P3                   @ 
 Sl$ l0C)HCC WhL/h
`CmCDi D`
R@W        1   	 Y0@R	!L`
	 D
CD
         )16 CS S) C)D1 p p	  0C9DI pCDL~CiCDi D`        HAYDEN SOFTWARE(   SCORE IMPROVEMENT SYSTEM FOR THE SATA. System User's Guide:B. Complementary, Supplementary Angles     And A        ngle SumsC. Parallel LinesD. Angles Of A Circle"E. Equalities Of Angles And SidesF. Pythagorean TheoremG. Special Triangles        H. Perimeters	I. AreasJ. Similar FiguresK. Volumes"L. All Examples Without TUTORMODEA. Using The GEOMETRY MODULEB. Organizat        ion Of The System
MAIN MENUDetail Menu PROGRAMREVIEW 'M = Main Menu                         Q = Quit               R = Res        tart M = Menu | = Back'R = Restart M = Menu O = Omit | = BackP = Paragraph Q = Quit                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            Px 
X`H232435;	1;		hh@ 2e1i 1L;Hҍ
00) 08109hh@         2e1i 1232435ޥ< <8                                                                                               = LxLLLLLLL	LL:	LT	LL_	L	L	L	L	 !h`LL6
LLL
LhLJLTLLLLLLL&LN
LL      !v         
5 7
 h` Ltv 
Wh`h  `v 
5v 
h` v 
	v 
v 
v 
Lx 
v 
[v         
h`t v 
8
v 
[v 
v 
h`v 
  7
Nh`Y sLw !v 
Lr	  7
A[08!        sh`v 
h` 
v 
5h` v 
Wv 
 	 
]h` v 
v 
 	 
h`
v 
v 
 v 
        v 
>v`h0BJ KՅԩ3D
ELVK:h0B HI V`hhhhhhh΢  e˅ː`˪8包        ˥卅̩ ***ΥeͥeΠˑȥ̑Ȋȩ `hh








i        w 	
 Ll    wv   `      ӠŠҠϠ] ] ] ] ] ] ] ] ] ] ] ] ] ] ] ] ] ] ] ]          ] ] ]      ҠӠͧϠנΠϠŠҠҺENTER YOUR NEW ANSWERϠŠҠҠĺENTER YO        UR PREVIOUS ANSWER   Ϡՠ٠ԠϠԿ٬ΩӠҧϠנŠӠէϠנĠ        ӠçϠӠӧϠ    ҠӠŠҠϠ      ŠҠϠ       Ԡ        ŠƠŠŠ           RAW       SAT        TIME    SCORE     SCORE     REMAININGENTER TOTAL OF PREVIOUSLY RECORDED DATA                  ĠӠ    ƠՠȠϠŠŠĠŠ     VERBAL NUMBER CORRECT   =             VERBAL NUMBER INCORRECT =     MATH   NUMBER CORRECT   =     MATH   NUMBER INCORRECT =   ҠԠӠΠϠ        hPͩΩ vw'u  vt$hPͩgΩ̢P vw`   ` vHwH)8        w hwhv` (wLO  ˑΥPɚ` v` w` t`  |`        έ͠ ɛLv LOL u k'u`   	 HL ` 8        Ii8HhIi	J`m8 Ln,8        `,`jnNN)P˩gȈi(ː`@ hhh
 h        HH` j  H h`v8`) 0i(mw  j


fjfj(j`vw( wv` έ)

&
&΅ͦν        i` w,I˥i(ː` ˮvʊ

mv JfJf˅̢ ,ue˅˽e̅`PPg        ѩi
PR          0m
`   0έͩP˩̩@Ѡ 075Ȍ2ˬȌ         e˅ːвˍ ȱˍ L  խ `  L/ =` ɛ        & ` !ɛέ `L ɛڭ...        eͭeΠȩ ͩ`ЩP
 K` `P  ` `ϩЩ`                                                    fff    1j$$$$ >`<|  fl0fF      0        p88p ``     ~       0   ~           0`@          <fnvf<  8~  <f0~  ~f<  <l~  ~`|f<  <`|ff<  ~00  <f<ff<  <f>8        0 @   ~  ~  `00`  <f             |    <ff~f  |f|ff|  <f``f<  xlfflx  ~`|``~  ~`|```  >``nf>  ff~fff  ~~  f<  flxxlf  `````~  cwkcc  fv~~nf  <ffff<          |ff|``  <fffl6  |ff|lf  <`<<  ~  fffff~  ffff<  cckwc  ff<<ff  ff<  ~0`~    @T88T HHHHHHH                 6>       8pp8?                                     ww            <~~~<              x`x`~  <~  ~<  0~0   ~                    <>f>  ``|ff|   <```<  >ff>   <f~`<  >   >ff>| ``|fff   8<   < ``lxlf  8<   fkc   |ffff   <fff<           |ff|``  >ff>  |f```   >`<|  ~   ffff>   fff<   ck>6   f<<f   fff>x  ~0~  $B 00  fff                 S`ѩ  Pυ˩Ѕ  0	 P
`L}) ȱˍ
0 ȑˍm         mː`˩ȱ)ȩˍ8`  ˍ
ȱˍ `  
ȭ `ˍ̍˭`˭`P        ̢ ѱϑ `P˩̩Т ѱˑ`` #mm  P        ˩  0	 P
`m 8J P˩  0	 P`        H232435;	1;		hh@ 2e1i 1L(Hҍ
00) 08109hh@Ҡ 2e1i 1232        435ޥ< <8hl˩l̠  e˅ː`hҢ ҩy ҍ* \*`h, \`&%8%0        %## #### #@#"!@!`! ! !0!!!!!!!!!!!!! ! !`!@        !""@#`#0### #$@%0%8%&                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                 7Ȍ!E`Ȍș`[ȌH șhL`LOAD "D:GEOM.BAS"RUN  h@}        
DR a   x 
X) L S`
i` A}          7Ȍ!E`Ȍș`[ȌH șhL`LOAD "D:GEOM.BAS"RUN  h         Pg p} @ЩѠ xeЅАeЅАѹkI (eЅАѹk 8аѹk (eЅАѹkC}         (eЅАѹk PeЅАѹk (eЅАѹk(eЅАѹlЈ(eЅАѹlЈ(eЅАD}        ѹlЈ(eЅАѹ.lЈ@ЩѠ(eЅАшeЅАѠ=lЈ eЅАйIlЈWeЅАѹjlЈ$eE}        Аѹ|lЈ3ЩѩͩΩMЩN͈Т  EͩΩBЩV(eЅА(e͐ͅ
ݠ؅Щ8nj Ј kjljmjF}        kjkjljlj} ЩѠk8 IЈѩйk8 IЈ Щѹvj8 󩰅Ѡ!fk8 ЈU
RG}        

$%& S0 S0	&%&$8D kjljkjljlj } b dpeH}        j k  Vbde hi  V0L^jL    GEOMETRY*   "SAT" AND "SCHOLASTIC APTITUDE TEST"ARE REGISTERED TRADEMARI}        KS OF THE COLLEGEENTRANCE    EXAMINATION   BOARD.   THESEMATERIALS  HAVE BEEN DEVELOPED BY  ARROWINSTRUCTIONAL  SYSTEMS, INC.J}          FOR HAYDENSOFTWARE COMPANY,  INC. WHICH BEARS SOLERESPONSIBILITY FOR THEIR CONTENTS.ATARI CONVERSIONBY ANDREW TAYLOR(!9$%.3K}        /&47!2%3#/2%NNNNNNNNNN)-02/6%-%.4NNNN3934%- &/2 4(%NHJ  HJ  
J  HJ  Y  ʀJ      Y   ʀJ   Y   ʀ     YL}          ʀ     Y  $%6%,/0%$ "9!22/7 ).3425#4)/.!, 3934%-3 ).##/092)'(4 #	 "9 (!9$%. 3/&47!2% #/ ).##).#3830 ;384M}        0 CHK2 STY TIMER ;timer =13850     JSR DSKINV ;do a read...3860     BMI ERROR  ;stop fo}r any err3870     LDA TIMER  ;wiN}        thin time limit?3880     CMP #8+13890     BCS CHK3  ; no!3900     DEC TESTCNT ;yes, dec t}est cnt3910 CHK3 DEC CURRCNTO}        3920     BNE CHK2 ;  test more intervals3930     LDA TESTCNT ;all intervals OK?3940     CLC  };      assume all is OK395P}        0     BEQ ACCEPT ;Yes, accept: PROTECTED3960     DEC RTRYCNT ;else, re-try Op3970     BNE CHK13}980 ERROR SEC  ;    GiveQ}         up!3990 ACCEPT BCC WAIT4000  LDA #$FF4001  STA $02444010  JMP $E4774020 WAIT LDA #04030  STA T}IM4040  STA TIM+140R}        50  TAX4060 WAIT1 INX4070  BNE WAIT14080  INC TIM4090  BNE WAIT14100  INC TIM+14110  LDA TIM+1412}0  CMP #184130  BS}        NE WAIT14140  LDA #04150  STA $02F04160  LDA #$7D4170  JSR $F6A44180  LDA #34190  STA IOCB+24200  D:AUTORUN.BAK210  STT}        A IOCB+44220PgЩѠ xeЅАeЅАѹkI (eЅАѹk 8аѹk (eЅАѹk            xY#k#MNDNFXABBLBSCDEFMPRRESCANF1F2F3FLINTTLTXDSSCSKTMCPPMERWRSRRTSWCPXYV}        YLNXXOORRRVRHRISPCASQQFMMFTMTVDMVDVRCT                                  W}                              	      
                                                                  X}                                      @      @      @                                    !       "       #       $    Y}           %       &       '       (       )       *       +       ,       -       .       /       0       1       2       3       4 Z}              5       6       7       8       9       :       ;       <       =       >       ?       @       A      B       C      [}         D       E       F       G       H       I       J       K       L       M       N       O       P       Q       R       S   \}            T       U       V        GEOMETRY ,A0   "F:@    , A(   ,
@     A"@    +F:@    ,&A(   ,AB7]}        t  +F:B7t  ,&A(   ,
 ss;@`    ,;@    ,;@@    ,;@    ,;@    ,;@    ,;@    ,;@    ,;@  ^}          ,;@    , ss;@    ,;@    ,;@    ,;@    ,;@'    ,;@    ,;@    ,;@    ,;@    ,;@ _}           , GG;@    ,;@    ,;@`    ,;@`    ,;@`    ,;@    , 669@    <@    ,9@E    <@    ,9@ `}           , 6-?:AQ   , )+@    @    )AV   @      Q0@    @	    @    70@          @    Q0@ a}           @	    @     "F:@    , A(   "
@%     A"@    +F:@    ,&A(   ,AB7t  +F:B7t  ,&A(   , 
b}        A"P    A!0   ! T6-?:,6-!6-?:A a   ,+A    J?:A!Q   <C:,<B:,,@    T
@0    ( 	6.A   2 c}        '	"@%    $6-?:A d   ,'$< 7<,0~Q
@x    = $6-%68<,-$6-A:7<,,> #8<,"68<,-d}        #
@u    @ '8<,"6-?:A d   ,'
@u    B "8<,Au   "
@u    D -6-%A	    #7<,4~S-
@h  e}          G 6-&K 	
@    N @    
@P    P 6-%A	    R 	"6-$T 7<,4~
A    Z 	f}        "$n ,	6.A   "6-?:A!E   ,,
@    x 4~RA
AP    4~RC
A     7<,4~Q6-?:,g}        $ 0~SD
Ap    $6-?:A d   ,6-$6-       "8<,%68<,- 
@     0~SB
A`   h}         !	6-"8<,%!68<,- 38<,%6-6-?:,)Ay   3
@     Au   
@     /	6-6-?:,i}        6-8<,)")"%/6- 		6- 6-%A	     "	46-?:A    ,"6.~SP !4~SPA    !
