÿà ð       e whose volume›is 64 cu. in.››(a) 4 in.  (b) 8 in.  (c) 10 2/3 in.››(d) 21 1/3 in.  (e) 2 in.›~RCA›3. (a) 4 in. Ans.››`      B }ÿà ð                  3›`            V = e››`                 3›`           64 = e››`        4 in. = e  Ans.›~RA›~Q›4. A sphere is inscrB}ÿà ð       ibed in a cube 6}›on an edge. The difference in volume›between the cube and the sphere in›cubic inches is~››(a) 432#  (b) 180B}ÿà ð       #  (c) 216  (d) 36#››(e) 216 - 36#›~RCE›4. (e) 216 - 36# Ans.››`        Cube          Sphere››`        e = 6         r = 3››`B}ÿà ð                    3            4  3›`        V = e         V = -#r›`                          3››`             3            4›`   B}ÿà ð            V = 6         V = -(27)#›`                          3››`        V = 216       V = 36#››Difference = 216 - 36#  Ans.›~RA›B}ÿà ð       ~Q›5. A rectangular swimming pool is 50›ft. by 120 ft. Water is entering at the›rate of 1750 gallons per minute. How›many houB}ÿà ð       rs will it take to fill the›pool to a depth of 7 feet?››(7 1/2 gallons = 1 cu. ft.)››(a) 1 4/5  (b) 3  (c) 3 1/5  (d) 180››(eB}ÿà ð       ) 2 1/2›~RCB›5. (b) 3 Ans.››V = 1wh››V = (120) (50) (7)››V = 42&000 cu. ft. x 7 1/2 gal/cu.ft.››` = 315&000 gal››1750 Gal/minB}ÿà ð        x 60 min/hr.=›`                      105&000 gal./hr.››`  315&000 gal.›`  -------------- = 3 hr. Ans.›`  105&000 gal/hr›~ET›B	}ÿà ð       ~ET›0 min/hr.=›`                      105&000 gal./hr.››`  315&000 gal.›`  -------------- = 3 hr. Ans.›`  105&000 gal/hr›~ET›@ ÿà ð       ›~Q›1. The wheel of a bicycle is 28 inches›in diameter. Compute the number of feet›that will be covered in six turns of›the wF}ÿà ð       heel of this bicycle.›(use # = 22/7)››(a) 14 ft  (b) 44 ft  (c) 88 ft››(d) 308 ft  (e) 528 ft›~RCB›~Q›2. A circle has a diameF}ÿà ð       ter that is 20›inches longer than the diameter of›another circle. If the radius of the›smaller circle is r& the circumferenceF}ÿà ð       ›of the smaller circle is to the›circumference of the larger as~››`     2r          2r           r›(a) ------  (b) ------  (cF}ÿà ð       ) -------›`   r + 10      r + 20      2r + 20››`     r           r›(d) ------  (e) ------›`   r + 20      r + 10›~RCE›~Q›3. TF}ÿà ð       he angles of {LMN are y*& y/2 + 10*›and 3y - 10*. Find y.››(a) 20  (b) 30  (c) 40  (d) 110››(e) 120›~RCC›~Q›4. The two legs oF}ÿà ð       f a right triangle are›.6} and .8}& respectively. The number›of inches in the hypotenuse is~››(a) .14}  (b) 1}  (c) .10}  (d)F}ÿà ð        14}››(e) .2}›~RCB›~Q›~SB›~SP196024259024259070196070›~SP196070196024197024210070›~SH0328D›~SH0338C›~SH1028A  E›~SH1038B›~SF›F}ÿà ð       5. The area of rectangle›ABCD is equal to 48 sq.›in. and AE~EB as 1~3.›The number of sq. in.›in {ADE is~››(a) 6  (b) 12  (c) F}ÿà ð       16››(d) 24  (e) 8›~RCA›~SD›~Q›6. The radius of a circular playground›is 4 times the radius of a circular›pool with the same cF}ÿà ð       enter. The ratio of›the area of the pool to the area of the›playground is~››(a) 2~1  (b) 1~2  (c) 4~1  (d) 16~1››(e) 1~16›~RCF}ÿà ð       E›~Q›7. A boat sailed 35* due east along a›parallel of latitude. If it started›from a port situated on 22* west›longitude& itF}ÿà ð       s new position is~››(a) 57* west long.›(b) 57* east long.›(c) 13* west long.›(d) 13* east long.