ZXTape! 0Created with Ramsoft MakeTZXPYTHAG1  ^  N0:0:0::236099\,502:0:0:0 q$="y"x -z$="  " K12 ,10 ;1;4;"PLEASE WAIT";0;0 "" .a=07:a,0;1;z$:a (H12 ,6;4;1;" LOADING PROGRAM ";:0 -*18,5;" " .13 ,0; / 0 2 "file" <*:15,0;"Leave tape running" F"" P*15,0;"Stop and rewind tape" R"clear (y/n) ?";q$ Tq$="y""address ? ";x ZJ17,0;"Type in file name in ";1;"UPPER CASE";0 n"Filename",a$ o015,0;"Recording " p517,0;" " x a$10 "logo"16384@,2304  '10 ,5;"Load main program" )12 ,5;"SAVE""file"" LINE 2" Z&  QnAPYTHAG1Hlogo @ ??80?߀????>?#|?~???|?|~?8???1???<???????<?0~|??~?~|?|~?>??#???8??????1???????~<?|??|~?|?|~?|?|~???9??>?? ?????8<0p??>?x?~ file ;&;;d 10808 80P .3,10 O1;21,31;8;9 ;"*":30:=""0180 #21,31;8;" ": **clean lines** 0120x Ec=v2v1-1:c,0;" ";:c  @30:.3,30 Tz=0:z$="" h,21,z;4;0;1;"N" | 1 a$= a$=12 460 5a$=13 z0a=z$:.05|L,20: &a$<480ůa$>579360h z=31360h ^a$>47/Ưa$<58:4;0;21,z;a$:.05|L,20:z=z+1 ̕a$=12 z>021,z-1;1;4;0;"N":21,z;" ":.05|L,20:z=z-1:z$=z$(z):0360h !a$=13 z0a=z$:  z$=z$+a$  0360h  D0:0:2::10 :.1}L,10 :0,11 ;" " X.1}L,15 l4,1;" " .1}L,20 \10 ,13 ;" " .1}L,25 15,1;" " .1}L,30  L 05600 VG21,11 ;6;"LESSON ONE":.1}L,35#:20 ` 0120x t?7::1,10 ;2;6;"LESSON ONE" y3,0;" Many of the major discoveriesin mathematics have come fromidle observation and the suddennoticing of a pattern. Althoughwe do not know for certain, itis possible that the theorem ofPythagoras was first suggestedto a Greek some 2500 years agowhile he was gazing at thepatterns in a floor made upfrom triangular tiles like this-"  2500 318,2;5;"It is a RIGHT ANGLED" "5;" ISOSCELES triangle." %v1=2:v2=20:0220 R0:0:6::16,1;"These tiles were laid in the" %" pattern shown above.":2620< &v1=16:v2=17:0220 (?6:16,0;" The Greek noticed that if he " <$" combined two tiles, he could " P$" make up a larger right angled " d" triangle.":2560 x&v1=16:v2=19:0220 416,1;"Then he noticed that a square" #" could be made up on the longer" 2" side of the triangle.":2980 :2560 &v1=16:v2=19:0220 S16,1;7;"How many tiles make up the square? ":0300, A=81620T gv1=20:v2=21:0260:.353333,0:19,1;2;"No, count them!"  0300, ,A=81620T @T.353333,0:19,1;2;"No, there are 8 tiles.":1640h Tev1=20:v2=21:0260:19,0;5;"Yes, 8 tiles make up the square." h[21,0;" ":v1=16:v2=21:0220 | 6 416,1;"He then noticed that two" #" squares could be made up on" `" the other two sides.":1:2860, :2920h :v1=16:v2=19:0220 s16,1;7;"What is the total number of tiles in the two smaller squares?":0300, =v1=20:v2=21:0260:A=818400 W.353333,0:19,1;2;"No, count them carefully.":0300, A=818400 e.353333,0:19,1;2;"No,there are 8 tiles,4 in each square.":1860D 0<19,1;5;"Yes, again there are 8 tiles." D[21,0;" ":v1=16:v2=21:0220 X 5 l416,1;"Now the Greek wondered if this" #" would work in the case of" H" other right angled triangles.":v1=16:v2=20:0220  7 416,1;"Turn to chapter five in the" #" booklet where you will find" #" three different right angled" #" triangles with squares drawn"  :" on their sides.":v1=16:v2=21:0220  416,1;"Measure the sides of the" 4#" squares and calculate their" H1" areas.":v1=16:v2=20:0220 \416,1;"In each case answer the ques-" p/" tion: ";5;"Is