Ay   j}         0~SG
A0    A    (-@    $6-?:<<<<,(	 (-@    $6-?:<<<<,(	 /6-?:k}        <<<<,%6-?:<<<<,/
Ay    56-+A:7<%,,,$'56-A:7@    <@    %,, G%6-+A:7@    <@    %,,,$l}        'G6-+A:7@    <@    %,,,$' ;6-A:7@    <@    %,,;6-A:7@    <@    %,, $ A    6-$m}        '6-$' A%6-+A:7@"    <@"    %,,,$'A6-A:7@%    <@%    %,, 6-?:<<<<<<<<,$ 0~SH
n}        A@    )6-?:<A:7<,,,)6-?:<A:7<,,, .$6-?:A!	   <C:7,,<<B:7,,,.
Ay    0~SF
Ay    26-?o}        :,6-?:A!9   ,&6-?:<,26-@     	"6-8<, $4~ET6-
AP   @
A0   ^      p}        6-?:A!9   ,c.6-?:<@!    ,6.$A   .A0   h	4
A    |4*+)",
A    !4Q)4)q}        4)4!
A   A   
AP   6-?:A!W   ,$A   
A@   4	"6-?:<@#    ,+6-?:<@0    r}        ,46-?:,$	6-      $6-?:A!9   ,8<,!
A2   
A@   @$6.YOUR PREVIOUS ANSWER WAS '@67Bs}        :,%@    ,.>:8<,,467B:,%@    ,.'*6-?:<@     ,4A   ?6.7<,6-?:<@!    ,?6.SELECT THE COt}        RRECT ANSWER A   A0   " $4)4EEA
A    4*"
A   0O
A    		6-u}        	"6-	"6-?:<,A    )6-?:,68<,-8<,%)68<,-
@    2)3
A   
v}        	"6-		6-	"6-?:<,A    68<,-@:,0	0
A   : 68<,- 68<,-8<,%D26w}        -?:<@     ,26.CORRECT, THE ANSWER IS IA67B:,%@    ,.-67B:,%@    ,.'7A   A
AP   N 68<,- 6x}        8<,-8<,%X-6-?:<@     ,-6.THE CORRECT ANSWERY167B:,%@    ,. IS '167B:,%@    ,.Z-67B:,%@y}            ,.'#A   -
AP   b#+2Q)3,*0*0#
A   v		6-0Q
A    A    "
AP   z}        
A    	0
A   	6-6-?:,$6-&"6-?:,$
@    4A	    4~SB)4~SD.6-?:A {}        d   ,46-7<,0~Q
A`   %6-A:7<+B:,&,%,,%6-8<,	"
AV   
AX   +68<,-8<|}        ,& 8<, +68<,-6-&6-7<,0~R
A    	"6-?:,
@    	6-
A    	0}}        
A    '6-?:A d   ,6-"$6-'$A    56-?:,-@E    "--68<,-1	5	,-~}        @    -$68<,-(	,	0	6-6-6-6-!6-'6--6-0$ 4)4~ET6-*	"6-?:,}        $46-?:,
@    >Ad   AU   R"F:Ad   ,"AU   "
AP   \6.>:?:A!   ,,$;'?:A!   <<@$  }          <C:,,@    -6-86.~   ;$B:,"6.      B:,"6.      $B:,!7<,4~Q$6.~Q0}        B:, @    6.   $6.7<,4~ET6-$16-?:,6--%8<,"+6-16-	}        "$6-?:A p   ,-A0   0U*0#A   -
A	F   	4
A    @	6-6-6-6-!6-'6}        -66-?:A a   ,@A    /6-?:A d   ,6- -+8<,!/		!
A    :6-8<,A	    #6-?:}        <,:6-A:7<+B:,&,%,,+68<,-8<,& 8<, +68<,-%6-?:A d   ,@d    "4%$	6-6-}        	$6-?:A d   ,6-?:,!6-$$V06-B:,!      06-?:A!$   <C:,<<,W$W	6-6-6.6.}        !6.'6--6-36-?6-@    K6-@    Q6-W6-56-?:,6-?:A X   ,&6-?:<,56-?:<<<,	6.}        A   9A   6-?:<<<,%6.7,/A   9A   F6-?:<<<,6."A   ,A   76-?:<,F6-?:}        A X   ,56-?:<,-%6-?:<<<,+6.5A   '6-?:A!E   ,0'6-?:A!E   ,1	6-?:<@"    ,}        '6-?:A X   ,1A    6-?:A    ,A0   #2)3QA   #
A@   	4
A   4QA  }          
A@   (	3
A0   -[	46-@    6-!6-'6.36-@    96-E6-@    Q6-@    [
A    2}        3L
A   <'2L6.6.6.'
AP   F	6.6.6.
A`   	3
A   
A@   9	}        6.67B:,%@    ,.#6-)6-/6-9@0    6.7<,6.
A    K	6.67B:,%@    ,.167B:,%@ }           ,.76-=6-E2GK6-	36-@0    
A   1	6.67B:,%@    ,.167B:,%@    ,./	6}        -6-6-@0    %4/
A   A	@   4
Ae   A 0   	6.
