Top 10k strings from Theorem of Pythagoras, The (1984)(Griffin Software)(Part 2 of 3).z80 in <root> / bin / z80 / software / Sinclair Spectrum Collection TOSEC.exe / Sinclair ZX Spectrum - Utilities & Educational / Sinclair ZX Spectrum - Utilities & Educational - [Z80] (TOSEC-v2007-01-01) /

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   6 ;" No, try again ":
   3 ;"three questions."
   3 ;"WELL DONE!":
   3 ;"Question ";j
   3 ;" You obtained ";m;" marks out of 3,"
   3 ;" Not too good, you need to work":
   3 ;" No. Try again ":
   3 ;" No, try  again ":
   3 ;" In a right angled"
   3 ;"  Now, try a short exercise of" 
   3 ;"  In a right angl-"
   3 ;"  Find  the length"
   3 **the exercise**
   3 **set up question array**
   3 " questions.":
   3 " ed triangle  ABC,"
   2 ;"That was not bad!":
   2 ;"LESSON TWO"
   2 ;". Therefore"
   2 ;" to one decimal place,":
   2 ;" to calculate the length of the"
   2 ;" AC=";h;",so AC
   2 **teaching section**
   2 **example**
   2 ***CORE***
   2 " through another exercise of  3":
   2 " the  area  of  the"
   2 " square root of ";h^2
   2 " hypotenuse AC.":
   2 " formula-"
   2 " decimal place."
   2 " correct   to  one"
   2 " and the  hypoten-"
   1 s$=s$+A$(k)
   1 =64+36=100, so AC=10.":
   1 ;"we can use the"
   1 ;"was equal"
   1 ;"When you are ready to go on to":
   1 ;"The computer uses the notation"
   1 ;"That was quite good!":
   1 ;"That completes Lesson  Two."
   1 ;"PYTHAG3";
   1 ;"PRESS S"
   1 ;"PRESS R":
   1 ;"NEW  followed by":
   1 ;"LESSON TWO":
   1 ;"In a right angled triangle ABC"
   1 ;"If you would like to go over":
   1 ;"Bye for now!":
   1 ;"BC=    cm.":
   1 ;"AB=    cm.":
   1 ;", then correct to"
   1 ;", then BC is the"
   1 ;", then  AB is the"
   1 ;"(this is equal"
   1 ;" were  introduced  and  used to"
   1 ;" we are  left  with"
   1 ;" to mean  5
   1 ;" to find the length of the side"
   1 ;" to find the length of the base"
   1 ;" to calculate the length of AB,"
   1 ;" the square  on the"
   1 ;" from the"
   1 ;" You still do not  seem to have":
   1 ;" We can write this as a formula"
   1 ;" We can now rewrite the theorem"
   1 ;" 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."
   1 ;" 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."
   1 ;" 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."
   1 ;" Then we  can call"
   1 ;" The other rectangle":
   1 ;" Still not  very good,  you had":
   1 ;" Still  not  good, but  you had":
   1 ;" So, if a square  has a side of"
   1 ;" Similarly  we can  produce the"
   1 ;" One rectangle":
   1 ;" Now we will put the theorem to"
   1 ;" In this section you will learn" 
   1 ;" In this case AB=8 and BC=6, so"
   1 ;" In the last section  you  were" 
   1 ;" In  Lesson Three  you will see"
   1 ;" IN A RIGHT  ANGLED  TRIANGLE,"
   1 ;" BC=6, so BC
   1 ;" BC=4cm,so BC
   1 ;" BC=";q;"cm,so BC
   1 ;" BC=";q;",so BC
   1 ;" AC=12,so AC
   1 ;" AC=11,so AC
   1 ;" AB=8, so AB
   1 ;" AB=6cm,so AB
   1 ;" AB=";p;"cm,so AB
   1 ;" AB=";p;",so AB
   1 ;"  You learned to use this form-"
   1 ;"  So if we subtract"
   1 ;"  Now we will use the formula:-"
   1 ;"  Let's put  these  formulae to"
   1 ;"  Just  as  we  could  use  the"
   1 ;"  In this section the formulae-"
   1 ;"  In this lesson you will learn"
   1 ;"  In Lesson One  you were shown" 
   1 ;"  First an  example  using  the"
   1 ;"  Do  you  remember"
   1 ;"  As you  know  the  area  of a" 
   1 ;"   So AB
   1 ;"   If  we  draw  a"
   1 ;"    Therefore BC=8.9cm.":
   1 ;"    Therefore AC=7.2cm.":
   1 ;"    Therefore AB=9.2cm.":
   1 **summary of lessons 3**
   1 **summary of lessons 1 or 2**
   1 **question find side**
   1 **question find hypotenuse**
   1 **question find base**
   1 **number string to 1 dec pl **
   1 **lesson summary**
   1 **example two**
   1 **ending routine**
