+i capmSVSz;0/H $IKM 8l&^_1. Look at the lengths of the two legs in the triangle above. What do you notice? Move the points on the triangle around and see if your conjecture always works. 2. How can you classify the triangle above by its sides? by its angles? 3. What is the relationship between the legs and the hypotenuse? (Hint: use a whole number or a square root) 4. Is this relationship always true? Move the points on the triangle around to test your conjecture. ,b 3A 3BEA EB*|0 " !!"PdHH45-45-90 degree triangle I!$JJ(!!"* "!  BDC h"m!!  AdJ(HH*CC ~ !  c1II / CCC!?5%F?5%Fg" !   second leg2`//CCDC? )$8 m2%%;% ,d second leg = .4```@@0$y`48T`@p@HH48`B@tL 7PTX@<Length(Segment second leg) = ( Ĵ V0P4LP DRVR@0`4 ~@@w,O"$ x\ Dh0JF00 C@8@0@<? 0 H( BHD@T 0 p0"4 (g'm Doc!  kFile Apps Graphics Apps Math Apps Network Apps UCCNCC?|hmH!   C$HICB {g" !   hypotenuse2`//CBDC?L+ {g"m !   first leg.2`//CBCC?>E0n 9M m3 first leg = .4```@@0$y`48T`@p@HH48`B@tL 7PTX@<Length(Segment first leg) = ( Ĵ V0P4LP DRVR@0`4 ~@@w,O"$ x\ Dh0JF00 C@8@0@<? 0 H( BHD@T 0 p0"4 ( K_% m1 $H hypotenuse = .4```@@0$y`48T`@p@HH48`B@tL 7PTX@<Length(Segment hypotenuse) = ( Ĵ V0P4LP DRVR@0`4 ~@@w,O"$ x\ Dh0JF00 C@8@0@<? 0 H( BHD@T 0 p0"4 (