the area of the"  5 #" big square the same as the sum" #" of the areas of the other two" 3" squares?":v1=16:v2=21:0220 416,1;"The Greek found this was true" #" for all the right angled tri-" #" angles he studied, so he made" 8" up a THEOREM.":v1=16:v2=21:0220 $+15,11 ;5;"THE THEOREM" 8/:6:" THE AREA OF THE SQUARE ON THE " L$" LONG SIDE OF A RIGHT ANGLED " `$" TRIANGLE IS EQUAL TO THE SUM " t$" OF THE AREAS OF THE SQUARES ON " I" THE OTHER TWO SIDES. ":v1=0:v2=21:0220 5000  **draw single tile** ]c=032 :6;176+c,16+c:6;64@-2*c,0:c  **two tile triangle** ^c=031:2;112p+c,96`+c:1;2;31-c,0:c ( <**draw tile pattern** P<c=213 :6;c,10 ;" ":c d2:2700 x **draw tile net** v=15995_-32 u=80P14432 ȉu,v:32 ,0:-32 ,-32 :32 ,0:-32 ,32 :0,-32 :32 ,0:0,32 u v ]79O,63?:98b,0:0,97a:-98b,0:0,-97a  ,**fill in side square** @Q2:c=031:144,96`+c:1;32 ,0:c T h**fill in base square** |Q2:c=031:112p,95_-c:1;32 ,0:c  **fill in hyp square** 2 Uc=031:112p-c,96`+c:1;2*(c+1),0:c ^c=031:80P+c,128+c:1;64@-2*(c+1),0:c   7400  120x J **clean lines**  5400 Ec=v2v1-1:c,0;" ";:c  **input a string** V.25~,24:"Type in y(es) or n(o) ";A$:A$<1űA$>35820  **triangle** M128,88X:47/,0:0,35#:-47/,-35#    **h-square** 4d127,88X:480,36$:-35#,480:-480,-36$:35#,-480 H \**b-square** p^127,87W:491,0:0,-480:-491,0:0,480  **s-square** ^176,87W:0,37%:36$,0:0,-37%:-36$,0   **c-line** 1175,88X:1;-524,72H  **fill rect1** Pc=011 :158-.75@*c,113q+c:2.08R*c,0:c Lc=034":149-.75@*c,125}+c:24,0:c (\c=011 :124|+1.33*=p*c,160+c:2.08R*(11 -c),0:c < P**fill rect2** dOc=021:128-.75@*c,90Z+c:2.08R*c,0:c xLc=024:111o-.75@*c,112p+c:45-,0:c [c=021:94^+1.33*=p*c,137+c:2.08R*(21-c),0:c  **shear 1a** c=021:139,171:1;-15+.75@*c,-12 -c:1;0,c*1.57H:174,122z:-15+.75@*c,-12 -c:0,c*1.57H:c  **shear 1b** c=044,:173,125}:1;-32 ,44,-c:1;.75@*c,0:173,89Y:-32 ,44,-c:.75@*c,0:c  ,**shear 2a** @c=021:94^,137:1;29+.75@*c,21-c:1;-2.08R*c,0:128,89Y:29+.75@*c,21-c:-2.08R*c,0:c T h**shear 2b** |c=034":127,90Z:1;-34"+c,46.:1;0,-1.33*=p*c:174,90Z:-34"+c,46.:0,-1.33*=p*c:c 10:127,88X:-35#,480  **fill in s-square** ?c=135#:176+c,88X:0,36$:c  **fill in b-square** ?c=047/:128,87W-c:47/,0:c  **ending routine** 0:6: 35,1;"That completes Lesson One, in" #" the next lesson we shall see" #" how the theorem can be put to"  " use."  10 ,1;"If you would like to go over":" Lesson One again, ":13 ,13 ;2;6;"PRESS R"::" otherwise:-":15,13 ;6;1;"PRESS S" 4=""72204 H="R"Ŧ="r"7360 \="S"Ŧ="s"7300 p 72204  :z=0 7,1;"When you are ready to go on to":" LESSON TWO type:-":10 ,8;"NEW followed by":13 ,9 ;1;7;"LOAD ";34";"PYTHAG2";34" E17,10 ;4;"Bye for now!":23636T\,255:  z=1   ***CORE***  :5 <4:9 ,0;" In this section you will beshown a demonstration or PROOFof the theorem for any rightangled triangle." x5400:  8420  0: 4:7,0;" Turn to chapter five in theworkbook where you will find anoutline of a right angledtriangle, squares and construct-ion lines. Find the areas of thetwo rectangles and the squaresand so confirm the proof in thatparticular case." @%v1=2:v2=21:5560 T6:2,0;" We name the theorem after theGreek thinker named Pythagoras,though it is doubtful whether itwas he who actually discoveredits truth." | 5400 4:9 ,0;" The long side of a right angledtriangle is usually called theHYPOTENUSE of the triangle, sowe can restate the theorem ofPythagoras:"  5400 2:16,0;" IN A RIGHT-ANGLED TRIANGLE, THESQUARE ON THE HYPOTENUSE ISEQUAL TO THE SUM OF THE SQUARESON THE OTHER TWO SIDES." l%v1=0:v2=21:5560 7080:z=1  B5:c=017:c,10 ;" ":c  **the proof** 0::8360 ! 