A   A   
A@   }        l6-?:A y   ,$6-?:A    ,,6-?:<,6-?:,,6-@!    &+B:,',36-?:<,6-A(   $A   06-}              3$M6-?:<,6.] A   16-?:<@9    ,;A   J6-?:A!E   ,M$V6-?:<,6-?:<@#    ,.6-?:}        <<<,46.>A   G6-?:,V6-?:A!E   ,Z6-?:<@$    ,#6-?:<<<,)6.3A   <6-?:,K6-?:A!E   ,Z6}        -?:A!E   ,$G6-?:<,)6.NUMBER CORRECT    = =67B:,,.=:8<,,GA   M6-?:<@    ,/6.NUMBER IN}        CORRECT  = C67B:,,.=:8<,,MA   M6-?:<@    ,/6.NUMBER UNANSWERED = C67B:,,.=:8<,,MA   6}        -?:A R   ,$R"6.G"67B:,%@    ,.=:,\2	"(?:A!Q   <C:,<B:,,"@    2
A!`   fBA"    ,?:A!Q   <C}        :,<B:,,@    86-@	    B
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D:GEOM.BASTXDSSCSKTMCPPMERWRSRRTSWCPXY K        Welcome to the MATH MODULE of theHAYDEN SCORE IMPROVEMENT SYSTEM FORTHE SAT& one of three modules designedto help you rai}        se your SAT scores.This Geometry Section is an effectivetool to begin your preparation for theMathematical section of the }        ScholasticAptitude Tests.The system is easy tooperate so that you can concentrate onits content. All of the informationyo}        u need to answer questions appearson the screen& as do instructions formoving from one part of the program toanother. More}         detailed informationfollows in the User's Guide.~RAThis GEOMETRY SECTION providesinstruction and practice in solvingthe}         entire range of geometry problemsof the types found on the SAT. Allfigures needed to solve the problemsare illustrated on}         the screen.~RAThis section includes eleven parts inaddition to the User's Guide~`  Complementary and Supplementary`  A}        ngles  and Angle Sums`  Parallel Lines`  Angles of a Circle`  Equalities of Angles and Sides`  Pythagorean Theorem`  Spe}        cial Triangles`  Perimeters`  Areas`  Similar Figures`  Volumes`  All Examples Without Tutormode~RAOther areas in the}         Mathematicalsection of the SAT are Algebra& andQuantitative Comparisons and WordProblems; two separate sections withinth}        is Math Module provide reviewmaterial in these areas.This section contains a two-sided diskwhich has a Program Disk side a}        nd aReview Disk side.  Always load theProgram Disk side first.  Instructionson the screen will prompt you toinsert the ot}        her side at theappropriate time.~RAMENUSThe MAIN MENU lets you move easilyfrom one section of the program toanother. S}        imply press the keycorresponding to the letter next tothe section you wish to see.~RAThe first ten topics on the MAIN MEN}        Ucombine two key aspects of the HaydenSystem~ DEFINITIONS& ANALYSIS ANDSTRATEGIES and EXAMPLES WITHTUTORMODE. First& DEFI}        NITIONS providesthe background information you willneed to tackle the problem type on theSAT and demonstrates problem-solv}        ingmethods& including valuable tricks andshortcuts.~RASecond& }TUTORMODE} gives you adetailed& step-by-step explanation }        ofhow to arrive at the correct answer.By reviewing and practicing& youdevelop more efficient problem-solvingtechniques.~}        RAThe ALL EXAMPLES WITHOUT TUTORMODEoption provides quick drill andpractice in all the problem typeslisted on the MAIN ME}        NU so that youcan improve speed and accuracy. If youanswer incorrectly& you are shown thecorrect answer& but no detailede}        xplanation is provided. At the end ofthe section& the computer tallies thenumber of questions answered correctlyand incorr}        ectly& providing anindication of how well you havemastered the material.~RAWhen you make a selection from theMAIN MENU& }        the System startspresenting the material or asks forthe other side of the disk to beloaded into the drive.~RAFUNCTION KE}        YSA function key is a key which has aspecific effect on the program'soperation each time it is pressed.Whenever a menu }        is on your screen thefollowing function keys areoperational~`    M (Main Menu)`    Q (Quit)Pressing }M} always brings }        you back tothe MAIN MENU. Pressing }Q} causes thecomputer to ask if you really want toquit. If you answer }Y}& you end the}        program. If you answer }N}& youcontinue where you left off.~RAWhile text is on the screen pressing}R} restarts the secti}        on (erasing anyprevious answers that you may haveentered)& pressing }M} takes you tothe main menu and pressing }Q} enables}        you to quit.~RAThe left-arrow key lets you pagebackwards through the text one screenat a time until the first screen of}        the section is reached. When aquestion appears on the screen& yourprevious answer& if any& is shown. Youcan replace that a}        nswer by enteringanother one& or you can leave youranswer undisturbed by pressing theleft-arrow again.~RAPressing  the l}        etter }O} leaves thecurrent question temporarilyunanswered and displays the nextquestion. At the end of the sectionyou ha}        ve a chance to review all theunanswered questions.~ET~ETYSTEM FORTHE SAT& one of three modules designedto help you rai =        The HAYDEN SCORE IMPROVEMENT SYSTEMFOR THE SAT is organized into threemodules. It includes both simulatedSAT exams and com}        plete reviews of theareas typically covered by the Verbaland Mathematical sections of the SAT.In addition to this Math Mod}        ule, thefollowing modules are available:~RAThe PRACTICE TESTS MODULE contains anAnalysis of the SAT, a Pre-Test, andtwo }        Practice Tests.The ANALYSIS OF THE SAT gives youinsight into the workings of theactual exam -- its organization andscori}        ng& plus test-taking strategiesand tips for raising your scores.~RAThe PRE-TEST is adiagnostic/prescriptive tool fordete}        rmining your strengths andweaknesses in the areas typicallycovered by the SAT. It is a two-hourtest consisting of a mix of}         Math andVerbal questions similar to that on anactual SAT. After you complete thetest your computer will provide scoresin}         each of sixteen subjects whichcontribute to your Math and Verbalscores. This profile of yourperformance indicates which a}        dditionalmodules in the Hayden System will beuseful in your preparation.~RAThe PRACTICE TESTS are two-hoursimulated exam}        s with completeMathematical and Verbal sections timedand formatted to be representative ofthe latest SATs and scored on th}        e SATscale. After reviewing your weakareas& take these Practice Tests andsee how your performance would measureup on the }        actual exam.~RAThe VERBAL MODULE provides tutorials,drill and analysis in the Verbal areasnormally covered on the SAT.T}        he VOCABULARY SECTION provides athorough review of antonyms& analogiesand sentence completions& as well asan on-screen dic}        tionary with 1000words.~RAThe READING COMPREHENSION SECTIONoffers strategies and practice inresponding to questions abou}        t thematerial just read. Working withpassages drawn from the mostup-to-date sources in a variety offields will help you i}        mprove yourability to determine main ideas& torecognize logical implications and toextract factual information from whaty}        ou read.~RAEach topic in a given section can beapproached in three ways~`acquiring background with`DEFINITIONS& ANALYSI}        S AND STRATEGIES`gaining practice and instruction with`EXAMPLES WITH TUTORMODE`drilling with`EXAMPLES WITHOUT TUTORMODE}        *NOTE~  This Geometry Section combinesDefinitions& Analysis and Strategiesand Examples With Tutormode.~ET~ETs and com t        ~SB~SH1127A  D  y    B~SH0431C~SH0537E~SH0932k~SP189089210089210032210032~SP258040210089254089254089~SFB. Complement}        ary and Supplementary`      Angles and Angle Sums.Make sure you know thesetheorems& definitions andassumptions.1. Two}         angles whose sumis a right angle arecomplementary.`    <k + <y = 90*2. Two angles whose sum is180* are supplementary.}        `  <ADE + <y = 180*~RA~SB~SH0828A   a  b B~SH0536C~SH1230D~SH1030d  c~SP196064252064252064252064~SP2470402240642100}        88210088~SF3. The sum of the anglesabout a point is 360*.<a + <b + <c + <d = 360*4. The sum of the anglesabout a poin}        t on one sideof a straight line is180 degrees.`   <a + <b = 180*&`   <c + <d = 180*~RA~SB~SH0927A           C~SH033}        3B~SP189064224028259064189064~SF5. The sum of the anglesof a triangle is 180*.<A + <B + <C = 180*~RA~SB~SH0827A~SH0}        735B~SH0939C~SH1432D~SP189064244056266072217104~SP217104189064189064189064~SF6. The sum of the angles of aquadrilatera}        l is 360*.<A + <B + <C + <D = 360*~RA~SB~SP196024210016245032252056~SP252056217064196048196024~SH0328A~SH0231B~SH043}        6C~SH0737D~SH0932E~SH0628F~SF7. The sum of the anglesof a polygon of n sidesis 180 (n - 2) degrees.Note~ In the diag}        ram(hexagon ABCDEF)& thesum of the angles is~180(6 - 2) degrees = 720*~RA~SB~SH0829A   E   B~SH0431C~SH0636D~SP2030}        55252055252055252055~SP217032224055245048245048~SF~Q1. Given~ CE[ED;<AEC = 70*. Find~ <BED(a) 20*  (b) 70*  (c) 90*(}        d) 110*  (e) 180*~RCA1. (a) 20* Ans.The sum of the anglesabout a point on one sideof a straight line is 180*therefore~}        `<BED + <DEC + <AEC = 180*Since   CE[ED& <DEC =  90*`<BED +  90* +  70* = 180*`              <BED =  20* Ans.