Question

Teorema de Pitágoras

Answer 1

Boa noite, vamos lá resolver essas questões, lembrando que o teorma de pitágoras é:

$a^{2}=b^{2} +c^{2}$

a) $x^{2}=21^{2} +28^{2} \\ x^{2} =441+784\\ x^{2} =1225\\x=\sqrt{1225}\\ x=35$

b) $25^{2} =x^{2} +24^{2} \\ 625=x^{2} +576\\ x^{2} =625-576 \\ x^{2} =49 \\ x=\sqrt{49} \\ x=7$

c) $11^{2} =5^{2} +x^{2} \\ 121=25+x^{2} \\ x^{2} =96 \\ x=\sqrt{96} \\ x=9,80$

d) $x^{2} =10+10 \\ x=\sqrt{100} \\ x=10$

e) $29=5^{2} +x^{2} \\ x^{2} =4 \\ x=\sqrt{4} \\ x=2$

f) $x^{2} =24^{2}+ 32^{2} \\ x^{2} =576+1024 \\ x^{2} =1600 \\ x=\sqrt{1600} \\x=40$