0:4:3,0;" A PROOF":4,1;"of the":5,1;"THEOREM":6,4;"of" ! 7,0;"PYTHAGORAS" !4 5400 !H>5:18,0;" First, we draw a right angled" !\T" triangle...":0:5860:v1=18:v2=21:0:5560 !p>4:18,0;" Then we draw a square on the" !e" long side of the triangle...":0:5920 :v1=18:v2=20:0:5560 !>6:18,0;" Now we draw squares on the" !p" other two sides...":5:0:6000p:6060:0:v1=18:v2=20:5560 !>7:18,0;" Next we draw in a construction" !#" line which divides the large" !o" square into two rectangles...":5:0:6320:v1=18:v2=21:0:5560 !>4:18,0;" We paint the smaller rect-" "^" angle red...":5:2:6380:v1=18:v2=21:0:5560 "$>5:18,0;" This red rectangle has the" "8W" same area as the red parallel- ogram, for it has the same length and height." "L5:2:6580 "t/v1=18:v2=21:0:5560 "G6:0:18,0;" This parallelogram has the" "A" same area as the square...":5:2:6640 "/0:v1=18:v2=21:5560 "518,0;" But this square has the same" "#" area as the square on the side" "b" of the triangle.":4:6:6840:v1=18:v2=21:0:5560 #Q8360 :0:5860:5920 :5980\:6040:6320 #G0:7:18,0;" Now look at the other rect-" #(p" angle. First we colour it red.":2:5:6480P:v1=18:v2=21:0:5560 #<>4:18,0;" In this case this rectangle" #P#" has the same area as the para-" #d/" llelogram...":5:2:6700, #x/0:v1=18:v2=21:5560 #>5:18,0;" The parallelogram has the same" #T" area as the square...":2:5:6760h:0:5 #10:v1=18:v2=21:5560 #>6:18,0;" This square is the same as the" #n" yellow square...":5:6:6900:v1=18:v2=21:0:5560:8360 #N5:0:5860:5920 :5980\:6040:6320 $G0:4:18,0;" So, we can divide the big" $|" square into two rectangles ";2;6;"A":4,18;6;2;1;"A":6:0 $,20,0;4;" and ";2;6;"B";4;0;".":6,15;2;6;1;"B":0:5560 $@l5:18,0;" Rectangle ";2;6;"A";0;5;" has the same area" $T:" as the yellow square...":5:6:6840 $h20:5:19,22;"Rectangle" $|N" ";2;6;"B";0;5;" has the same area as the " $`" blue square...":5:1:6900:v1=18:v2=21:0:5560 $J7:18,0;" We can see then, that ";4;"THE AREA" $74:" OF THE LARGE SQUARE IS THE " $#" SAME AS THE SUM OF THE AREAS" $D" OF THE OTHER TWO SQUARES.":v1=18:v2=21:5560 $D5:19,3;"That completes the proof.":5560 $T20,1;4;"Would you like to see the proof again? (press Y or N)" %=""9475% %="y"Ŧ="Y"8420 % "n"Ʀ"N"9475% % %0 & 20 &"file"9899&: '"" ?_  zZA PYTHAG2  L  N0:0:0::236099\,502:0:0:0 q$="y"x -z$="  " K12 ,10 ;1;4;"PLEASE WAIT";0;0 "" .a=07:a,0;1;z$:a (H12 ,6;4;1;" LOADING PROGRAM ";:0 -*18,5;" " .13 ,0; / 0 2 "file" <*:15,0;"Leave tape running" F"" P*15,0;"Stop and rewind tape" R"clear (y/n) ?";q$ Tq$="y""address ? ";x ZJ17,0;"Type in file name in ";1;"UPPER CASE";0 n"Filename",a$ o015,0;"Recording " p517,0;" " x a$10 "logo"16384@,2304  '10 ,5;"Load main program" )12 ,5;"SAVE""file"" LINE 2" Z&  QnAPYTHAG2logo @D ??80?߀????>?#|?~???|?|~?8???1???