Question

utiliza el teorema de pitágoras

Answer 1

Partiendo del Teorema de Pitágonas y siendo "a" y "b" los cateto y "c" la hipotenusa:

$Hipotenusa^{2} = Cateto_{1} ^{2} +Cateto_{2} ^{2}\\c^{2} =a^{2} +b^{2}$

a)

$c^{2} =2^{2} +8^{2} \\c^{2} =4+64\\c^{2} =68\\c=\sqrt{68} =8,246$

b)

$19^{2} = 13^{2} +b^{2} \\361=169+b^{2}\\361-169=b^{2}\\192=b^{2}\\\sqrt{192} =b\\b=13,856$

c)

$14^{2} =5^{2} +b^{2}\\196=25+b^{2}\\196-25=b^{2}\\171=b^{2}\\b=\sqrt{171} =13,077$

Espero haberte ayudado.

Answer 2

1)

${c}^{2} = {a}^{2} + {b }^{2} \\ {c}^{2} = {2}^{2} + {8}^{2} \\ c = \sqrt{4 + 64} = \sqrt{68} = \sqrt{4 \times 17 } \\ c = 2 \sqrt{17}$

2)

${b}^{2} = {c}^{2} - {a}^{2} \\ {b}^{2} = {19}^{2} - {13}^{2} =361 - 169 \\ b = \sqrt{192} = 8 \sqrt{3}$

3)

${b}^{2} = {c}^{2} - {a}^{2} \\ {b}^{2} = {14}^{2} - {5}^{2} = 196 - 25 \\ b = \sqrt{171} = 3 \sqrt{19}$