~RA~Q}        ~SB~SH0828A         B~SH0338C~SH1336D~SH0930E~SP198060257060257060257060~SP254016210060245096245096~SF2. Given~ <CED }        is a right angle. <CEBis 50* Find~ <AED(a) 40*  (b) 90*  (c) 130*(d) 140*  (e) It cannotbe determined from theinformat}        ion given.~RCD2. (d) 140* Ans.` <CEB + <BED =  90*`  50* + <BED =  90*`        <BED =  40*Since the sum of theangl}        es about a pointon one side of astraight line is 180*&` <BED + <AED = 180*`  40* + <AED = 180*`        <AED = 140* An}        s.~RA~Q3. <CED is a right angle& and <AEC =<AED. Find the numberof degrees in <CEB.(a) 45  (b) 90  (c) 135(d) 180  (}        e) 20~RCA3. (a) 45 Ans.The sum of the anglesaround a point is 360*.Therefore~<CED + <AEC + <AED = 360*`90* + <AEC +}         <AED = 360*`      <AEC + <AED = 270*`                    270*`      <AEC = <AED = ----`                     2`      }                    = 135*~RA`       <AEC + <CEB = 180*`       135* + <CEB = 180*`              <CEB =  45* Ans.~RA~SD~Q4. }        In {ABC& <A~<B~<C = 1~2~3. Find thenumber of degrees in <B.(a) 15  (b) 30  (c) 45  (d) 60  (e) 90~RCD4. (d) 60 Ans.`  }        <A + <B + <C = 180*`   y + 2y + 3y = 180`            6y = 180`             y = 30`       <B = 2y = 60*  Ans.~RA~Q5}        . 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~RCB5. (b) 20 }        Ans.`      <A + <B + <C = 180*`       6y + 2y + y = 180`                9y = 180`                 y = 20`          }          <C = y = 20*  Ans.~RA~Q~SB~SP196055266055217032210055~SP266055279055279055279055~SH0828D  A      B E~SH0432C~SF6. }        If <DAC = 120* and <EBC = 150*& then{ABC is~(a) equiangular(b) equilateral(c) isosceles(d) right(e) It cannot be`   d}        etermined from the`   information given~RCD6. (d) right Ans.Since the sum of anglesabout a point on one sideof a strai}        ght line is 180*`   <DAC + <CAB = 180*`   120* + <CAB = 180*`          <CAB =  60*and`   <EBC + <ABC = 180*`   150}        * + <ABC = 180*`          <ABC =  30*~RAThe sum of the angles ofa triangle is 180*.Therefore~<CAB + <ABC + <ACB = 180}        *`60* +  30* + <ACB = 180*`             <ACB =  90*&making {ABC a righttriangle. Ans.~RA~Q~SB~SH0330A~SH1127D  C }               B~SH1238E~SP203079203024273096273096~SP182079256079256079256079~SF7. If <DCA = 2<A& and<CBE = 3<A& then thenumb}        er of degrees in<A is~(a) 15  (b) 30  (c) 45(d) 74  (e) 75~RCC7. (c) 45* Ans.The sum of the anglesof a triangle is }        180*`<A + <B + <C = 180*Since the sum of theangles about a pointon one side of astraight line is 180*~<C = (180* - 2}        A)<B = (180* - 3A)<A + (180* - 3A) + (180* - 2A) = 180*<A = 45* Ans.~RA~Q~SB~SP189064244056266072217104~SP21710418}        9064189064189064~SH0827A~SH0735B~SH0939C~SH1432D~SF8. In quadrilateral ABCD& <A = y*& <B =2y*& <C = y - 30*& and <D = }        y + 30*Find the number of degrees in <A.(a) 36  (b) 43  (c) 72(d) 102  (e) 144~RCC8. (c) 72 Ans.<A + <B + <C + <D = }        360*y + 2y + y - 30 + y + 30`                 = 360`              5y = 360`               y = 72*`     <A = y = 72*}          Ans.~RA~SD~Q9. Each angle of an equiangularpentagon (a polygon of 5 sides) is~(a) 72*  (b) 90*  (c) 108*  (d) 180*}        (e) 540*~RCC9. (c) 108* Ans.Note~ The sum of the angles of apolygon = (N - 2)180`         (5 - 2)180`           = (3}        )180`           = 540* in all`     540/5 = 108* each  Ans.~RA~Q~SB~SH0528A        B~SH0928C        D~SH0739E~SP196}        036245036266052245068~SP245068196068196068196068~SF10. Given~ AB ^ CD; <B = 140*;<D = 150*. Find~ <E.(a) 10*  (b) 30*  }        (c) 40*(d) 50*  (e) 70*~RCE10. (e) 70* Ans.NOTE~ Draw a lineperpindicular to lines ABand CD.The sum of the angles i}        npolygon ABEDC =(5 - 2) x 180* = 540*Both <A and <C = 90*90* + 140* + <E + 150* + 90* = 540*`                       }         <E = 70* Ans.~ET~ETB~SH0431C~SH0537E~SH0932k~SP189089210089210032210032~SP258040210089254089254089~SFB. Complement         ~SB~SC226039210056233056233056~SH0534C~SH0735B~SH0830O~SFD. Angles of a Circle.Make sure you know these theorems&de }        finitions and assumptions.1. A central angle ismeasured by itsintercepted arc. Statedotherwise& a centralangle is equal! }         in degreesto its intercepted arc.<COB = degrees in BC~RA~SB~SC210032186056226038226038~SH0726A~SH0534B~SH0430C~SF!}        2. An inscribed angle is equal indegrees to one-half its interceptedarc.<CAB = 1/2  degrees in BC3. In a circle& adiam!}        eter is twice theradius and divides thecircle into twosemicircles.~RA~SB~SC186056235056225038186056~SH0726A~SH0735B~!}        SH0534C~SF4. An angle inscribed in a semicircleis a right angle.<ACB = 90*~RA~SB~SC210056192071226071226071~SH0731O!}        ~SH1027A~SH1034B~SFExample~Given~ Circle 0; AB = 60*&OA = 6} Find~ Length of AB.1. Draw radius OB.2. Since AB = 60*&!}        `  <AOB = 60*.3. Since OA = OB&`  <A = <B.4. But <A + <B`  = 180* - <O`  = 180* -60* = 120*5. Then& <A = <B = 60* = <O!}        6. Since { AOB is equiangular& it isequilateral; OA = OB = AB = 6} Ans.~RA~Q~SB~SC186055236055210055214033~SH0726A~SH!}        0735B~SH0431C~SH0831O~SH0730y~SF1. Given~ Circle O&diameter AB& radius OC&BC = 80*.Find~  Number ofdegrees in <y.!}        (a) 40  (b) 80  (c) 100(d) 140  (e) 0~RCC1. (c) 100 Ans.The central angle of acircle is equal to itsintercepted arc.!	}        Therefore~`  <COB = 80*Since y + <COB are onthe same side of astraight line about acentral point& theytotal 180*`y !
}        + 80* = 180*`      y = 100* Ans.~RA~Q~SB~SC193038226038214078214033~SP214033192038193038193038~SH0527A~SH1131B~SH04!}        31C~SH0534D~SF2. Given~ Circle O& <A = 30*. Find~Number of degrees in <B.(a) 15  (b) 30  (c) 60(d) 90  (e) 10~RCB2.!}         (b) 30 Ans.Angle A and Angle Bintercept the same arcCD and are thereforeequal.`   <B = <A = 30* Ans.~RA~Q~SB~SC18!}        6056236056210056206033~SP186056206033206033206033~SH0726A~SH0735B~SH0430C~SH0831O~SF3. Given~ Circle O& diameter AB&r!}        adius OC& <COB = 100*. Find~ Numberof degrees in <A.(a) 50  (b) 90  (c) 100(d) 200  (e) 150~RCA3. (a) 50 Ans.An insc!}        ribed angle isequal in degrees to onehalf its interceptedarc. Since <A interceptsthe same arc as <COB&`<A = 1/2 <COB`!}           = 1/2 (100*)`   = 50* Ans.~ET~ET~SH0735B~SH0830O~SFD. Angles of a Circle.Make sure you know these theorems&de  )        F. The Pythagorean TheoremMake sure you know these theorems&definitions and assumptions.~RA~SB~SP205024251070205070205%}        024~SH1029B   b  A~SH0329C~SH0729a~SH0534c~SFThe Pythagorean Theoremstates~In any right triangle&the square of the%}        hypotenuse equals thesum of the squares ofthe other two sides.In symbols this theory is expressed~`    2    2    2` %}          c  = a  + bwhere c is always the hypotenuse.~RAExample~The perimeter of righttriangle ABC is 3 ft. Thesides forming%}         the rightangle are in a 3~4 ratio.Find the length of each ofthe three sides.Let p = number of feet inperimeter of tria%}        ngle ABC.Let c = number of feet inhypotenuse AB.`    Let 3y = side a.`    Let 4y = side b.~RAIn right triangle ABC~%}        `2    2    2c  = a  + b`2       2      2c  = (3y) + (4y)`2     2      2c  = 9y  + 16y`2      2c  = 25yc  = 5yp %}         = a + b + c3  = 3y + 4y + 5y~RA~SB~SP205024251070205070205024~SH1029B   b  A~SH0329C~SH0729a~SH0534c~SF`   3  = 1%}        2y`   y  = 1/4`  Therefore~side a = 3y = 3/4ft.side b = 4y`      = 4/4`      = 1ft.side c = 5y`      = 5/4`%}              = 1 1/4ft.~RA~SDNote the following multiplecombinations of the sides of a fewtypical right triangles(integersonly%}        ).-----------------------------------`          LEG   LEG   HYPOTENUSE-----------------------------------`           3  %}           4        5`   A       6     8       10`           9    12       15-----------------------------------`           5    %}        12       13`   B      10    24       26-----------------------------------`           8    15       17`   C      16    30%}               34-----------------------------------~RA~Q1. A ladder 20 ft. long is placedagainst a building so that it justrea%}        ches a window 16 ft. high. How farfrom the wall is the foot of theladder?(a) 4ft. (b) 12 ft. (c) 18 ft.`      ___(d) 2\% }        /139 ft.(e) It cannot be determined from theinformation given.~RCB1. (b) 12 ft. Ans.`     2    2    2`    a  + b  = c%!}        `     2    2     2`   16  + b  = 20`          2`         b  = 400 - 256`          2`         b  = 144`          b %"}        = 12 ft. Ans.~RA~Q2. How much distance is saved inwalking diagonally across arectangular vacant lot 30 ft. by 40ft& ins%#}        tead of walking around itslength and width?(a) 10 ft. (b) 20 ft. (c) 40 ft.(d) 70 ft. (e) 0 ft.