   1 **draw triangle-hyp&side**
   1 **draw triangle-hyp&base**
   1 **draw triangle**
   1 **draw triangle and squares**
   1 **concatenation of a string**
   1 **clean lines**
   1 **check input of a string**
   1 **check input a number 2 digits**
   1 **bookwork**
   1 **bookwork 2**
   1 **bookwork 1**
   1 **Exercise**
   1 **Exercise 2**
   1 **Exercise 1**
   1 " work....":
   1 " work through this Lesson again":
   1 " where you will  be  shown  an-  other example.":
   1 " where angle  ABC  is the right"
   1 " using  letters  to  label  the"
   1 " use AC=12.   Find"
   1 " use AC=11.   Find"
   1 " use AC correct to"
   1 " understood, so  I suggest  you":
   1 " ula to calculate the length of"
   1 " triangle ABC,BC=6"
   1 " triangle ABC,AB=8"
   1 " triangle ABC,AB=6"
   1 " triangle AB,  the"
   1 " to the  square on the side AB.":
   1 " through another exercise of  5":
   1 " theorem and it  was  then  re-"
   1 " the side BC.":
   1 " the length of the"
   1 " the length of one of its sides.":
   1 " the length of BC.":
   1 " the length of AB.":
   1 " the hypotenuse  and the  other"
   1 " the base AB.":
   1 " the base  of this"
   1 " the  hypotenuse  of  a   right"
   1 " stated as follows:-":
   1 " square on the side"
   1 " square on the base"
   1 " square  is found  by  squaring"
   1 " side using a  formula  derived"
   1 " side BC in a right angled tri-"
   1 " side  BC  and the"
   1 " shown in Lesson 1,":
   1 " right angled triangles.":
   1 " right angled triangle.":
   1 " right angled tri-"
   1 " right   angled  triangle   ABC"
   1 " rectangles.":
   1 " otherwise:-":
   1 " oras  can  be  expressed  as a"
   1 " one decimal place,AC is
   1 " on the  hypotenuse"
   1 " of the other  two  sides  of a"
   1 " of the  square  on"
   1 " of Pythagoras  was discovered."
   1 " of  the  side  BC"
   1 " of  the  hypoten-"
   1 " of  the  base  AB"
   1 " may  be   applied   to   solve"
   1 " lengths of its other two sides.":
   1 " length  5cm, then its  area is"
   1 " learned.":
   1 " its corners  A, B"
   1 " hypotenuse given  the  lengths"
   1 " hypotenuse AC,":
   1 " hypotenuse AC, the  side BC or"
   1 " hypotenuse AC  was"
   1 " how to work  out the length of"
   1 " how these formulae and methods"
   1 " how it is  thought the theorem"
   1 " how  to  use  the  theorem  to"
   1 " how  in the  proof"
   1 " given a more  convenient  form"
   1 " given  the   lengths  of   the"
   1 " from the  one  you   have just"
   1 " formulae-"
   1 " formula:-
   1 " formula:"
   1 " for the  theorem of Pythagoras"
   1 " find the  base or  side  of  a"
   1 " equal  to the area"
   1 " divided  into  two"
   1 " different  problems  involving"
   1 " demonstration  proof  for  the" 
   1 " corners of the triangle:-":
   1 " calculate the  length  of  the"
   1 " better work through the lesson":
   1 " better  go on with the  lesson":
   1 " as follows:-"
   1 " area of the square"
   1 " angled  triangle   given   the"
   1 " angle.":
   1 " angle, the  theorem of Pythag-"
   1 " angle  and  label"
   1 " and C...":
   1 " and  BC=4.   Find"
   1 " again.":
   1 " a side  given  the  lengths of"
   1 " You   were   then   given    a"
   1 " THE OTHER TWO SIDES.
   1 " THE AREA OF THE  SQUARE ON THE"
   1 " SQUARE ROOT of 85, that is ":
   1 " SQUARE ROOT of 80, that is ":
   1 " SQUARE ROOT of 52,which is ":
   1 " OF THE AREAS OF THE SQUARES ON"
   1 " LESSON THREE type:-":
   1 " LESSON  TWO again, ":
   1 " HYPOTENUSE IS EQUAL TO THE SUM"
   1 " AC=";h;"cm, BC=";q;"cm.":
   1 " AC=";h;"cm, AB=";p;"cm.":
   1 " AC  (equal to AC
   1 " AB=";p;"cm, BC=";q;"cm.":
   1 " AB (equal to AB
   1 " 1 decimal place."
   1 " - but not straight away!":
   1  is 52, then AC  is the" 
   1   which is 5x5=25cm
   1   is 85, then AB  is the" 
   1   is 80, then BC  is the" 
   1   is ";h^2