-kcapm SVSziN B3\1. Look at the lengths of the sides above. What could you do to the short leg to get the length of the hypotenuse? What could you do to the short leg to get the length of the hypotenuse?(Hint: try using a square root or a whole number!) 2. Move the points around on the 30-60-90 degree right triangle above. Does your conjecture always work?05!  BCC 0=5B#!  AX`//CC/<5 !   short leg.2`//CCCC? %Nb m8 short leg = .4```@@0$y`48T`@p@HH48`B@tL 7PTX@<Length(Segment short leg) = ( Ĵ V0P4LP DRVR@0` ~@@w,O"$ x\ Dh0JF00 C@8@0@<? 0 H( BHD@T 0 p0"4 (g5 !  j"dIHH" CChcICC?<'BRe!  kcscsNI{|CCCC@?sx>!   Ce PadRecent ApplicationsRecent DocumentsRecent ServCMC| S't%!  l"@$I !(CMC|CC?=B!  DI / /J /CCƔ  9M m5 m{!:A}DAB = .4```@@0$y`48T`@p@HH48`B@tL 7PTX@< Angle(DAB) = t short leg) = ( Ĵ V0P4LP DRVR@0` ~@@w,O"$ x\ Dh0JF00 C@8@0@<? 0 H( BHD@T 0 p0"4 ( $8 m4R >AR @ m{!:A}BDA = .4```@@0$y`48T`@p@HH48`B@tL 7PTX@< Angle(BDA) = t short leg) = ( Ĵ V0P4LP DRVR@0` ~@@w,O"$ x\ Dh0JF00 C@8@0@<? 0 H( BHD@T 0 p0"4 ( #F m38 m{!:A}DBA = .4```@@0$y`48T`@p@HH48`B@tL 7PTX@< Angle(DBA) = t short leg) = ( Ĵ V0P4LP DRVR@0` ~@@w,O"$ x\ Dh0JF00 C@8@0@<? 0 H( BHD@T 0 p0"4 ( <5 !   hypotenuse2`//CCƔCC?m`+ <5Bs !  long leg.2`//CCƔCC?dm  ey m7 hypotenuse = .4```@@0$y`48T`@p@HH48`B@tL 7PTX@<Length(Segment hypotenuse) = ( Ĵ V0P4LP DRVR@0` ~@@w,O"$ x\ Dh0JF00 C@8@0@<? 0 H( BHD@T 0 p0"4 (| m6 long leg = = .4```@@0$y`48T`@p@HH48`B@tL 7PTX@<Length(Segment long leg) = = ( Ĵ V0P4LP DRVR@0` ~@@w,O"$ x\ Dh0JF00 C@8@0@<? 0 H( BHD@T 0 p0"4 (