~RCB2. (b) 20 ft. Ans.%$}        `       40 + 30 = 70  walking around`        2    2    2`       a  + b  = c`       2     2    2`     30  + 40  = c` %%}                         2`    900 + 1600 = c`                  2`          2500 = c`            50 = c   walking across70 ft%&}        . - 50 ft. = 20 ft. saved Ans.~RA~Q3. Find the altitude of an isoscelestriangle if its base is 10 inches longand each of%'}         its equal sides is 13inches long.(a) 8 in. (b) 12 in. (c) 24 in.`     ___(d) \/269 in. (e) 0 in.~RCB3. (b) 12 in. Ans%(}        .`      2    2    2`     a  + b  = c`      2    2     2`     5  + h  = 13`           2`          h  = 169 -25`    %)}               2`          h  = 144`           h = 12 in.  Ans.~RA~Q4. The hypotenuse of a right triangleis 25 in. The other t%*}        wo sides are inthe ratio of 3~4. Find the length ofeach side in inches.(a) 3&4  (b) 6&8  (c) 9&12  (d) 12&16(e) 15&20~%+}        RCE4. (e) 15&20 Ans.`        2    2`       a  + b  = c`     2       2       2` (3y)  + (4y)  = (25)`      2      2`%,}            9y  + 16y  = 625`             2`          25y  = 625`             2`            y  = 25`             y = 5There%-}        fore~ 3y = 15 in.   4y = 20 in.~RA~Q5. A certain right triangle has aperimeter of 4 ft. The sides formingthe right angle%.}         are in a 3~4 ratio.Find the length of each of the threesides in inches.(a) 3&4&5  (b) 1&4/3&5/3(c) 36&48&60  (d) 12&16%/}        &20(e) 15&20&13~RCD5. (d) 12&16&20 Ans.`    3y + 4y + c = 4 ft.`      2       2    2`  (3y)  + (4y)  = c`       2 %0}             2    2`     9y  + 16y  = c`              2    2`           25y  = c`             5y = c~RA`          4 ft. = 48 %1}        in.`   3y + 4y + 5y = 48`            12y = 48`              y = 4Therefore~    3y = 12 in.`             4y = 16 in.%2}        `             5y = 20 in. Ans.~ET~ET know these theorems&definitions and assumptions.~RA~SB~SP205024251070205070205$ *        C. Parallel Lines`    (AB^CD)Make sure you know these theorems&definitions and assumptions.~RA~SB~SH11311   3~SH0933)4}        2~SH13334~SP196052259116259116259116~SP259052196116196116196116~SF1. If two straight lines intersect&the pairs of oppos)5}        ite anglesVertical Angles~`    angle <1 = <3`          <2 = <4Supplementary Angles~`     <1 + <2 = 180*`     <2 + <)6}        3 = 180*`     <3 + <4 = 180*`     <4 + <1 = 180*~RA~SB~SH0927l~SH1327m~SH08361  2~SH10334   3~SH12315  6~SH14298  7)7}        ~SP193068266068266068266068~SP193100266100266100266100~SP265050196120196120196120~SF2. If two straight lines are paralle)8}        l(never meeting) and are cut by atransversal (a line that touches eachof the parallel lines)~a. the pairs of alternatei)9}        nterior angles are equal;b. the pairs ofcorresponding anglesare equal;c. the pairs of interiorangles on the same side):}        of the transversal aresupplementary.~RAIf line l is parallel to line m~a. Alternate interior angles`    <3 = <5`    <);}        4 = <6b. Corresponding angles`    <1 = <5`    <4 = <8`    <2 = <6`    <3 = <7c. Interior angles on`  the same side )<}        of`  the transversal`  <4 + <5 = 180*`  <3 + <6 = 180*~RA~SB~SH0927l~SH1327m~SH08361  2~SH10334   3~SH12315  6~SH)=}        14298  8~SP193068266068266068266068~SP193100266100266100266100~SP265050196120196120196120~SFNotice that there are many p)>}        airs ofvertical angles and supplementaryangles.Vertical Angles`  <1 = <3`  <2 = <4`  <5 = <7`  <6 = <8Supplementar)?}        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)@}        ~SB~SP198028264028264028264028~SP198070264070264070264070~SP244012217082217082217082~SH0428A~SH0439B~SH0928C~SH0939D~)A}        SH0236E~SH1133F~SH0533a~SH0834b~SH0336c~SH0535d~SH1031e~SF1. If <d = 150* andAB^CD& find the numberof degrees in <e)B}        .(a) 30  (b) 60  (c) 150(d) 180 (e) It cannot bedetermined from theinformation given.~RCA1. (a) 30 Ans.Because <a a)C}        nd <dare supplementary<a = 180* - <d<a =  30*<a and <e arecorresponding anglesand therefore equal<e = <a = 30* Ans)D}        .~RA~Q2. Given~ <a = 3y + 5*&<b = 2y + 11*. Find thenumber of degrees in <a.(a) 6  (b) 16  (c) 23(d) 43  (e) 0~RCC)E}        2. (c) 23 Ans.Since <a and <b arealternate interiorangles&`    <a = <b`3y + 5 = 2y + 11`     y = 6*`     a = 3 (6)F}        *) + 5*`       = 23* Ans.~RA~Q~SB~SP198028264028264028264028~SP198070264070264070264070~SP244012217082217082217082~S)G}        H0428A~SH0439B~SH0928C~SH0939D~SH0236E~SH1133F~SH0533a~SH0834b~SH0336c~SH0535d~SH1031e~SF3. Angle b = 4m* andang)H}        le d = 5m*. Find thevalue of m.(a) 10  (b) 20  (c) 40(d) 50  (e) 80~RCB3. (b) 20 Ans.Angle b and <d areinterior ang)I}        les onthe same side of thetransversal andtherefore total 180*` 4m* + 5m* = 180*`        m  =  20* Ans.~RA~Q~SB~SH0)J}        528A           B~SH0828C           D~SH0434y~SH0932140*~SP198036275036275036275036~SP198060275060275060275060~SP2520242)K}        03072203072203072~SF4. AB ^ CD. Find the number of degreesin angle y.(a) 40  (b) 50  (c) 90(d) 140  (e) 180~RCD4. (d)L}        ) 140 Ans.y is a correspondingangle to the verticalangle equal to 140*and is therefore alsoequal to 140* Ans.~RA~Q~S)M}        B~SH0427A      x    B~SH0627C           D~SH0827E    G~SH0234F    H~SH0734y~SP189032266032266032266032~SP1890472660472)N}        66047266047~SP231016189056189056189056~SP266016224056224056224056~SF5. AB ^ CD& EF ^ GH& and <x = 30*.Find the number of)O}        degrees in angle y.(a) 30  (b) 60  (c) 90(d) 150  (e) 180~RCD5. (d) 150 Ans.y is a correspondingangle to thesupple)P}        mentary angleequal to 30* therefore~`  y + 30* = 180*`        y = 150* Ans.~ET~ETtions.~RA~SB~SH11311   3~SH0933( Z        ~SB~SH0432a b~SP208008245048245048245048~SP245008208048208048208048~SFE. Equalities of Angles and Sides.Make sure you-R}         know thesetheorems& definitions andassumptions.1. Vertical angles are equal.`    <a = <b~RA~SB~SH1828A        B~SH-S}        0933C~SP196136252136224072193136~SF2. An isosceles triangle has two sidesequal in length. The angle between theequal sid-T}        es is called the vertexangle. The other two angles are calledbase angles and are equal.If AC = BC& then <A = <B.3. If t-U}        wo angles of atriangle are equal& thesides opposite are equal.If <A = <B& then AC = BC.~RA~SB~SH1531A       B~SH0735C-V}        ~SH14311~SP196112266112243062217112~SF4. An equilateral triangle (ABC) isequiangular& and an equiangulartriangle is equ-W}        ilateral.If AB = BC = AC&then <A = <B = <C.5. An exterior angleof a triangle is thesum of the non-adjacentinterior an-X}        gles.<1 = <B + <C.~RA~Q~SB~SH0627A      B    D~SH0332C~SH0739E~SP266039189039220024266048~SF1. Given~ Isosceles {A-Y}        BCwith AC = BC& and <DBE =20*. Find <C.(a) 20*  (b) 70*  (c) 140*(d) 160*  (e) 150*~RCC1. (c) 140* Ans.Since <DBE a-Z}        nd <B of {ABCare vertical angles& theyare equal.`  <B = <DBE = 20*Since AC = BC the anglesopposite them are equal.` -[}                 <A = <B`<A + <B + <C = 180*`20*+ 20*+ <C = 180*`          <C = 140* Ans.~RA~Q~SB~SH1328A      B~SH0532C~-\}        SH0230D~SP217046196096238096206016~SF2. Given~ Isosceles {ABCwith <ACD = 130*. Find~ <B.(a) 50*  (b)  60*  (c) 65*(d)-]}         130*  (e) 75*~RCC2. (c) 65* Ans.<ACD and <ACB aresupplementary.`<ACD + <ACB = 180*`130* + <ACB = 180*`       <ACB-^}         =  50*~RAThe angles of triangletotal 180*`  A + <B + <C = 180*`<A + <B + 50* = 180*`      <A + <B = 130*In the is-_}        oscelestriangles& <A = <B`<B + <B = 130*`     <B =  65* Ans.~RA~Q~SB~SH0435C~SH0928 y~SH1028D   A     B~SH0729E~-`}        SH1236F~SP189071259071241038224071~SP196056252088252088252088~SF3. Given~ {ABC is equilateral and <CAFis a right angle. -a}        Find~ <y.(a) 30*  (b) 60*  (c) 90*(d) 150*  (e) 45*~RCA3. (a) 30* Ans.Since ABC is equilateral&<CAB = 60*<BAF = <-b}        CAF - <CAB<BAF = 90* - 60*`    = 30*<y is a vertical angle to <BAF` <y = <BAF = 30* Ans.~RA~Q~SB~SH1032A     B~S-c}        H0531D   C   E~SP224072259072241038224072~SP215039266039266039266039~SF4. Given~ {ABC with AB = 10in.; <B =60*; DCE ^ AB-d}        ; <DCA = 60*. Find~Length of BC.(a) 4}  (b) 10}  (c) 20}(d) 50} (e) 8}~RCB4. (b) 10} Ans.<A = <DCA because theyare -e}        alternate interiorangles`            <A =  60*`  <A + <B + <C = 180*` 60* + 60*+ <C = 180*`            <C =  60*Th-f}        erefore {ABC is equilateral and BC =AB = 10}. Ans.~RA~Q~SB~SP224071259071241038224071~SP279032259071279071279071~SH103-g}        2A     B D~SH0435C    E~SF5. BE^AC& and BEbisects <CBD. Then&(a) AB = AC  (b) AB = BC(c) AC = BC  (d) AC = BE(e) AB-q}        c                                                                                                                        b7  PRNTS   SYSB > G0         B ? AUTORUN BAKB B AUTORUN SYSBG U GEOM    BASB!  AA         B  AB         B.  BBB                B  DBB        B" FBB        B 3CBB        B QEBB        B' uDOS     SYSB G1         B4 GBB        B JBB                B KBB        B) 
