Question

PRECISO DE AJUDAA AQUI
1) Determine o valor de x nos triângulos retângulos:

Answer 1

Resposta:

a)

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

$20^{2} = 4x^{2} + 3x^{2}$

$400 = 7x^{2}$

$x^{2} = 400/7$

$x = \sqrt{400/7}$

$x = 20/ \sqrt{7}$

racionalizando:

$20/\sqrt{7} . \sqrt{7} / \sqrt{7}$ = x

$20\sqrt{7} / \sqrt{49}$ = x

$20\sqrt{7} / 7$ = x

b)

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

$3\sqrt{5}^{2} = 6^{2} + x^{2}$

$9\sqrt{25} = 9.5 =45$

$45 = 36 + x^{2}$

$x^{2} = 9$

$x= \sqrt{9}$

$x = 3$

Espero ter ajudado!!

Answer 2

Caso esteja pelo app, e tenha problemas para visualizar esta resposta, experimente abrir pelo navegador https://brainly.com.br/tarefa/41565374

                                                       

Teorema de Pitágoras

$\large\boxed{\begin{array}{l}\sf a^2=b^2+c^2\\\sf a\longrightarrow hipotenusa\\\sf b\longrightarrow cateto\\\sf c\longrightarrow cateto\\\sf 20^2=(4x)^2+(3x)^2\\\sf400=16x^2+9x^2\\\sf 25x^2=400\\\sf x^2=\dfrac{400}{25}\\\sf x^2=16\\\sf x=\sqrt{16}\\\sf x=4\end{array}}$

$\boxed{\begin{array}{l}\tt b)~\sf(3\sqrt{5})^2=6^2+x^2\\\sf 45=36+x^2\\\sf x^2=45-36\\\sf x^2=9\\\sf x=\sqrt{9}\\\sf x=3\end{array}}$