›(e) 0* east long.›~RCD›~Q›8. F}ÿà ð       The diagonal of a square divides it›into two triangles each 32 sq. in. in›area. The length of the diagonal in›inches is~›`   F}ÿà ð           _                 _›(a) 16\/2  (b) 16  (c) 8\/2›`       _         _›(d) 32\/2  (e) 4\/2›~RCC›~Q›9. The scale of a map is F}ÿà ð       1 in. = 50›miles. The actual area in square miles›of a circle 2 inches in diameter is~››(a) 50#  (b) 100#  (c) 200#  (d) 2500F}ÿà ð       #››(e) 10&000#›~RCD›~Q›10. A rectangle is 4 inches longer than›it is wide. Its perimeter is 20 inches.›The number of square iF}ÿà ð       nches in its area›is~››(a) 96  (b) 32  (c) 21  (d) 12  (e) 36›~RCC›~Q›11. Through how many degrees does the›hour hand of a clF}ÿà ð       ock move in 150›minutes?››(a) 2 1/2*  (b) 25*  (c) 75*  (d) 150*››(e) 256/#*›~RCC›~Q›~SB›~SH0927A           B›~SH0533C›~SH033F}ÿà ð       7D›~SP259024189064266064227040›~SF›12. AC = 10}& <A = 30*&›and <BCD = 60*. The›number of inches in the›length of BC is~›`    F}ÿà ð                _         _›(a) 5  (b) 5\/2  (c) 5\/3››(d) 10  (e) 20›~RCD›~Q›~SB›~SP161070217024273070161070›~SP2170242170702170702F}ÿà ð       17070›~SH1023A›~SH1039B›~SH0332C›~SH1032D›~SF›13. AC = 10}& CD [ AB&›CD = 6}& and <A = <B.›The number of inches›in AB is~››(aF }ÿà ð       ) 6  (b) 8  (c) 10››(d) 12  (e) 16›~RCE›~Q›~SB›~SH0631t  p›~SH0735q›~SH0929s   r›~SP189056256056224056256032›~SP2030882240562F!}ÿà ð       27024227024›~SF›14. <p + <q + <r = 210*. <t = 2<s.The›number of degrees in <s is~››(a) 10  (b) 30  (c) 50››(d) 100  (e) 150›~F"}ÿà ð       RCC›~SD›~Q›15. A cylindrical can of oil with a›circular base has a height h and holds›30 quarts. If its height is to remain›cF#}ÿà ð       onstant& by what number must its base›radius be multiplied so that it will›hold 60 quarts?›`              _›(a) 2h  (b) h\/2 F$}ÿà ð        (c) 4  (d) 16›`     _›(e) \/2›~RCE›~Q›16. How many rods of steel 4 inches by›4 inches by 9 feet can be made from 144›cubic fF%}ÿà ð       eet of steel?››(a) 1  (b) 4  (c) 16  (d) 144  (e) 122›~RCD›~Q›~SB›~SP210024252024252096210096›~SP210096210024210024210024›~SPF&}ÿà ð       206020256020256100206100›~SP206100206020206020206020›~SH12336}›~SH08349}›~SF›17. A picture 6} x 9} is enclosed in a›frame 1/2F'}ÿà ð       } wide. The outside perimeter›is~››(a) 16}  (b) 17}  (c) 32}››(d) 34}  (e) 27}›~RCD›~SD›~Q›18. If the length of a rectangle iF(}ÿà ð       s›decreased by 10%& and its width is›increased by 50%& by what percent is›its area changed?››(a) 35  (b) -35  (c) 65  (d) -65F)}ÿà ð       ››(e) 40›~RCA›~Q›19. The supplement of an angle is 10*›less than 3 times its complement. Find›the number of degrees in the anF*}ÿà ð       gle.››(a) 5  (b) 40  (c) 45  (d) 50  (e) 20›~RCB›~Q›20. Given~ Trapezoid KLMN with NM`[`LM&›KN = 10}& LM = 16} and <L = 45*. F+}ÿà ð        Find~›The number of square inches in the area›of KLMN.›`               _›(a) 48  (b) 48\/2  (c) 78›`       _›(d) 78\/2  (e) F,}ÿà ð       96›~RCC›~Q›21. Water flows 2 feet deep in a›rectangular channel 4 feet wide. If the›water flows at 3 miles an hour& what›weigF-}ÿà ð       ht of water (in tons) will pass a›given point in 6 hours? (A cubic yard›of water weighs 3/4 ton).››(a) 21&120  (b) 37&545  (cF.