<???????<?0~|??~?~|?|~?>??#???8??????1???????~<?|??|~?|?|~?|?|~???9??>?? ?????8<0p??>?x?~file v&uhOvdha=07:b:"a"+a,b:a:24,36$,8,16,60<,0,0,0 x$p(5):q(5):h(5)  2280 n$=(n+.05|L):l=n$ c=1̱n$ n$(c)="."l=c+1 c n$=n$(1l)  ,621,0;" " @80P:.3,10 TO1;21,31;8;9 ;"*":30:=""0340T h#21,31;8;" ": | **clean lines**  0320@ Ec=v2v1-1:c,0;" ";:c  Bf=1100d:f:21,0;" " #**check input a number 2 digits** 30:.3,30 z=0:z$="":fs=0 ,21,z;4;0;1;"N" & 3 0a$= Da$=12 0740 N.a$=13 z=1fs=15600 X5a$=13 z0a=z$:.05|L,20: l'a$<46.ůa$>5790540 a$=47/0540 %fs=1Ưa$=46.0540 z=310540 a$=46.fs=1 O4;0;21,z;a$:.05|L,20:z=z+1:08204 z=00540 a$=12 z>021,z-1;1;4;0;"N":21,z;" ":.05|L,20:z$(z)=46.fs=0  $z=z-1:z$=z$(z):0540  !a$=13 z0a=z$: 4 z$=z$+a$ H 0540 \ p**draw triangle** 7:146,173:101e,0:0,-108l:-101e,0:0,108l:149,170:95_,0:0,-102f:-95_,0:0,102f 7:160+16*(p<6),80P+16*(q<6):p*8-1,0:0,q*8-1:-(8*p-1),-(8*q-1) 6:12 -2*(q<6),19+2*(p<6);"A":12 -2*(q<6),20+p+2*(p<6);"B":11 -2*(q<6)-q,20+p+2*(p<6);"C" 3:12 -2*(q<6),20+(p/2)+2*(p<6);p:11 -(q/2)-2*(q<6),20+p+2*(p<6);q  **question find hypotenuse**  :w=0 )2:0,1;"Question ";j $04:2,0;" In a right angl-" 8" ed triangle ABC," L(" AB=";p;"cm, BC=";q;"cm.":0880p `06:6,0;" Find the length" t" of the hypoten-" " use AC correct to" " 1 decimal place." =5:11 ,0;" AB=";p;"cm,so AB=":b=0  0480 &(a-p^2)<.05|L1340< $b=b+1:b>11320( .4L,-10 :21,0;7;" No. Try again ":f=1100d:f:21,0;" ":0480  1240 (!.4L,-10 :w=w+1 <11 ,16;0;" ":11 ,17-((p^2));7;p^2:21,0;" " P412 ,0;" BC=";q;"cm,so BC=":b=0 d 0480: x&(a-q^2)<.05|L1500 $b=b+1:b>11480 .4L,-10 :7;21,0;" No, try again ":f=1100d:f:21,0;" ":0480  1400x !.4L,-10 :w=w+1 ܏12 ,16;0;" ":12 ,17-((q^2));7;q^2:21,0;" " T120x,68D:15,0:14,0;" AB+BC =":b=0  0480 2(a-(p^2+q^2))<.05|L1660| ,$b=b+1:b>11640h @.4L,-10 :7;21,0;" No, try again ":f=1100d:f:21,0;" ":0480 T 1560 h!.4L,-10 :w=w+1 |14,17-((p^2+q^2));7;p^2+q^2:112p,535:23,0:112p,502:23,0:21,0;" " R7:17,0;" If AC is ";p^2+q^2;", then correct to" $" one decimal place,AC is cm." W200,21:47/,0:200,18:47/,0:b=0  0480 3(a-(p^2+q^2))<.05|L1860D $b=b+1:b>118400 .4L,-10 :7;21,0;" No, try again ":f=1100d:f:21,0;" ":0480  1760 0!.4L,-10 :w=w+1 Dn=(p^2+q^2):180:18,(29-(n$));4;n$:21,0;" ":v1=0:v2=21:0400 X l **Title** 0:0:2::10 :.1}L,10 :0,11 ;" " .1}L,15:5,1;" " r.1}L,20:11 ,13 ;" " .1}L,25:16,1;" " .1}L,30   ***CORE*** U1900l:6;21,11 ;"LESSON TWO":.1}L,35#:0320@  2500 $ 2920h 8 3520 L 3840 ` 4440X 5500| 14,1;"NB"  **summary of lessons 1 or 2** 4:0,11 ;2;6;"LESSON TWO" >5:2,0;" In Lesson One you were shown" #" how it is thought the theorem" #" of Pythagoras was discovered." (#" You were then given a" <$" demonstration proof for the" P#" theorem and it was then re-" d$" stated as follows:-":0300, x=4:10 ,1;" IN A RIGHT ANGLED TRIANGLE," #" THE AREA OF THE SQUARE ON THE" #" HYPOTENUSE IS EQUAL TO THE SUM" #" OF THE AREAS OF THE SQUARES ON" /" THE OTHER TWO SIDES. ":0300, >6:16,0;" In this lesson you will learn" #" how to use the theorem to" #" calculate the length of the" #" hypotenuse given the lengths" ,#" of the other two sides of a" @@" right angled triangle.":v1=0:v2=21:0400 T h**teaching