Question

Determine a hipotenusa de um triângulo retângulo cujo perímetro é 12 e a altura relativa a hipotenusa vale 14/5.

Answer 1

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

$\boxed{\begin{array}{l}\tt Dados:\\\sf a+b+c=12\\\sf h=\dfrac{14}{5}\\\sf a=?\end{array}}$

$\sf a+b+c=12\\\sf a^2=b^2+c^2\\\sf a\cdot h=b\cdot c\\\sf a\cdot\dfrac{14}{5}=b\cdot c\\\sf a+b+c=12\implies b+c=12-a\\\sf a^2=b^2+c^2\\\sf b^2+c^2=(b+c)^2-2bc\\\sf a^2=(12-a)^2-2\cdot\dfrac{14a}{5}$

$\sf \diagup\!\!\!a^2=144-24a+\diagup\!\!\!a^2-\dfrac{28a}{5}\\\sf0=720-120a-28a\\\sf 120a+28a=720\\\sf148a=720\\\sf a=\dfrac{720\div4}{148\div4}\\\huge\boxed{\boxed{\boxed{\boxed{\sf a=\dfrac{180}{37}}}}}\blue{\checkmark}$