Question

em cada caso, calcule a area do setor circular de raio r e angulo central de medida teta A) r=4m e teta=30 grausB) r=9dm e teta=120 graus

Answer 1

A=πR²θ/360
a)
A= π.(4)².30/360
A= π.16.30/360
A=480π/360 : 10
A=48π/36
A=(12.4)π/(12.3)
A=4π/3 m² 

b) A=πr²θ/360
  A=π.(9)².(120)/360
A=81π.120/360
A=81π.1/3
A=81π/3
A=27π dm

C) A=$\pi$.(12)².45/360
A=144$\pi$.45/360
A=6480$\pi$/360
A=18$\pi$m²

Espero ter ajudado.

Answer 2

$\text{Área do setor circular:}$

$\text{A} = \frac{\pi \cdot {r}^{2} \cdot \theta }{360°} \\ \\ \text{A} = \text{Área} \\ \pi = \text{pi} \\ r = \text{raio} \\ \theta = \text{ângulo central}$

$\bf{Resolução:}$

$a) \: r = 4 \: m \: e \: \theta = 30° \\ \\ \text{A} = \frac{\pi \cdot {r}^{2} \cdot \theta }{360°} \\ \\\text{A} = \frac{\pi \cdot {4}^{2} \cdot30 }{360} \\ \\ \text{A} = \frac{\pi \cdot16 \cdot30 }{360} \\ \\\text{A} = \frac{ {480\pi}^{ \color{red} { \div 120}} }{ {360}^{ \color{red} { \div 120}} } \\ \\ \bf{A} = \frac{4\pi}{3} \: {m}^{2}$

$b) \: r = 9\: dm \: e \: \theta = 30° \\ \\ \text{A} = \frac{\pi \cdot {r}^{2} \cdot \theta }{360°} \\ \\\text{A} = \frac{\pi \cdot {9}^{2} \cdot120 }{360} \\ \\ \text{A} = \frac{\pi \cdot81 \cdot30 }{360} \\ \\\text{A} = \frac{ {9720} \pi }{ {360} } \\ \\ \bf{A} = 27\pi \: {dm}^{2}$