}ÿà ð       ) 63&360››(d) 190&080  (e) 17&320›~RCA›~Q›~SB›~SH073810›~SH10274    y›~SH09292›~SP252072182072196064196080›~SP196064252032252F/}ÿà ð       080252080›~SF›22. Find the length of y.››(a) 20  (b) 16  (c) 12  (d) 5  (e) 10›~RCB›~SD›~Q›23. A bridge 3500 feet long and 16F0}ÿà ð        feet›wide is designed to withstand a traffic›weight of 50 pounds per square foot.›Assuming that the two lanes of the›bridge F1}ÿà ð       are filled with cars standing›bumper to bumper and each 210} long&›what should be the maximum average›weight per car in poundF2}ÿà ð       s?››(a) 200  (b) 400  (c) 583 1/3››(d) 7000  (e) 14&000›~RCD›~ET›~ET››t should be the maximum average›weight per car in poundD Fÿà ð       ›~SB›~SP205024259070205070205024›~SH0729a›~SH1033b›~SH0534c›~SH1632s     s›~SH1935s›~SP217143266143243106217143›~SF›H. PerimeJ4}ÿà ð       ters››Make sure you know these›theorems& definitions and›assumptions.››A perimeter is the measure›in linear units of the›distJ5}ÿà ð       ance around an object.››1. Triangle››`  Perimeter = a + b + c››2. Equilateral Triangle››`  Perimeter = 3s›~RA›~SD›3. RectanglJ6}ÿà ð       e››`   Perimeter = 2l + 2w = 2 (l + w)››4. Square››`   Perimeter = 4s››5. Circle››`   Perimeter = Circumference (C)››`   C = J7}ÿà ð       2#r = #d›~RA›~SB›~SC192071210056226071226071›~SH0828r›~SH0833r›~SH0930n*›~SH1133L›~SF›6. Arc (length) is a fraction of the›ciJ8}ÿà ð       rcumference of a circle.››`         n          n#r›Length = --- x 2#r = ---›`        360         180››››7. Irregular Figures.J9}ÿà ð       ››To find the perimeter of an›irregular figure~››a. Compute the perimeters of figures›for which formulas exist.››b. DetermineJ:}ÿà ð        their sum& difference& or›fractional parts to obtain the›required answer.›~RA›~Q›~SB›~SP189071217016252071189071›~SH0232B›~SJ;}ÿà ð       H1027A          C›~SF›1. The perimeter of {ABC›is 36}. AB = X + 3&›BC = 3X - 9& AC = 2X›Find the longest side›of the triangleJ<}ÿà ð       .››(a) 5}  (b) 7}  (c) 14}››(d) 10} (e) 12}›~RCC›1. (c) 14} Ans.››x + 3 + 3x - 9 + 2x = 36››`            6x - 6 = 36››`      J=}ÿà ð                 6x = 42››`                 x = 7››AB = 10}& BC = 12& AC = 14››longest is AC = 14} Ans.››Note~ Perimeter is not = 18J>}ÿà ð       0*›~RA›~SD›~Q›2. The distance around a rectangular›swimming pool is 250 feet. The pool is›50 feet wide. Find its length.››(a)J?}ÿà ð        5'  (b) 75'  (c) 100'  (d) 200'››(e) 150'›~RCB›2. (b) 75' Ans.››Let l = length of pool.››`  50 + l + 50 + l = 250››`        J@}ÿà ð        100 + 2l = 250››`               2l = 150››`                l = 75'  Ans.›~RA›~Q›3. Find the diameter of a quarter-mile›circuJA}ÿà ð       lar track. (use # = 22/7 and›express your answer in yards)››(a) 20 yds. (b) 40 yds. (c) 70 yds.››(d) 140 yds. (e) 240 yds.›~RJB}ÿà ð       CD›3. (d) 140 yds. Ans.››Note~ 1/4 mile = 440 yds.››`            c = #d››`                22›`          440 = --d›`          JC}ÿà ð              7››`      140 yds. = d  Ans.›~RA›~Q›4. Find the cost of framing a picture›3 feet by 4 feet at $2.00 per foot.››(a) $12JD}ÿà ð       .00  (b) $14.00  (c) $24.00››(d) $28.00  (e) $48.00›~RCD›4. (d) $28.00 Ans.››Note~ Framing is perimeter.››`         p = 3 + 4JE}ÿà ð        + 3 + 4››`         p = 14 ft.››Cost = 14 x 2 = $28.00  Ans.›~RA›~Q›5. Find the side of an equilateral›triangle whose perimetJF}ÿà ð       er is equal to›that of a square of side 6}.