section** | 5: 53,0;" As you know the area of a" #" square is found by squaring" 0" the length of one of its sides.":0300, =6:8,0;" So, if a square has a side of" #" length 5cm, then its area is" 34;'" 5 which is 5x5=25cm":0300, $5:14,1;"NOTE:" 416,1;"The computer uses the notation" 017;" 5^2 ";5;" to mean 5. " D%v1=0:v2=21:0400 X05;2,0;" If we draw a" l" right angled tri-" " angle and label" " its corners A, B" /" and C...":p=8:q=6:0880p 04:8,0;" Then we can call" " the base of this" " triangle AB, the" " side BC and the" " hypotenuse AC.":0300, >7:15,0;" We can now rewrite the theorem" 4" as follows:-" H:5:18,8;"AC = AB + BC":0300, \>6:20,0;" In this case AB=8 and BC=6, so" pC" AC=64+36=100, so AC=10.":v1=20:v2=21:0400 >4:20,0;" Now we will put the theorem to" 2" work....":v1=1:v2=21:0400  **example** 7: 2,3;"AC=AB+BC" 06:4,0;" In a right angled" " triangle ABC,AB=6" $" and BC=4. Find" 8" the length of the" L5" hypotenuse AC.":p=6:q=4:0880p `4:11 ,0;" AB=6cm,so AB=":0300,:11 ,16;5;1;"36":502:11 ,16;5;0;"36" t12 ,0;" BC=4cm,so BC=":300,:12 ,16;5;1;"16":502:12 ,16;5;0;"16" &128,68D:15,0 14,0;5;" So AB+BC=":0300,:14,16;7;1;"52":128,535:15,0:128,502:15,0:502:14,16;7;0;"52" >6:16,0;" If AC is 52, then AC is the" Ă" SQUARE ROOT of 52,which is ":17,28;1;5;"7.2":502:17,28;5;0;"7.2" خ20,0;5;" Therefore AC=7.2cm.":7:136,6:39',0:136,3:39',0:v1=1:v2=21:0400  **Exercise** /7::3,6;"AC = AB + BC" (=4:7,0;" Now, try a short exercise of" <%9 ,9 ;"three questions." P 0300, d**set up question array** x d=0  m=3 c=13 6p(c)=(7*)+3:q(c)=(7*)+3 "k=0:c=14120 k=k+1:k=c4120 !p(c)=p(k)q(c)=q(k)4020  4060 #h(c)=(p(c)^2+q(c)^2) ,c @**the exercise** T hj=13 |$p=p(j):q=q(j):h=h(j):1000 w>0m=m-1 j  6: 84,0;" You obtained ";m;" marks out of 3," "m=14,20;" " @m=36,11 ;1;"WELL DONE!":44000 @m=26,5;"That was quite good!":44000 m<2d=16,0;" You still do not seem to have":" understood, so I suggest you":" work through this Lesson again":" - but not straight away!":44000 m<2d=06,0;" Not too good, you need to work":" through another exercise of 5":" questions.":d=1:0300,:3980 0 0300, D X**bookwork** l 4:  4,0;" Turn to chapter five in the workbook where you will find exercise 2a which contains questions similar to the ones you have just been doing. The example will be displayed again on the screen to help you in the setting out."  300,   |$p(5):q(5):h(5)   7320 **concatenation of a string** s$="" &k=1̱A$:A$(k)=" "5640  s$=s$+A$(k) k A$=s$  & **number string to 1 dec pl ** 0n$=(n+.05|L):l=n$ :c=1̱n$ Dn$(c)="."l=c+1 Nc Xn$=n$(1l) b >**check input of a string** H z=0 R5A$<1űA$>10 z=1:6020 \ 5600 fi=1̱A$ p,A$(i)<43+ůA$(i)>94^z=1 zi  **draw triangle-hyp&base** 7:146,173:106j,0:0,-108l:-106j,0:0,108l:149,170:100d,0:0,-102f:-100d,0:0,102f &q=((h^2-p^2)+.5) 160+16*(p<6),80P+16*(q<6):p*8-1,0:0,q*8:-8*p-1,-8*q 6:12 -2*(q<6),19+2*(p<6);"A":12 -2*(q<6),20+p+2*(p<6);"B":11 -q-2*(q<6),20+p+2*(p<6);"C" 3:12 -2*(q<6),20+(p/2)+2*(p<6);p:10 -(q/2)-2*(q<6),20-(h)+(p/2)+2*(p<6);h  **draw triangle-hyp&side** 7:146,173:106j,0:0,-108l:-106j,0:0,108l:149,170:100d,0:0,-102f:-100d,0:0,102f &p=((h^2-q^2)+.5) 