Question

sendo O ,o centro da circunferência e o triângulo aob e equilátero com altura igual a 3 cm, calcule a área colorida da figura ​

Answer 1

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Área do setor circular com  o ângulo em graus

$\huge\boxed{\boxed{\boxed{\boxed{\sf A_{setor}=\dfrac{\pi\cdot r^2\cdot\alpha}{360^\circ}}}}}$

Altura do triângulo equilátero

$\huge\boxed{\boxed{\boxed{\boxed{\sf h=\dfrac{\ell\sqrt{3}}{2}}}}}$

Área do triângulo equilátero

$\huge\boxed{\boxed{\boxed{\boxed{\sf A=\dfrac{\ell^2\sqrt{3}}{4}}}}}$

$\underline{\rm c\acute alculo~do~lado~do~tri\hat angulo:}\\\sf h=\dfrac{\ell\cdot\sqrt{3}}{2}\\\sf3=\dfrac{\ell\sqrt{3}}{2}\\\sf\ell\sqrt{3}=6\\\sf\ell=\dfrac{6}{\sqrt{3}}\\\sf\ell=\dfrac{6\sqrt{3}}{3}\\\sf\ell=2\sqrt{3}~cm$

$\sf\underline{c\acute alculo~da~\acute area~do~tri\hat angulo:}\\\sf A=\dfrac{\ell^2\sqrt{3}}{4}\\\sf A=\dfrac{(2\sqrt{3})^2\cdot\sqrt{3}}{4}\\\sf A=\dfrac{\diagup\!\!\!4\cdot3\cdot\sqrt{3}}{\diagup\!\!\!4}\\\sf A=3\sqrt{3}~cm^2$

$\underline{\rm c\acute alculo~da~\acute area~do~setor:}\\\sf r=\ell=2\sqrt{3}~~~\alpha=60^\circ\\\sf A=\dfrac{\pi\cdot r^2\cdot\alpha}{360^\circ}\\\sf A=\dfrac{\pi\cdot(2\sqrt{3})^2\cdot\diagup\!\!\!6^1\diagup\!\!\!0}{\diagup\!\!\!\!\!\!36_6\diagup\!\!\!0}\\\sf A=\dfrac{\pi\cdot4\cdot3}{6}=\dfrac{12\pi}{6}=2\pi$

$\sf A_{colorida}=A_{setor}-A_{tri\hat angulo}\\\sf }A_{colorida}=2\pi-3\sqrt{3}\\\sf adote~\pi=3,14~e~\sqrt{3}=1,73\\\sf A_{colorida}=2\cdot3,14-3\cdot1,73\\\sf A_{colorida}=6,28-5,19\\\huge\boxed{\boxed{\boxed{\boxed{\sf A_{colorida}=1,09~cm^2}}}}\blue{\checkmark}$