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Dedução da altura do triângulo equilátero
[tex]\underline{\sf observe~a~figura~que~anexei.}\\\sf no~tri\hat angulo~equil\acute atero~a~mediana,a~bissetriz~e~a~mediatriz~coincidem.\\\sf usando~o~teorema~de~Pit\acute agoras~no~\triangle AHC~temos:\\\sf \ell^2=\red{\sf h}^2+\bigg(\dfrac{\ell}{2}\bigg)^2\\\sf\red{\sf h}^2+\dfrac{\ell^2}{4}=\ell^2\\\sf\red{\sf h^2}=\ell^2-\dfrac{\ell^2}{4}\\\sf\red{\sf h}=\dfrac{4\ell^2-\ell^2}{4}\\\sf\red{\sf h^2}=\dfrac{3\ell^2}{4}\\\sf h=\dfrac{\sqrt{3\ell^2}}{\sqrt{4}}[/tex]
[tex]\huge\boxed{\boxed{\boxed{\boxed{\sf\red{\sf h}=\dfrac{\ell\sqrt{3}}{2}}}}}\blue{\checkmark}[/tex]
