Answer:
[tex]\pi (5\: \sf in)^2\left(\dfrac{30^{\circ}}{360^{\circ}}\right)[/tex]
π(5in)² 30 over 360
Step-by-step explanation:
[tex]\textsf{Area of a sector of a circle}=\left(\dfrac{\theta}{360^{\circ}}\right) \pi r^2[/tex]
(where [tex]\theta[/tex] is the angle and r is the radius)
Given:
- [tex]\theta[/tex] = 30°
- r = 5 in
Substituting these values into the equation:
[tex]\begin{aligned}\implies\textsf{Area} &=\left(\dfrac{30^{\circ}}{360^{\circ}}\right) \pi \cdot (5\: \sf in)^2\\\\ & = \pi (5\: \sf in)^2\left(\dfrac{30^{\circ}}{360^{\circ}}\right)\\\\ & = \pi \cdot 25\:(\sf in^2) \cdot \dfrac{1}{12}\\\\ & = \dfrac{25}{12} \pi \:(\sf in^2) \\\\ & = 6.54\: \sf in^2\:(nearest\:hundredth) \end{aligned}[/tex]
