Question

determine as incógnitas
alguém indica um app para estes cálculos? preciso das contas. ​

Answer 1

Resposta:

Explicação passo-a-passo:

Sendo sincero, app não vai te ajudar.

Tem que fazer as contas para memorizar melhor como fazer.

Veja a imagem abaixo sobre relações métricas do triangulo retângulo

espero ter ajudado!

Answer 2

$\large\boxed{\begin{array}{l}\underline{\sf Relac_{\!\!,}\tilde oes~m\acute etricas~no~tri\hat angulo~ret\hat angulo}\\\rm 1)~\red{\sf a^2}=\blue{\sf b^2}+\green{\sf c^2}\\\rm 2)~\blue{\sf b^2}=\red{\sf a}\cdot\pink{\sf m}\\\rm 3)~\green{\sf c^2}=\red{\sf a}\cdot\pink{\sf n}\\\rm 4)~\sf h^2=\pink{\sf m\cdot n}\\\rm 5)~\red{\sf a}\cdot h=\blue{\sf b}\cdot\green{\sf c}\end{array}}$

$\boxed{\begin{array}{l}\tt a)~\sf y=3+9=12\\\sf h^2=3\cdot9\\\sf h=\sqrt{3\cdot9}\\\sf h=3\sqrt{3}\\\sf z^2=12\cdot3\\\sf z^2=36\\\sf z=\sqrt{36}\\\sf z=6\\\sf y\cdot h=z\cdot x\\\sf 12\cdot3\sqrt{3}=6\cdot x\\\sf 6x=36\sqrt{3}\\\sf x=\dfrac{36\sqrt{3}}{6}\\\sf x=6\sqrt{3}\end{array}}$

$\boxed{\begin{array}{l}\tt b)~\sf 8^2=16\cdot n\\\sf 16n=64\\\sf n=\dfrac{64}{16}\\\sf n=4\\\sf m+n=16\\\sf m+4=16\\\sf m=16-4\\\sf m=12\end{array}}$

$\boxed{\begin{array}{l}\tt c)~\sf (2\sqrt{6})^2=x\cdot3\\\sf3x=24\\\sf x=\dfrac{24}{3}\\\sf x=8\\\sf a~outra~projec_{\!\!,}\tilde ao~vale~8-3=5\\\sf y^2=5\cdot3\\\sf y^2=15\\\sf y=\sqrt{15}\end{array}}$

$\boxed{\begin{array}{l}\tt d)~\sf x^2+20^2=25^2\\\sf x^2+400=625\\\sf x^2=625-400\\\sf x^2=225\\\sf x=\sqrt{225}\\\sf x=15\\\sf 25\cdot y=20\cdot x\\\sf 25y=20\cdot15\\\sf 25y=300\\\sf y=\dfrac{300}{25}\\\sf y=12\end{array}}$

$\boxed{\begin{array}{l}\tt e)~\sf a=64+36=100\\\sf h^2=36\cdot 64\\\sf h=\sqrt{36\cdot64}\\\sf h=6\cdot8\\\sf h=48\\\sf b^2=a\cdot36\\\sf b^2=100\cdot36\\\sf b=\sqrt{100\cdot36}\\\sf b=10\cdot6\\\sf b=60\\\sf a\cdot h=b\cdot c\\\sf 100\cdot48=60\cdot c\\\sf 60c=4800\\\sf c=\dfrac{4800}{60}\\\sf c=80\end{array}}$

$\boxed{\begin{array}{l}\tt f)~\sf 12^2=9\cdot x\\\sf 9x=144\\\sf x=\dfrac{144}{9}\\\sf x=16\\\sf a~hipotenusa~mede~16+9=25\\\sf y^2=25\cdot 9\\\sf y=\sqrt{25\cdot9}\\\sf y=5\cdot3\\\sf y=15\end{array}}$

$\boxed{\begin{array}{l}\tt g)~\sf a~hipotenusa~mede~16-\backslash\!\!\!y+\backslash\!\!\!y=16\\\sf 8^2=16\cdot y\\\sf 16y=64\\\sf y=\dfrac{64}{16}\\\sf y=4\\\sf a~outra~projec_{\!\!,}\tilde ao~vale~16-4=12\\\sf x^2=12\cdot4\\\sf x^2=3\cdot4\cdot4\\\sf x=\sqrt{3\cdot16}\\\sf x=4\sqrt{3}\\\sf 16\cdot x=8\cdot z\\\sf 8z=16\cdot4\sqrt{3}\\\sf 8z=64\sqrt{3}\\\sf z=\dfrac{64\sqrt{3}}{8}\\\sf z=8\sqrt{3}\end{array}}$

$\boxed{\begin{array}{l}\tt h)~\sf n=14-5=9\\\sf h^2=5\cdot9\\\sf h=\sqrt{5\cdot9}\\\sf h=3\sqrt{5}\\\sf c^2=14\cdot9\\\sf c=\sqrt{14\cdot9}\\\sf c=3\sqrt{14}\end{array}}$

$\boxed{\begin{array}{l}\tt i)~\sf a~outra~projec_{\!\!,}\tilde ao~vale~ 10-7=3\\\sf x^2=3\cdot7\\\sf x^2=21\\\sf x=\sqrt{21}\end{array}}$

$\boxed{\begin{array}{l}\tt j)~\sf o~cateto~maior~do~tri\hat angulo~mede\\\sf 18-x.\\\sf x^2=6^2+(18-x)^2\\\sf \diagup\!\!\!\!\!x^2=36+324-36x+\diagup\!\!\!\!\!x^2\\\sf 36x=360\\\sf x=\dfrac{360}{36}\\\sf x=10\end{array}}$