LCC        B 3HBB        B" NIBB                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                         = AC = BC~RCB5. (b) AB = BC Ans.Since <ACB and <CBE arealternate interior angles&they are equal. The sum ofthe angles-r}         around a pointon one side of a straightline is 180*.<CBA + <CBE + <EBD = 180*`<CBE = <EBD`<CBA = 180* - 2 <CBE`<CB-s}        A = 180* - 2 <ACB~RAThe sum of the angles ofa triangle is 180*<CAB + <ACB + <CBA = 180*<CAB = 180* - <ACB - <CBA<CAB-t}         = 180* - <ACB`      - (180* - 2<ACB)<CAB = <ACBSince sides opposite equal angles areequal& AB = BC Ans.~ET~ETe you, x        
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Special TrianglesMake sure you know these theorems&definitions and assumptions.~RA~SB~SP266095224095224031266095~S9}        H1335s~SH0736h~SF`A. The 30*& 60*& 90* Triangle.1. The side(s) oppositethe 30* angle equalsone-half the hypotenuse (h)9}        .`    s = h/22. The side(s) oppositethe 60* angle equalsone-half the hypotenuse (h)`       _times \/3.`       h  9}          _`   s = -  \/3`       2~RA~SB~SH1433s~SH0729s      s~SH0934a~SP182095266095224031182095~SP224031224095224095224099}        5~SF3. The altitude (a) ofan equilateral triangleequals one-half the`               _side(s) times \/3.`       s    _9}        `   a = -  \/3`       2~RA~SB~SH0733h    s~SH1134s~SP210079258079258031210079~SF`B. The 45*& 45*& 90* Triangle1. T9}        he side(s) opposite the45* angle equals one-half`                          _the hypotenuse (h) times \/2.`       h    _9}        `   s = -  \/2`       22. The hypotenuse (h)       _equals the side (s) times \/2.`           _`    h = s\/2~RA~SB~S9}        H1134s~SH0730s  d    s~SH0434s~SP210079258079258031210079~SP258031210031210079210079~SF3. The side of a square(s)equal9}        s half the diagonal (d)`       _times \/2.`        d   _`    s = - \/2`        24. The diagonal (d) of asquare equa9}        ls the side(s)`       _times \/2.`            _`    d = s \/2~RA~SD~Q1. Find the altitude of an equilateraltriangle 9}        whose side is 10.`             _         _(a) 5  (b) 5\/2  (c) 5\/3`       _          _(d) 10\/2  (e) 10\/3~RCC`       9}          _1. (c) 5\/3 Ans.Note~ Draw an altitude (a) to create a30*& 60*& 90* triangle.`          1    _`      a = -10\/3`  9}                2`             _`      a = 5\/3  Ans.~RA~Q~SB~SH0729s      s~SH09336~SH1326A      s     B~SH0333C~SP18209529}        66095224024182095~SP224024224096224095224095~SF2. Find the side s ofequilateral triangle ABCwhose altitude is 6.`      9}               _         _(a) 3  (b) 2\/3  (c) 3\/3`      _         _(d) 4\/3  (e) 6\/3~RCD`         _2. (d) 4\/3 Ans.Altitud9}        e of an equilateral`               S  _triangle equals -\/3`               2`  S  _`  -\/3 = 6`  2`     _` S \/3 = 19}        2`             _`    3S = 12\/3`            _`     S = 4\/3 Ans.~RA~Q~SB~SH0330A~SH1329B~SH1029D     C~SP2030242039}        096238078203024~SP203078238078238078238020~SF3. In {ABC& AC = 8} and <A = 30*. Findthe length of altitude CD.`          9}            _          _(a) 4}  (b) 4\/2}  (c) 4\/3}`      _          _(d) 8\/2}  (e) 8\/3}~RCA3. (a) 4} Ans.The side opposite9}         the 30*angle of a 30*& 60*& 90*triangle equals h/2Altitude CD = h/2`             8}`           = --`             29}        `           = 4}~RA~Q~SB~SH0733D~SH1130A       C~SH0438B~SP210080259080249031210080~SP234055259080259080259080~SF4.9}         Find the length ofaltitude CD to the hypotenuseof isosceles right triangleABC each of whose equalsides is 6 inches long.9}        `              _(a) 6}  (b) 6\/2}  (c) 3}`      _          _(d) 3\/2}  (e) 3\/3}~RCD`         _4. (d) 3\/2} Ans.The 9}        original hypotenuse`     _                 _is 6\/2& each part = 3\/2.~RA~SD~Q5. Find the diagonal of a square whosesi9}        de is 10.`             _         _(a) 5  (b) 5\/2  (c) 5\/3`       _          _(d) 10\/2  (e) 10\/3~RCD`          _5. 9}        (d) 10\/2 Ans.Note~ The diagonal creates a 45*& 45*&90* triangle. According to the`                            _formula 9}        the hypotenuse is s\/2. Since`                         _s = 10 the diagonal = 10\/2.  Ans.~RA~Q~SB~SH1130A       B~SH09}        430D       C~SP210080259080259031210080~SP259031210031210080210080~SF6. Find to the nearest footthe distance saved bywa9}        lking along diagonal AC ofsquare lot ABCD& instead ofalong its sides AB and BC&each of which is 100 feet`           _lon9}        g (use \/2 = 1.414.)(a) 58 (b) 58.6  (c) 59(d) 117  (e) 117.2~RCC6. (c) 59 Ans.Note~ The diagonalcreates a 45*& 45*&9}        90* triangle.According to theformula the hypotenuse`     _is s\/2. Therefore`          _AC = 100 \/2= 141.4AB + BC9}         = 200 along sides200 - 141.4 = 58.6To nearest foot = 59 ft. Ans.~RA~Q~SB~SP252036210036189079273079~SP27307925203629}        52079252079~SH1125  A      F  E  D~SH0430B       C~SH0836y~SF7. Given~ BC^AD& BC = 4}& AD = 8}& <A =<D = 60*& CE[AD. Fi9}        nd y in inches.Note~ Draw CF^AB&thus completingparallelogram ABCF.`             _(a) 2  (b) 2\/2`      _(c) 2\/3  (d)9}         4`      _(e) 4\/3~RCC`         _7. (c) 2\/3  Ans.Drawing CF^AB creates{CFD which isequilateral since <D= 60*. In p9}        arallelogramABCF&`    AF = BC = 4`    FD = AD - AF`       = 8} - 4}`       = 4}~RAIn right triangle CED y is oppos9}        ite the60* angle`         h  _`     y = -\/3`         2The hypotenuseCD = FD = 4}`         4  _`     y = -\/3`   9}              2`            _`     y = 2\/3  Ans.~RA~Q~SB~SH0733D~SH1130A       C~SH0438B~SP210080259080259031210080~SP23409}        55259080259080259080~SF8. Given~ Right isosceles triangle ABC&with <C = 90* andAC = BC; CD bisects <C;AB = 20}. Find the9}         lengthof CD.`                _(a) 10}  (b) 10\/2}`       _(c) 10\/3}  (d) 5}`      _(e) 5\/2}~RCA8. (a) 10} Ans.S9}        ince AC = BC& <A = <B`    <A + <B + <C = 180*`   <A + <B + 90* = 180*`         <A + <B = 90*`            2 <A = 90*9}        `              <A = 45*`              <B = 45*~RASince CD bisets <C& <ACD= <BCD = 45* DC = AD andDC = BD because they a9}        resides opposite equal angles.Therefore&`        AD = BD`   AD + BD = AB`   AD + BD = 20}`      2 AD = 20}`      9}          AD = 10} Ans.~RA~Q~SB~SH1117A   E               B~SH1917C                   D~SH1723F~SP112088252088119144259144~SP9}        119144140088154128154128~SF9. Given~ AB^CD& EF[BC& <AEC = 75*&<BCD = 30*& EB = 2}. Find~ Length ofEC.(a) 1}  (b)  2}  (9}        c)  3}(d) 4}  (e)  6}~RCB9. (b) 2} Ans.Since the angles of a triangle total180*<FEB + <EBF + <BFE = 180*<FEB +  30*9}         +  90* = 180*`             <FEB =  60*{FEB is therefore a 30*& 60*& 90*{~RA`               h`          EF = -`       9}                2`               2}`          EF = --`               2`             = 1}~RAThe sum of the angles about a poin9}        t onthe same side of a straight line is180*`<AEC + <CEF + <FEB = 180*`  75* + <CEF + 60* = 180*`              <CEF = 9}         45*~RATherefore& {CEF is an isosceles righttriangle in which the sides are 1}. Thehypotenuse  EC = 2s`                9}          = 2 (1})`                  = 2} Ans.~ET~ETngle in which the sides are 1}. Thehypotenuse  EC = 2s`                8 3        ~SB~SP182063224063196016182063~SP196016196063196063196063~SP238063259063245040238063~SP245040245063245063245063~SH0927A=}        'D' B'A D B~SH0229C'~SH0629h'~SH0536C~SH0836h~SFJ.Similar Figures.Make sure you know thesetheorems& definitions and=}        assumptions.1. Corresponding lines&sides& and perimeters (p)of similar polygons arein proportion.`      h    s    pth=}        en&  -- = -- = --`      h'   s'   p'~RA~SD2. Areas (A) of similar polygons are toeach other as the squares ofcorrespond=}        ing lines& sides& andperimeters.`   {ABC is similar to {A'B'C'`   then~`        2      2     2`   A   h      s     p=}        `   - = --- = --- = ---`         2     2     2`   A'  h'    s'    p'~RA3. In any two circles~`    C    r    d`   -- = =}        -- = --`   C'   r'   d'`         2     2     2`   A    r     d     C`   - = --- = --- = ---`         2     2     2`   =}        A'  r'    d'    C'~RAExample~Given~ Circle O is 4 times as large ascircle O'. Radius OP = 10}.Find~ Length of radius O'=}        P'.Let O' P' = ya) In circles O and O'&`              2`       A    OP`       - = -----`               2`       A' =}         O'P'~RA`              2`        4   10b) then~ - = ---`              2`        1    y`        2`      4y  = 100` =}               2`       y  = 25`       y  = 5} Ans.~RA~Q1. The sides of two similar polygonsare in the ratio 4~1. Theircorres=}        ponding perimeters are in theratio~(a) 4~1  (b) 8~1  (c) 12~1  (d) 16~1(e) 2~1~RCA1. (a) 4~1 Ans.`  p     s`  -- = =}        --`   1    1`  p    s`   p   4`  -- = - Ans.