Matemática, perguntado por kaylanegames2134, 6 meses atrás

determine as medidas .
se alguém souber do assunto, me ajude.​

Anexos:

Soluções para a tarefa

Respondido por CyberKirito
3

\large\boxed{\begin{array}{l}\underline{\rm 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{m\cdot n}\\\rm 5)~\red{\sf a}\cdot\sf h=\blue{\sf b}\cdot\green{\sf c}\end{array}}

\large\boxed{\begin{array}{l}\rm 1)~\sf 6^2=10n\\\sf 10n=36\\\sf n=\dfrac{36}{10}=3,6\\\sf  m=10-n\\\sf m=10-3,6\\\sf m=6,4\\\rm 2)~\sf a~outra~projec_{\!\!,}\tilde ao~mede~ 20-7,2=12,8\\\sf b^2=20\cdot12,8\\\sf b^2=256\\\sf b=\sqrt{256}\\\sf b=16\\\sf 20\cdot h=12\cdot b\\\sf 20\cdot h=12\cdot16\\\sf h=\dfrac{\backslash\!\!\!4\cdot3\cdot16}{\backslash\!\!\!4\cdot5}\\\sf h=\dfrac{48}{5}\\\sf h=9,6\end{array}}

\large\boxed{\begin{array}{l}\rm 3)~\sf h^2=1\cdot3\\\sf h=\sqrt{3}\\\sf a=1+3=4\\\sf c^2=4\cdot1\\\sf c^2=4\\\sf c=\sqrt{4}\\\sf c=2\\\sf a\cdot h=b\cdot c\\\sf 4\cdot\sqrt{3}=b\cdot 2\\\sf 2b=4\sqrt{3}\\\sf b=\dfrac{4\sqrt{3}}{2}\\\sf b=2\sqrt{3}\end{array}}

Anexos:

kaylanegames2134: vlw, mn me ajudou muito.
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