X!),~;K,FD/X ( @ (//(  0/80@0?/8 0( ?G@POWOXOP?H0@/7@O_``h_hPX 0(P_oppxow_g@O/8'' 7p`o?G/7 0??H/8`/o?GOX@H0?/pOW?H/8 7OP?G/7( P0@PX@O0@ 0O @H/@' 00@@P0?'_OP@H( 7/7' ^^^^^^^^^^^^^^^^^^^^^^^^^^TeeBBBBeTT^^^^^^^eBDJ33333JJDBeZ^^^^TD33==#####==3JDBT^^^^eJ=###=3DeZ^^TB3# #=3JBZ^^T3#  #=JBT^^J# Sll #=JBZ^T# %,68??86,/@A*< ##45('.6787-9%:;*<=#$+&,--.,/01)2 3 $%&&'%($!)* ##  !"#     ??dm-^@object0 [oshow_info() x1=200 /*baseline left*/ x2=300 /*baseline right*/ y1=300 /*baseline height*/ y2=225 /*vertex*/00000002cx2mouse_x000000cy2mouse_y000000]base0000000action_draw_linex1y1x2y10000]vertical0000000action_draw_linex2y2x2y10000] hypotenuse0000000action_draw_linex1y1x2y20000[for (i=x1;i<=x2;i+=25) /* baseline */ { draw_line(i,y1,i,y1+x2-x1) /*base verticals */ draw_line(x1,i+y1-x1,x2,i+y1-x1) /*base horizontals */ } for (i=y1;i>=y2;i-=25) /*vertical */ { draw_line(x2-i+y1,y1,x2-i+y1,y2) /*verticals*/ draw_line(x2,i,x2+y1-y2,i) /*horizontals*/ } hyp=sqrt(sqr(y1-y2)+sqr(x2-x1)) alpha=arctan2((y1-y2),(x2-x1)) room_caption="Pythagoras' Theorem horiz.="+string(sqr(x1-x2)/625)+ "squares vert.="+string(sqr(y1-y2)/625)+"squares hyp=" +string((sqr(x1-x2)+sqr(y1-y2))/625)+"squares" for(i=0;i<=hyp/25;i+=1) { draw_line(x1-i*25*sin(alpha),y1-i*25*cos(alpha), /*parallel*/ x2-i*25*sin(alpha),y2-i*25*cos(alpha)) draw_line(x1+i*25*cos(alpha),y1-i*25*sin(alpha), /*perp*/ x1-hyp*sin(alpha)+i*25*cos(alpha),y1-hyp*cos(alpha)-i*25*sin(alpha)) } 0000000room0&        XGame InformationX{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fnil\fcharset0 Arial;}{\f1\fnil Arial;}} {\colortbl ;\red0\green0\blue0;} \viewkind4\uc1\pard\cf1\b\fs32 Pythagoras' Theorem \par \fs24 Tony Forster January 2006 \par \b0 May be copied with acknowledgement \par \par The statement of the Theorem was discovered on a Babylonian tablet circa 1900-1600 B.C. Pythagoras (c.560-c.480 B.C.) is claimed as the first to discover its proof. \par \par \i "\f1 In any right triangle, the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares of the other two sides.\f0 " \par \par \i0 Or as it is most commonly expressed a\'b2 + b\'b2 = c\'b2 where a, b and c are the lengths of the sides. \par \par In this demonstration, use the mouse to move the triangle vertex. This is not a proof, it is only a demonstration. A large number of elegant proofs can be found on the internet.\i\f1 \par } SpritesSounds BackgroundsPathsScripts Fonts Time LinesObjectsobject0Roomsroom0 Game Information Global Game Settings