Question

Por favor, alguém pode resolver essas questões? É urgente

Answer 1

Explicação passo-a-passo:

1)

a)

$\Large \bf \: Teorema \: de \: Pitágoras : \\ \\ \Large \boxed{ \bf \: {a}^{2} = {b}^{2} + {c}^{2} } \\ \\ {x}^{2} = {1}^{2} + {3}^{2} \\ \\ {x}^{2} = 1 + 9 \\ \\ {x}^{2} = 10 \\ \\ \Large \boxed{ \green{x = \bf \: \sqrt{10} }}$

b)

${x}^{2} = ( {2 \sqrt{3}) }^{2} + {2}^{2} \\ \\ {x}^{2} = 4 \times 3 + 4 \\ \\ {x}^{2} = 12 + 4 \\ \\ {x}^{2} = 16 \\ \\ x = \sqrt{16} \\ \\ \Large \boxed{ \green{ x \bf \: = 4}}$

c)

${x}^{2} = {1}^{2} + (2 \sqrt{2} ) {}^{2} \\ \\ {x}^{2} = 1 + 4 \times 2 \\ \\ {x}^{2} = 1 + 8 \\ \\ {x}^{2} = 9 \\ \\ x = \sqrt{9} \\ \\ \Large \boxed{ \green{x = \bf 3}}$

d)

${( \sqrt{17} )}^{2} = {x}^{2} + {1}^{2} \\ \\ 17 = {x}^{2} + 1 \\ \\ 17 + 1 = {x}^{2} \\ \\ {x}^{2} = 18 \\ \\ {x} = \sqrt{18} \\ \\ x = \sqrt{9 \times 2} \\ \\ x = \sqrt{9} \times \sqrt{2} \\ \\ \Large \boxed{ \green{x \bf = 3 \sqrt{2} }}$

e)

${10}^{2} = {x}^{2} + {6}^{2} \\ \\ 100 = {x}^{2} + 36 \\ \\ 100 - 36 = {x}^{2} \\ \\ {x}^{2} = 64 \\ \\ x = \sqrt{64} \\ \\ \Large \boxed{ \green{x = \bf 8}}$

f)

${x }^{2} = {6}^{2} + ( {2 \sqrt{7} )}^{2} \\ \\ {x}^{2} = 36 + 4 \times 7 \\ \\ {x}^{2} = 36 + 28 \\ \\ {x}^{2} = 64 \\ \\ x = \sqrt{64} \\ \\ \Large \boxed{ \green{x = \bf \: 8}}$

$\Large \boxed{ \underline{ \blue{ \bf \: Bons \: Estudos!}}} \\ \\ \Large \boxed{ \underline{ \bf \: 23/05/2021}}$

Answer 2

01 - Teorema de Pitágoras

h² = c1² + c2²

( hipotenusa o quadrado = cateto 1 ao quadrado + cateto 2 ao quadrado).

a)

${x}^{2} = {1}^{2} + {3}^{2} \\ {x}^{2} = 1 + 9 \\ {x}^{2} = 10 \\ \large x = \sqrt{10}$

b)

${x}^{2} = (2 \sqrt{3} ) {}^{2} + {2}^{2} \\ {x}^{2} = 4 \times 3 + 4 \\ {x}^{2} = 16 \\ x = \sqrt{16} \\ \large \: x = 4$

c)

${x}^{2} = {1 }^{2} + (2 \sqrt{2} ) {}^{2} \\ {x}^{2} = 1 + 4 \times 2 \\ {x }^{2} = 9 \\ x = \sqrt{9} \\ \large \: x = 3$

d)

$( \sqrt{17} ) {}^{2} = {x}^{2} + {1}^{2} \\ 17 = {x}^{2} + 1 \\ {x}^{2} = 17 - 1 \\ {x}^{2} = 16 \\ x = \sqrt{16} \\ \large x = 4$

e)

${10}^{2} = {x}^{2} + {6}^{2} \\ 100 = {x}^{2} + 36 \\ {x}^{2} = 100 - 36 \\ {x}^{2} = 64 \\ x = \sqrt{64} \\ \large \: x = 8$

f)

${x}^{2} = {6}^{2} + (2 \sqrt{7}) {}^{2} \\ {x}^{2} = 36 + 4 \times 7 \\ {x}^{2} = 64 \\ x = \sqrt{64} \\ \large \: x = 8$