`   1`  p    1~RA~Q2. The areas of two similar trianglesare in the ra=}        tio 4~9. Theircorresponding sides are in the ratio~(a) 16~81  (b) 4~9  (c) 2~9/2  (d) 2~3(e) 3~2~RCD2. (d) 2~3 Ans.`=}           2`  s     A` ---- = --`    2`   1     1` (s )   A`   2`  s     4` ---- = -`    2`   1` (s )   9~RA`        =}         _`   s   \/4`  -- = ---`   1     _`  s    \/9`   s   2`  -- = -  Ans.`   1`  s    3~RA~Q3. The circumferences o=}        f two circlesare 4#} and 9#}& respectively. Theirareas are in the ratio~(a) 4~9  (b) 2~3  (c) 8~18  (d) 16~81(e) 16#~81=}        ~RCD3. (d) 16~81 Ans.`            2`           c    a`       -----  = --`           2`       (c')     a'`          =}          2`       (4#r)    a`       ------ = --`            2`       (9#r)    a'`          2 2`       16# r    a`       ----=}        -- = --  Ans.`          2 2`       81# r    a'~RA~Q~SB~SP217016217095252095217016~SP238064238095238095238095~SH0831y=}        ~SH13333 2~SH10344~SF4. Find y in the diagram~(a) 10  (b) 6  (c) 2.5(d) 1.5  (e) 5~RCA4. (a) 10 Ans.`  y     4`--=}        --- = -`(3+2)   2`    y`    - = 2`    5`    y = 10 Ans.~ET~ET(d) 1.5  (e) 5~RCA4. (a) 10 Ans.`  y     4`--< I        K. Volumes.Make sure you know these theorems&definitions and assumptions.Volume is the measure in cubic units ofthe spA}        ace contained in or occupied byan object.1. Rectangular Solid.`      Volume = Bh = lwh2. Cube.`                3`   A}           Volume = e~RA3. Right Circular Cylinder.`                      2`      Volume = Bh = #r h~RA4. Sphere`            A}                3        3`      Volume = 4/3#r  = 1/6#d~RAExample~A cylindrical bar of metal having acircular base of radius 5A}         inches and aheight of 3 inches weighs 1000 grams.What would a cylindrical bar of thesame metal weigh if its circular baseA}        has a radius of 3 inches and its heightis 5 inches?~RA`            2                  2`      V = #r h          V' = #r A}        h`           2                  2`      V = 5  x 3#       V' = 3  x 5#`      V = 75#           V' = 45#`          DireA}        ct proportion`            V    W`            -- = --`            V'   W'`           75#   1000`           --- = ----`A}                   45#     y`           75y = 45000`             y = 600 grams  Ans.~RA~Q1. A tank 8' x 6' x 5' is filled with A}        aliquid weighing 120 lbs. What would bethe weight of a quantity of this liquidwhich fills a tank 4' x 3' x 1'?(a) 20 lbsA}        . (b) 12 lbs. (c) 10 lbs.(d) 6 lbs.  (e) 4 lbs.~RCD1. (d) 6 lbs. Ans.`    V     W`    -- = --`     1    1`    V    WA}        ` 8x6x5   120 lbs.` ----- = --------`             1` 4x3x1      W`   240   120`   --- = ---`           1`    12    A}        W`     1`    W  = 6 Ans.~RA~Q2. A cylinder with base diameter of 10}and height of 6} weighs 15 lbs. Whatwould a cylinA}        der of base diameter 6}and height of 10} weigh if it is madeof the same material?(a) 90# lbs. (b) 360# lbs.  (c) 25 lbs.A}        (d) 9 lbs.  (e) 3# lbs.~RCD2. (d) 9 lbs. Ans.`           V    W`          -- = --`           1    1`          V    WA}        `     2` #(5)   (6)    15` -----  ---- = --`     2          1` #(3)   (10)   W`        150#   15 lbs.`        ---- = A}        -------`                  1`         90#     W`           1`          W  =  9 lbs. Ans.~RA~Q3. Find the edge of a cubA}        e whose volumeis 64 cu. in.(a) 4 in.  (b) 8 in.  (c) 10 2/3 in.(d) 21 1/3 in.  (e) 2 in.~RCA3. (a) 4 in. Ans.`      B }                   3`            V = e`                 3`           64 = e`        4 in. = e  Ans.~RA~Q4. A sphere is inscrB}        ibed in a cube 6}on an edge. The difference in volumebetween the cube and the sphere incubic inches is~(a) 432#  (b) 180B}        #  (c) 216  (d) 36#(e) 216 - 36#~RCE4. (e) 216 - 36# Ans.`        Cube          Sphere`        e = 6         r = 3`B}                     3            4  3`        V = e         V = -#r`                          3`             3            4`   B}             V = 6         V = -(27)#`                          3`        V = 216       V = 36#Difference = 216 - 36#  Ans.~RAB}        ~Q5. A rectangular swimming pool is 50ft. by 120 ft. Water is entering at therate of 1750 gallons per minute. Howmany houB}        rs will it take to fill thepool 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}        ) 2 1/2~RCB5. (b) 3 Ans.V = 1whV = (120) (50) (7)V = 42&000 cu. ft. x 7 1/2 gal/cu.ft.` = 315&000 gal1750 Gal/minB}         x 60 min/hr.=`                      105&000 gal./hr.`  315&000 gal.`  -------------- = 3 hr. Ans.`  105&000 gal/hr~ETB	}        ~ET0 min/hr.=`                      105&000 gal./hr.`  315&000 gal.`  -------------- = 3 hr. Ans.`  105&000 gal/hr~ET@         ~Q1. The wheel of a bicycle is 28 inchesin diameter. Compute the number of feetthat will be covered in six turns ofthe wF}        heel of this bicycle.(use # = 22/7)(a) 14 ft  (b) 44 ft  (c) 88 ft(d) 308 ft  (e) 528 ft~RCB~Q2. A circle has a diameF}        ter that is 20inches longer than the diameter ofanother circle. If the radius of thesmaller circle is r& the circumferenceF}        of the smaller circle is to thecircumference of the larger as~`     2r          2r           r(a) ------  (b) ------  (cF}        ) -------`   r + 10      r + 20      2r + 20`     r           r(d) ------  (e) ------`   r + 20      r + 10~RCE~Q3. TF}        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~Q4. The two legs oF}        f a right triangle are.6} and .8}& respectively. The numberof inches in the hypotenuse is~(a) .14}  (b) 1}  (c) .10}  (d)F}         14}(e) .2}~RCB~Q~SB~SP196024259024259070196070~SP196070196024197024210070~SH0328D~SH0338C~SH1028A  E~SH1038B~SFF}        5. The area of rectangleABCD 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}        16(d) 24  (e) 8~RCA~SD~Q6. The radius of a circular playgroundis 4 times the radius of a circularpool with the same cF}        enter. The ratio ofthe area of the pool to the area of theplayground is~(a) 2~1  (b) 1~2  (c) 4~1  (d) 16~1(e) 1~16~RCF}        E~Q7. A boat sailed 35* due east along aparallel of latitude. If it startedfrom a port situated on 22* westlongitude& itF}        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~Q8. F}        The diagonal of a square divides itinto two triangles each 32 sq. in. inarea. The length of the diagonal ininches is~`   F}            _                 _(a) 16\/2  (b) 16  (c) 8\/2`       _         _(d) 32\/2  (e) 4\/2~RCC~Q9. The scale of a map is F}        1 in. = 50miles. The actual area in square milesof a circle 2 inches in diameter is~(a) 50#  (b) 100#  (c) 200#  (d) 2500F}        #(e) 10&000#~RCD~Q10. A rectangle is 4 inches longer thanit is wide. Its perimeter is 20 inches.The number of square iF}        nches in its areais~(a) 96  (b) 32  (c) 21  (d) 12  (e) 36~RCC~Q11. Through how many degrees does thehour hand of a clF}        ock move in 150minutes?(a) 2 1/2*  (b) 25*  (c) 75*  (d) 150*(e) 256/#*~RCC~Q~SB~SH0927A           B~SH0533C~SH033F}        7D~SP259024189064266064227040~SF12. AC = 10}& <A = 30*&and <BCD = 60*. Thenumber of inches in thelength of BC is~`    F}                 _         _(a) 5  (b) 5\/2  (c) 5\/3(d) 10  (e) 20~RCD~Q~SB~SP161070217024273070161070~SP2170242170702170702F}        17070~SH1023A~SH1039B~SH0332C~SH1032D~SF13. AC = 10}& CD [ AB&CD = 6}& and <A = <B.The number of inchesin AB is~(aF }        ) 6  (b) 8  (c) 10(d) 12  (e) 16~RCE~Q~SB~SH0631t  p~SH0735q~SH0929s   r~SP189056256056224056256032~SP2030882240562F!}        27024227024~SF14. <p + <q + <r = 210*. <t = 2<s.Thenumber of degrees in <s is~(a) 10  (b) 30  (c) 50(d) 100  (e) 150~F"}        RCC~SD~Q15. A cylindrical can of oil with acircular base has a height h and holds30 quarts. If its height is to remaincF#}        onstant& by what number must its baseradius be multiplied so that it willhold 60 quarts?`              _(a) 2h  (b) h\/2 F$}         (c) 4  (d) 16`     _(e) \/2~RCE~Q16. How many rods of steel 4 inches by4 inches by 9 feet can be made from 144cubic fF%}        eet of steel?(a) 1  (b) 4  (c) 16  (d) 144  (e) 122~RCD~Q~SB~SP210024252024252096210096~SP210096210024210024210024~SPF&}        206020256020256100206100~SP206100206020206020206020~SH12336}~SH08349}~SF17. A picture 6} x 9} is enclosed in aframe 1/2F'}        } wide. The outside perimeteris~(a) 16}  (b) 17}  (c) 32}(d) 34}  (e) 27}~RCD~SD~Q18. If the length of a rectangle iF(}        sdecreased by 10%& and its width isincreased by 50%& by what percent isits area changed?(a) 35  (b) -35  (c) 65  (d) -65F)}        (e) 40~RCA~Q19. The supplement of an angle is 10*less than 3 times its complement. Findthe number of degrees in the anF*}        gle.(a) 5  (b) 40  (c) 45  (d) 50  (e) 20~RCB~Q20. Given~ Trapezoid KLMN with NM`[`LM&KN = 10}& LM = 16} and <L = 45*. F+}         Find~The number of square inches in the areaof KLMN.