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Jonatan Emmanuel Guerrero Nuñez
Kevin de Jesús González Martínez
Maestro: Orlando Ayala
3° B
 1) A= 2 B=5 C=?
22+52=h2
4+25=h2
29=h2 2 ?
√29=5.38
3
 2) A=3.4 B=1.2 C=?
3.42+1.22=h2
11.56+1.44=h2
13=h2 3.4 ?
√13=3.60
1.2
 3) B=1/2 C=3 A=?
.52+32=h2
.25-9=h2
8.75=h2 ? 3
√8.75=2.95
.5
 4) C=3/4 B=1/3 A=?
.752+.332=h2
.5625-.1089=h2
.4536=h2 ? .75
√.4536=0.6734
.33
 5) C=5.8 A=4.1 B=?
5.82+4.12=h2
33.64-16.81=h2
16.83=h2 4.1 5.8
√16.81=4.10
?

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Teorema de pitagoras

  • 1. Jonatan Emmanuel Guerrero Nuñez Kevin de Jesús González Martínez Maestro: Orlando Ayala 3° B
  • 2.  1) A= 2 B=5 C=? 22+52=h2 4+25=h2 29=h2 2 ? √29=5.38 3
  • 3.  2) A=3.4 B=1.2 C=? 3.42+1.22=h2 11.56+1.44=h2 13=h2 3.4 ? √13=3.60 1.2
  • 4.  3) B=1/2 C=3 A=? .52+32=h2 .25-9=h2 8.75=h2 ? 3 √8.75=2.95 .5
  • 5.  4) C=3/4 B=1/3 A=? .752+.332=h2 .5625-.1089=h2 .4536=h2 ? .75 √.4536=0.6734 .33
  • 6.  5) C=5.8 A=4.1 B=? 5.82+4.12=h2 33.64-16.81=h2 16.83=h2 4.1 5.8 √16.81=4.10 ?