5:160+16*(p<6),80P+16*(q<6):p*8-1,0:0,q*8:-8*p-1,-8*q 6:12 -2*(q<6),19+2*(p<6);"A":12 -2*(q<6),20+p+2*(p<6);"B":11 -2*(q<6)-q,20+p+2*(p<6);"C" 3:11 -(q/2)-2*(q<6),20+p+2*(p<6);q:10 -(q/2)-2*(q<6),20-(h)+(p/2)+2*(p<6);h  **question find base** $ :w=0 .)2:1,1;"Question ";j 806:3,0;" In a right angl-" B" ed triangle ABC," L(" AC=";h;"cm, BC=";q;"cm.":6100 V06:6,0;" Find the length" `" of the base AB" j" correct to one" t" decimal place." ~Q5:11 ,0;" AC=";h;",so AC";11 ,13 ;"=":b=0  480 &(a-h^2)<.05|L6340 $b=b+1:b>16330 M.3,-10 :21,0;7;" No. Try again ":470  6290 !.3,-10 :w=w+1 m11 ,17-((h^2));7;h^2:21,0;" " I12 ,0;" BC=";q;",so BC";12 ,13 ;"=";:b=0  480 &(a-q^2)<.05|L6420 $b=b+1:b>16410  N.3,-10 :21,0;7;" No, try again ":470  6370  !.3,-10 :w=w+1 m12 ,17-((q^2));7;q^2:21,0;" " U112p,68D:23,0:14,0;" AC-BC =":b=0 ( :480 22(a-(h^2-q^2))<.05|L6500d <$b=b+1:b>16490Z FM.3,-10 :21,0;7;" No, try again ":470 P 64502 Z!.3,-10 :w=w+1 d14,17-((h^2-q^2));7;h^2-q^2:112p,535:23,0:112p,502:23,0:21,0;" " nR4:16,0;" If AB is ";h^2-q^2;", then AB is the" x6" square root of ";h^2-q^2;". Therefore" 18,0;" to one decimal place,":19,21;"AB= cm.":7:192,13 :31,0:192,10 :31,0:b=0  480 3(a-(h^2-q^2))<.05|L6600 $b=b+1:b>16590 M.3,-10 :21,0;7;" No, try again ":470  6550 !.3,-20:w=w+1 Ⱥ21,0;" ":n=(h^2-q^2):5670&:19,(28-(n$));2;n$:21,0;" "  **question find side**  :w=0 )3:1,1;"Question ";j 06:3,0;" In a right angl-" " ed triangle ABC," (" AC=";h;"cm, AB=";p;"cm.":6030 06:6,0;" Find the length" "" of the side BC" ," correct to one" 6" decimal place." @Q5:11 ,0;" AC=";h;",so AC";11 ,13 ;"=":b=0 J 480 T&(a-h^2)<.05|L6790 ^$b=b+1:b>16780| hM.3,-10 :7;21,0;" No. Try again ":470 r 6740T |!.3,-10 :w=w+1 m11 ,17-((h^2));7;h^2:21,0;" " I12 ,0;" AB=";p;",so AB";12 ,13 ;"=";:b=0  480 &(a-p^2)<.05|L6870 $b=b+1:b>16860 N.3,-10 :7;21,0;" No, try again ":470  6820 !.3,-10 :w=w+1 m12 ,17-((p^2));7;p^2:21,0;" " f7:112p,68D:23,0:5:14,0;" AC-AB =":b=0  470 2(a-(h^2-p^2))<.05|L6950& $b=b+1:b>16940 M.3,-10 :7;21,0;" No, try again ":470  6900 !.3,-10 :w=w+1 &14,17-((h^2-p^2));7;h^2-p^2:7:112p,535:23,0:112p,502:23,0:21,0;" " 0R4:16,0;" If BC is ";h^2-p^2;", then BC is the" :6" square root of ";h^2-p^2;". Therefore" D18,0;" to one decimal place,":19,21;"BC= cm.":7:192,12 :47/,0:192,9 :47/,0:b=0 N 480 X3(a-(h^2-p^2))<.05|L7050 b$b=b+1:b>17040 lM.3,-10 :7;21,0;" No, try again ":470 v 7000X !.3,-10 :w=w+1 21,0;" ":n=(h^2-p^2):5670&:19,(28-(n$));2;n$:21,0;" "  **ending routine** F0:6::0,11 ;2;7;"LESSON TWO" 05,1;"That completes Lesson Two." 7,1;"If you would like to go over":" LESSON TWO again, ":10 ,13 ;2;6;"PRESS R"::" otherwise:-":12 ,13 ;6;1;"PRESS S" =""7160 ="R"Ŧ="r"  ="S"Ŧ="s"7200   7160  z=0: *7,1;"When you are ready to go on to":" LESSON THREE type:-":10 ,8;"NEW followed by":13 ,9 ;2;7;"LOAD ";34";"PYTHAG3";34" 4E17,10 ;4;"Bye for now!":23636T\,255: > z=1 H R**draw triangle and squares** \7:157,173:98b,0:0,-109m:-98b,0:0,109m:160,170:92\,0:0,-103g:-92\,0:0,103g:c f7:192,104h:31,0:0,23:-31,-23:9 ,23;"A":9 ,28;"B":5,28;"C" pe192,105i:30,23:-23,31:-31,-23:24,-31 z_224,104h:23,0:0,23:-23,0:0,-23 _192,103g:31,0:0,-31:-31,0:0,31   ***CORE***  7450  7640  8190  8620!  