››(a) 4}  (b) 8}  (c) 12}  (d) 24}››(e) 9}›~RCB›5. (b) 8} Ans.››Note~ The perimeteJG}ÿà ð       r of the square is›24}. Therefore the side of the›triangle = 24/3 = 8}  Ans.›~RA›~Q›6. The perimeter of an equilateral›triangJH}ÿà ð       le equals the perimeter of a›square. A side of the triangle is b&›and a side of the square is k. Express›b in terms of k.››(aJI}ÿà ð       ) 3b/4  (b) 4b/3  (c) 3k/4››(d) 4k/3  (e) 4b/3k›~RCD›6. (d) 4k/3 Ans.››` 3b = perimeter of triangle››` 4k = perimeter of squaJJ}ÿà ð       re››` 3b = 4k››`  b = 4k/3 Ans.›~RA›~Q›~SB›~SP224016189080252080224016›~SP189080189128252128252080›~SH062710}›~SH063610}›~SH1JK}ÿà ð       3266}›~SH13386}›~SH102860*   60*›~SF›7. Find the perimeter of the›accompanying geometric figure.››(a) 32}  (b) 38}  (c) 42}››JL}ÿà ð       (d) 44}  (e) 52}›~RCC›7. (c) 42}  Ans.››The triangle is equilateral›since two angles are 60*.›Therefore the common line›in thJM}ÿà ð       e triangle and the›rectangle is 10}& and the›base of the rectangle is›also 10}.››10}+10}+6}+10}+6} = 42}›~ET›~ET›››››››››n thH yÿà ð       ›I. Areas››Make sure you know these theorems&›definitions and assumptions.››Area is the measure in square units of›the surfacNO}ÿà ð       e of an object.››1. Triangle››`    Area = 1/2 bh››2. Equilateral Triangle››`            2›`           s   _›`    Area = --\/3NP}ÿà ð       ›`           4›~RA›3. Rectangle››`    Area = lw›~RA›~SB›~SP217032255032231063189063›~SP189063217032217063217063›~SH0633h›~SH0NQ}ÿà ð       932b›~SH1531h›~SH1833b'›~SH1333b›~SP203104254104266135196135›~SP196135203104203135203135›~SF›4. Parallelogram››`   Area = bh›NR}ÿà ð       ›››5. Square››`           2›`   Area = s››››6. Trapezoid››`          h›`   Area = -(b + b')›`          2›~RA›~SD›~SB›~SC19207NS}ÿà ð       1210056226071226071›~SH0828r›~SH0833r›~SH0930n*›~SH1133L›~SF›7. Circle››`            2›`   Area = #r›››››8. Sector››A sector NT}ÿà ð       is fraction of›the area of the whole circle.››`                    2›`   Area = n/360 x #r››`             2›`          n#r›` NU}ÿà ð              = ----›`          360›~RA›~SD›9. Ring››A ring is a figure with two concentric›circles (circles having the same›center)NV}ÿà ð       . To get the area of the ring -›that is& the outer portion - subtract›the area of the smaller circle from›the area of the larNW}ÿà ð       ger circle.›Area of Ring = Area of larger circle -›Area of smaller circle.››`                2     2›Area of Ring = #R  - #r›NX}ÿà ð       ›`                 2    2›Area of Ring = #(R  - r )›~RA›10. Irregular Figures.››To find the area of an irregular›figure~››a. NY}ÿà ð       Compute the areas of figures for›which formulas exist.››b. Determine their sum& difference& or›fractional parts to obtain theNZ}ÿà ð       ›required answer. In general~ Area›shaded region = area outside - area›inside.›~RA›~Q›~SB›~SP203024254024266055196055›~SP1960N[}ÿà ð       55203024203055203055›~SH03338}›~SH06314}›~SH083310}›~SP210085252085252128210128›~SP210128210085210085210085›~SH1430y›~SF›ExamN\}ÿà ð       ple~››1. The area of a square›is equal to the area of a›trapezoid with bases 8}›and 10} and altitude 4}.› Find a side of the N]}ÿà ð       square.››Area of square = Area of›trapezoid.›~RA›`2›y  = h/2 (b + b')››`2›y  = 4/2 (8 + 10)››`2›y  = 2 (18)››`2›y  = 36››y  =N^}ÿà ð        6} Ans.