`               _(a) 48  (b) 48\/2  (c) 78`       _(d) 78\/2  (e) F,}        96~RCC~Q21. Water flows 2 feet deep in arectangular channel 4 feet wide. If thewater flows at 3 miles an hour& whatweigF-}        ht of water (in tons) will pass agiven point in 6 hours? (A cubic yardof water weighs 3/4 ton).(a) 21&120  (b) 37&545  (cF.}        ) 63&360(d) 190&080  (e) 17&320~RCA~Q~SB~SH073810~SH10274    y~SH09292~SP252072182072196064196080~SP196064252032252F/}        080252080~SF22. Find the length of y.(a) 20  (b) 16  (c) 12  (d) 5  (e) 10~RCB~SD~Q23. A bridge 3500 feet long and 16F0}         feetwide is designed to withstand a trafficweight of 50 pounds per square foot.Assuming that the two lanes of thebridge F1}        are filled with cars standingbumper to bumper and each 210} long&what should be the maximum averageweight per car in poundF2}        s?(a) 200  (b) 400  (c) 583 1/3(d) 7000  (e) 14&000~RCD~ET~ETt should be the maximum averageweight per car in poundD F        ~SB~SP205024259070205070205024~SH0729a~SH1033b~SH0534c~SH1632s     s~SH1935s~SP217143266143243106217143~SFH. PerimeJ4}        tersMake sure you know thesetheorems& definitions andassumptions.A perimeter is the measurein linear units of thedistJ5}        ance around an object.1. Triangle`  Perimeter = a + b + c2. Equilateral Triangle`  Perimeter = 3s~RA~SD3. RectanglJ6}        e`   Perimeter = 2l + 2w = 2 (l + w)4. Square`   Perimeter = 4s5. Circle`   Perimeter = Circumference (C)`   C = J7}        2#r = #d~RA~SB~SC192071210056226071226071~SH0828r~SH0833r~SH0930n*~SH1133L~SF6. Arc (length) is a fraction of theciJ8}        rcumference of a circle.`         n          n#rLength = --- x 2#r = ---`        360         1807. Irregular Figures.J9}        To find the perimeter of anirregular figure~a. Compute the perimeters of figuresfor which formulas exist.b. DetermineJ:}         their sum& difference& orfractional parts to obtain therequired answer.~RA~Q~SB~SP189071217016252071189071~SH0232B~SJ;}        H1027A          C~SF1. The perimeter of {ABCis 36}. AB = X + 3&BC = 3X - 9& AC = 2XFind the longest sideof the triangleJ<}        .(a) 5}  (b) 7}  (c) 14}(d) 10} (e) 12}~RCC1. (c) 14} Ans.x + 3 + 3x - 9 + 2x = 36`            6x - 6 = 36`      J=}                  6x = 42`                 x = 7AB = 10}& BC = 12& AC = 14longest is AC = 14} Ans.Note~ Perimeter is not = 18J>}        0*~RA~SD~Q2. The distance around a rectangularswimming pool is 250 feet. The pool is50 feet wide. Find its length.(a)J?}         5'  (b) 75'  (c) 100'  (d) 200'(e) 150'~RCB2. (b) 75' Ans.Let l = length of pool.`  50 + l + 50 + l = 250`        J@}         100 + 2l = 250`               2l = 150`                l = 75'  Ans.~RA~Q3. Find the diameter of a quarter-milecircuJA}        lar track. (use # = 22/7 andexpress your answer in yards)(a) 20 yds. (b) 40 yds. (c) 70 yds.(d) 140 yds. (e) 240 yds.~RJB}        CD3. (d) 140 yds. Ans.Note~ 1/4 mile = 440 yds.`            c = #d`                22`          440 = --d`          JC}               7`      140 yds. = d  Ans.~RA~Q4. Find the cost of framing a picture3 feet by 4 feet at $2.00 per foot.(a) $12JD}        .00  (b) $14.00  (c) $24.00(d) $28.00  (e) $48.00~RCD4. (d) $28.00 Ans.Note~ Framing is perimeter.`         p = 3 + 4JE}         + 3 + 4`         p = 14 ft.Cost = 14 x 2 = $28.00  Ans.~RA~Q5. Find the side of an equilateraltriangle whose perimetJF}        er is equal tothat of a square of side 6}.(a) 4}  (b) 8}  (c) 12}  (d) 24}(e) 9}~RCB5. (b) 8} Ans.Note~ The perimeteJG}        r of the square is24}. Therefore the side of thetriangle = 24/3 = 8}  Ans.~RA~Q6. The perimeter of an equilateraltriangJH}        le equals the perimeter of asquare. A side of the triangle is b&and a side of the square is k. Expressb in terms of k.(aJI}        ) 3b/4  (b) 4b/3  (c) 3k/4(d) 4k/3  (e) 4b/3k~RCD6. (d) 4k/3 Ans.` 3b = perimeter of triangle` 4k = perimeter of squaJJ}        re` 3b = 4k`  b = 4k/3 Ans.~RA~Q~SB~SP224016189080252080224016~SP189080189128252128252080~SH062710}~SH063610}~SH1JK}        3266}~SH13386}~SH102860*   60*~SF7. Find the perimeter of theaccompanying geometric figure.(a) 32}  (b) 38}  (c) 42}JL}        (d) 44}  (e) 52}~RCC7. (c) 42}  Ans.The triangle is equilateralsince two angles are 60*.Therefore the common linein thJM}        e triangle and therectangle is 10}& and thebase of the rectangle isalso 10}.10}+10}+6}+10}+6} = 42}~ET~ETn thH y        I. AreasMake sure you know these theorems&definitions and assumptions.Area is the measure in square units ofthe surfacNO}        e of an object.1. Triangle`    Area = 1/2 bh2. Equilateral Triangle`            2`           s   _`    Area = --\/3NP}        `           4~RA3. Rectangle`    Area = lw~RA~SB~SP217032255032231063189063~SP189063217032217063217063~SH0633h~SH0NQ}        932b~SH1531h~SH1833b'~SH1333b~SP203104254104266135196135~SP196135203104203135203135~SF4. Parallelogram`   Area = bhNR}        5. Square`           2`   Area = s6. Trapezoid`          h`   Area = -(b + b')`          2~RA~SD~SB~SC19207NS}        1210056226071226071~SH0828r~SH0833r~SH0930n*~SH1133L~SF7. Circle`            2`   Area = #r8. SectorA sector NT}        is fraction ofthe area of the whole circle.`                    2`   Area = n/360 x #r`             2`          n#r` NU}               = ----`          360~RA~SD9. RingA ring is a figure with two concentriccircles (circles having the samecenter)NV}        . To get the area of the ring -that is& the outer portion - subtractthe area of the smaller circle fromthe area of the larNW}        ger circle.Area of Ring = Area of larger circle -Area of smaller circle.`                2     2Area of Ring = #R  - #rNX}        `                 2    2Area of Ring = #(R  - r )~RA10. Irregular Figures.To find the area of an irregularfigure~a. NY}        Compute the areas of figures forwhich formulas exist.b. Determine their sum& difference& orfractional parts to obtain theNZ}        required answer. In general~ Areashaded region = area outside - areainside.~RA~Q~SB~SP203024254024266055196055~SP1960N[}        55203024203055203055~SH03338}~SH06314}~SH083310}~SP210085252085252128210128~SP210128210085210085210085~SH1430y~SFExamN\}        ple~1. The area of a squareis equal to the area of atrapezoid with bases 8}and 10} and altitude 4}. Find a side of the N]}        square.Area of square = Area oftrapezoid.~RA`2y  = h/2 (b + b')`2y  = 4/2 (8 + 10)`2y  = 2 (18)`2y  = 36y  =N^}         6} Ans.~RA~SD~Q1. ABCD is a parallelogram& with AB =10}& AD = 6} and <A = 60*. Find thenumber of square inches in the N_}        area ofABCD.`               _                  _(a) 15  (b) 15\/3  (c) 30  (d) 30\/3`     __(e) \/30~RCD`          _1N`}        . (d) 30\/3 Ans.Using the 30*& 60*& 90* triangle`       6  _      _`   h = -\/3 = 3\/3        2Now A = bh`          Na}           _`   A = 10(3\/3)`           _`   A = 30\/3  Ans.~RA~Q2. If the area of a circle isquadrupled& by what percent is tNb}        heradius increased?(a) 100  (b) 200  (c) 300  (d) 400(e) 10~RCA2. (a) 100 Ans.`            2`    A      r`   ---- Nc}        = ----`             2`     1      1`    A     (r)`            2`    1      r`   ---- = ----`             2`        Nd}            1`    4     (r)~RA`      2      2`     1`   (r)  =  4r`     1`    r   =  2rwhich represents a 100% increase.Ne}        ~RA~Q3. Cruiser A patrols a circular areaof 3600# square miles. Cruiser Bpatrols a circular area of 900# squaremiles. WhNf}        at is the difference betweenthe radii of action of the cruisers?`                     _(a) 30 miles  (b) 30\/3 miles`    Ng}                                     __(c) 90 miles  (d) 60 miles  (e) \/90~RCA3. (a) 30 miles Ans.The radius of Cruiser A~`  Nh}          2`  #r  = 3600# sq. miles`    r = 60 milesThe radius of Cruiser B~`    2`  #r  = 900# sq. miles`    r = 30 milesNi}        The difference between the radii is~`60 miles - 30 miles = 30 miles Ans.~RA~Q4. A wire 22} long is to be bent intothe Nj}        shape of a square. Another pieceof wire of the same length is to beshaped into a rectangle whose lengthis 4}. Find the difNk}        ference in areabetween 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}        in.(e) 2 1/2 sq. in.~RCA4. (a) 2 1/4 sq. in. Ans.Perimeter of a square~`       4s = 22}`        s = 5 1/2}Area ofNm}         a square~`             2`        A = s`          = 30 1/4 sq. in.~RAPerimeter of a rectangle`  2l + 2w = 22}`2(4)Nn}         + 2w = 22}`       2w = 14}`        w =  7}Area of a rectangle~`        A = lw`          = 28 sq. in.30 1/4 sq.inNo}        . - 28 sq. in =`          2 1/4 sq. in. Ans.~ET~ETf a rectangle~`        A = lw`          = 28 sq. in.30 1/4 sq.inL 7                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           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 Sō
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 S LD1 p p	  0C9DI pCDL~CiCDi D`          @L |u}6CD l0C)HCC WhL/h
`CmCDi D`
R@W                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                          