8940"  8420  9300T$  9030F#  9120# #7120:z=17320  **summary of lessons 3** $ .>5:2,0;" In the last section you were" 8#" given a more convenient form" B#" for the theorem of Pythagoras" L#" using letters to label the" V)" corners of the triangle:-":320@ `<4:8,1;"In a right angled triangle ABC" j#" where angle ABC is the right" t#" angle, the theorem of Pythag-" ~#" oras can be expressed as a" #" formula:- " 67:14,10 ;"AC=AB+BC":320@ >6:16,0;" You learned to use this form-" #" ula to calculate the length of" #" the hypotenuse of a right" #" angled triangle given the" H" lengths of its other two sides.":v1=0:v2=21:400  **teaching section**  >6:4,0;" In this section you will learn" #" how to work out the length of" #" a side given the lengths of"  #" the hypotenuse and the other" #" side using a formula derived" #" from the one you have just" (1" learned.":v1=0:v2=21:400 214:2,0;" Do you remember" <" how in the proof" F#" shown in Lesson 1,":7250R P14:5,0;" the square on the" Z" hypotenuse AC was" d" divided into two" nB" rectangles.":7:211,119w:-23,32 xp7:10 ,0;" One rectangle":4,25;5;1;" ":10 ,16;"was" " equal to the area" " of the square on" K" the side BC.":c=02:6+c,28;5;" ":c |2:15,0;" The other rectangle":5,23;6;1;" ":15,22;"was equal" ^" to the square on the side AB.":c=03:9 +c,24;6;" ":c $v1=0:v2=21:400 >7250R:7:211,119w:-23,32 15:2,0;" So if we subtract" " the area of the" " square on the side" e" BC ":c=02:6+c,28;5;" ":c:5,5;"(this is equal" -" to BC),":6,10 ;" from the" " area of the square" " on the hypotenuse" R" AC (equal to AC)":c=03:9 +c,24;6;" ":c )10 ,0;" we are left with" "" the area of the" ," square on the base" 6"" AB (equal to AB).":320@ @>6:15,0;" We can write this as a formula" J+2:17,10 ;"AB=AC-BC" T>6:19,0;" Similarly we can produce the" ^ " formula-" hN2:21,9 ;"BC=AC-AB":v1=0:v2=21:400 r=6:2,0;" Just as we could use the" | " formula-" *5,10 ;4;"AC=AB+BC" 47,0;" to calculate the length of the" 7" hypotenuse AC,":8,17;"we can use the" " formulae-" +4:11 ,10 ;"AB=AC-BC" >6:13 ,0;" to calculate the length of AB," 15,6;"and:-" +4:17,10 ;"BC=AC-AB" >6:19,0;" to calculate the length of the" #" side BC in a right angled tri-" /" angle.":v1=1:v2=21:400   **example**  5: 46,0;" Let's put these formulae to"  " work:-" &>7:10 ,0;" First an example using the" 0 " formula:" :+2:12 ,10 ;"AB=AC-BC" D>7:14,0;" to find the length of the base" N," AB.":v1=0:v2=21:400 X*2,1;2;"AB=AC-BC": b04:4,0;" In a right angled" l" triangle ABC,BC=6" v" and the hypoten-" " use AC=11. Find" 9" the length of AB.":q=6:h=11 :6100 5:11 ,0;" AC=11,so AC=":320@:11 ,14;7;1;"121":502:11 ,14;0;"121" 12 ,0;" BC=6, so BC=":320@:12 ,15;7;1;"36":502:12 ,15;0;"36" &120x,68D:15,0 14,0;" So AC-BC=":320@:14,15;7;1;"85":120x,535:15,0:120x,502:15,0:502:14,15;0;7;"85" ?4:16,0;" If AB is 85, then AB is the" Ɓ" SQUARE ROOT of 85, that is ":17,28;1;7;"9.2":0:17,28;0;2;"9.2" Ф20,0;" Therefore AB=9.2cm.":7:136,5:39',0:136,2:39',0:v1=0:v2=21:400  **example two**  =4:7,0;" Now we will use the formula:-" !