›~RA›~SD›~Q›1. ABCD is a parallelogram& with AB =›10}& AD = 6} and <A = 60*. Find the›number of square inches in the N_}ÿà ð       area of›ABCD.›`               _                  _›(a) 15  (b) 15\/3  (c) 30  (d) 30\/3›`     __›(e) \/30›~RCD›`          _›1N`}ÿà ð       . (d) 30\/3 Ans.››Using the 30*& 60*& 90* triangle››`       6  _      _›`   h = -\/3 = 3\/3›        2››Now A = bh›`          Na}ÿà ð          _›`   A = 10(3\/3)›`           _›`   A = 30\/3  Ans.›~RA›~Q›2. If the area of a circle is›quadrupled& by what percent is tNb}ÿà ð       he›radius increased?››(a) 100  (b) 200  (c) 300  (d) 400››(e) 10›~RCA›2. (a) 100 Ans.››`            2›`    A      r›`   ---- Nc}ÿà ð       = ----›`             2›`     1      1›`    A     (r)›››`            2›`    1      r›`   ---- = ----›`             2›`        Nd}ÿà ð           1›`    4     (r)››~RA››`      2      2›`     1›`   (r)  =  4r››`     1›`    r   =  2r››which represents a 100% increase.›Ne}ÿà ð       ~RA›~Q›3. Cruiser A patrols a circular area›of 3600# square miles. Cruiser B›patrols a circular area of 900# square›miles. WhNf}ÿà ð       at is the difference between›the radii of action of the cruisers?›`                     _›(a) 30 miles  (b) 30\/3 miles›`    Ng}ÿà ð                                    __›(c) 90 miles  (d) 60 miles  (e) \/90›~RCA›3. (a) 30 miles Ans.››The radius of Cruiser A~››`  Nh}ÿà ð         2›`  #r  = 3600# sq. miles››`    r = 60 miles››The radius of Cruiser B~››`    2›`  #r  = 900# sq. miles››`    r = 30 miles›Ni}ÿà ð       ›The difference between the radii is~››`60 miles - 30 miles = 30 miles Ans.›~RA›~Q›4. A wire 22} long is to be bent into›the Nj}ÿà ð       shape of a square. Another piece›of wire of the same length is to be›shaped into a rectangle whose length›is 4}. Find the difNk}ÿà ð       ference in area›between the square and the rectangle.››(a) 2 1/4 sq. in. (b) 2 3/4 sq. in.››(c) 5 3/4 sq. in. (d) 16 1/4 sq. Nl}ÿà ð       in.››(e) 2 1/2 sq. in.›~RCA›4. (a) 2 1/4 sq. in. Ans.››Perimeter of a square~››`       4s = 22}››`        s = 5 1/2}››Area ofNm}ÿà ð        a square~››`             2›`        A = s››`          = 30 1/4 sq. in.›~RA›Perimeter of a rectangle››`  2l + 2w = 22}››`2(4)Nn}ÿà ð        + 2w = 22}››`       2w = 14}››`        w =  7}››Area of a rectangle~››`        A = lw››`          = 28 sq. in.››30 1/4 sq.inNo}ÿà ð       . - 28 sq. in =›`          2 1/4 sq. in. Ans.›~ET›~ET››f a rectangle~››`        A = lw››`          = 28 sq. in.››30 1/4 sq.inL 7ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð       îî
 SäÎÎ©Å�
êêê SäÀ�ÐþÎ
 SäÀÐþ© �LD�1 pˆð pÊ²¬	¢ © ˆ˜�˜0¥C�9¥D�I pèàÐâ¥C�ç¥D�èL~¥Ci€…C¥Di …D` ÿà ð         @L |u}Ë¬ð6­…C�­…D�­¬® l0¬±C)HÈCð±C¨ WhL/©ÀÐh
¨`¥Cm�…C¥Di �…D`�Œ
©R @�©W €�Œÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ïà ä                                                                                                                                   ÿà ïà ä                                                                                                                                   ïà ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       ÿà ð                                                                                                                                       