A circular sector has central angle pi/6 radians and radius 4 m. Find the arc length s and the sector area A, rounded to three decimal places.

Step-by-step solution

1. 1

   Write the two sector formulas. For a central angle $\theta$ in radians,

   $s=r\theta,\qquad A=\frac12 r^2\theta$

   They are simply the fraction $\theta/(2\pi)$ of the full circumference $2\pi r$ and the full area $\pi r^2$.

2. 2

   Compute the arc length.

   $s=r\theta=4\left(\frac{\pi}{6}\right)=\frac{2\pi}{3}\approx 2.094\ \text{m}$

3. 3

   Compute the sector area. Square the radius first, then halve:

   $A=\frac12 r^2\theta=\frac12(4)^2\left(\frac{\pi}{6}\right)=\frac12(16)\left(\frac{\pi}{6}\right)=\frac{4\pi}{3}\approx 4.189\ \text{m}^2$

4. 4

   State both rounded answers.

   $\boxed{s\approx 2.094\ \text{m},\qquad A\approx 4.189\ \text{m}^2}$

5. 5

   Check with the relation A = ½rs. Eliminating $\theta$ gives $A=\tfrac12 rs$, and $\tfrac12(4)\left(\tfrac{2\pi}{3}\right)=\tfrac{4\pi}{3}$, exactly the area found above.

6. 6

   Compare with the π/3, r = 2 ft case. There the arc was also $\tfrac{2\pi}{3}$, because halving $\theta$ and doubling $r$ leaves $r\theta$ unchanged. The area, however, depends on $r^2\theta$, so doubling $r$ and halving $\theta$ doubles it — from $\tfrac{2\pi}{3}$ to $\tfrac{4\pi}{3}$. Equal arcs do not mean equal sector areas.