Answer:
[tex]A_\text{sector} = \dfrac{49}{9}\pi\text{ units}^2\\\\ \text{ } \ \ \ \ \ \ \ \,\approx 17.1\text{ units}^2[/tex]
Step-by-step explanation:
We can find the area of the sector using the equation:
[tex]A_\text{sector} = \dfrac{\theta}{360\°} \cdot A_\text{circle}[/tex]
where:
- [tex]\theta[/tex] = central angle of the sector
We can also substitute in the area of a circle formula to get:
[tex]A_\text{sector} = \dfrac{\theta}{360\°} \cdot \pi r^2[/tex]
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Notice that the number:
[tex]\dfrac{\theta}{360\°}[/tex]
tells us what fraction of the circle's total area that the sector takes up.
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We are given the following information:
- [tex]\theta = \overset{\frown}{AB} = 40\°[/tex]
- [tex]r = BC = 7[/tex]
Plugging these into the sector area equation, we get:
[tex]A_\text{sector} = \dfrac{40\°}{360\°} \cdot \pi(7^2)[/tex]
[tex]A_\text{sector} = \dfrac{1}{9} \cdot 49\pi[/tex]
[tex]A_\text{sector} =\boxed{ \dfrac{49}{9}\pi\text{ units}^2}\\\\ \text{ } \ \ \ \ \ \ \ \,\approx \boxed{17.1\text{ units}^2}[/tex]