*5:9 ,10 ;"BC=AC-AB" ! >6:11 ,0;" to find the length of the side" !," BC.":v1=0:v2=21:400 ! )2,2;2;"BC=AC-AB" !*06:4,0;" In a right angled" !4" triangle ABC,AB=8" !>" and the hypoten-" !H" use AC=12. Find" !R9" the length of BC.":p=8:h=12 :6030 !\4:11 ,0;" AC=12,so AC=":320@:11 ,14;7;1;"144":502:11 ,14;0;"144" !f12 ,0;" AB=8, so AB=":320@:12 ,15;7;1;"64":502:12 ,15;0;"64" !p/7:112p,68D:23,0 !z4:14,0;" So AC-AB=":320@:14,15;7;1;"80":7:112p,535:23,0:112p,502:23,0:502:14,15;0;"80" !?5:16,0;" If BC is 80, then BC is the" !" SQUARE ROOT of 80, that is ":17,28;1;7;"8.9":502:17,28;2;0;"8.9" !20,0;" Therefore BC=8.9cm.":7:136,5:39',0:136,2:39',0:v1=0:v2=21:400 ! !**Exercise 1** !,6::3,10 ;"AB=AC-BC" !=7:7,0;" Now, try a short exercise of" !%9 ,9 ;"three questions." ! 320@ !**set up question array** ! d=0 ! m=3 !c=13 "q(c)=(8*)+3:h(c)=(7*)+q(c)+1:h(c)^2-q(c)^2>107kh(c)^2-q(c)^2<38710" " (q(c)>9 h(c)>13 8710" ""k=0:c=187608" "k=k+1:k=c87608" "$!q(c)=q(k)h(c)=h(k)8710" ". 8730" "8#p(c)=(h(c)^2-q(c)^2) "Bc "L**the exercise** "V "`j=13 "j$p=p(j):q=q(j):h=h(j):6170 "t$v1=0:v2=21:400 "~w>0m=m-1 "j " 6: "84,0;" You obtained ";m;" marks out of 3," ""m=14,20;" " "@m=36,11 ;1;"WELL DONE!":8920" "=m26,7;"That was not bad!":8920" "īm<2d=16,0;" Still not good, but you had":" better go on with the lesson":" where you will be shown an- other example.":8920" "Ξm<2d=06,0;" Not too good, you need to work":" through another exercise of 3":" questions.":d=1:5740l:8690! " 320@ " "**bookwork 1** " #5,0;" Turn to chapter five in the workbook where you will find exercise 2b which contains questions similar to the ones you have just been doing." #2$v1=0:v2=21:320@ #< #F**bookwork 2** #P #Z5,0;" Turn to chapter five in the workbook where you will find exercise 2c which contains questions similar to the ones you have just been doing." #$v1=0:v2=21:400 # #**lesson summary** # #=6:2,0;" In this section the formulae-" #*4,10 ;7;"AB=AC-BC" #6,14;"and" #*8,10 ;7;"BC=AC-AB" #510 ,0;" were introduced and used to" ##" find the base or side of a" ##" right angled triangle ABC" ##" given the lengths of the" $#" hypotenuse AC, the side BC or" $" the base AB.":320@ $>4:17,0;" In Lesson Three you will see" $"#" how these formulae and methods" $,#" may be applied to solve" $6#" different problems involving" $@@" right angled triangles.":v1=0:v2=21:400 $J $T**Exercise 2** $^,:7:3,10 ;"BC=AC-AB" $h47,0;" Now, try a short exercise of" $r%9 ,9 ;"three questions." $| 320@ $**set up question array** $ d=0 $ m=3 $c=13 $p(c)=(8*)+3:h(c)=(7*)+p(c)+1:h(c)^2-p(c)^2>107kh(c)^2-p(c)^2<39390$ $"k=0:c=19440$ $k=k+1:k=c9440$ $!p(c)=p(k)h(c)=h(k)9390$ $ 9410$ $#q(c)=(h(c)^2-p(c)^2) $c $**the exercise** $ %j=13 %$p=p(j):q=q(j):h=h(j):6620 %$v1=0:v2=21:400 %&w>0m=m-1 %0j %: 6: %N84,0;" You obtained ";m;" marks out of 3," %S"m=14,20;" " %X7m=36,11 ;"WELL DONE!":9600% %b=m26,7;"That was not bad!":9600% %lm<2d=16,0;" Still not very good, you had":" better work through the lesson":" again.":9600% %vm<2d=06,0;" Not too good, you need to work":" through another exercise of 3":" questions.":d=1:320@:9370$ % 320@ % & 20 &"file"9899&: '"" !qh p dmk`"wbzZ8A dX