Answer:
1) 3π inches
2) 299.7 square inches
Step-by-step explanation:
Question 1
To find the length of arc ML, we can use the arc length formula:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Arc length}}\\\\s= 2\pi r\left(\dfrac{\theta}{360^{\circ}}\right)\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$s$ is the arc length.}\\\phantom{ww}\bullet\;\textsf{$r$ is the radius.}\\\phantom{ww}\bullet\;\textsf{$\theta$ is the central angle in degrees.}\end{array}}[/tex]
In this case:
Substitute the given values into the formula and solve for s:
[tex]s=2\pi \cdot 18 \left(\dfrac{30^{\circ}}{360^{\circ}}\right)\\\\\\s=36\pi \cdot \dfrac{1}{12}\\\\\\s=3\pi[/tex]
Therefore, the length of arc ML is 3π inches.
[tex]\dotfill[/tex]
Question 2
To find the area of sector JWK, we can use the area of a sector formula:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Area of a sector}}\\\\A=\left(\dfrac{\theta}{360^{\circ}}\right) \pi r^2\\\\\textsf{where:}\\\phantom{ww}\bullet\;\;\textsf{$r$ is the radius.}\\\phantom{ww}\bullet\;\;\textsf{$\theta$ is the angle measured in degrees.}\end{array}}[/tex]
In this case:
Substitute the given values into the formula and solve for A:
[tex]A=\left(\dfrac{106^{\circ}}{360^{\circ}}\right) \pi \cdot 18^2\\\\\\A=\dfrac{53}{180} \pi \cdot 324\\\\\\A=\dfrac{477}{5}\pi\\\\\\A=299.70793915...\\\\\\A=299.7\sf \;(nearest\;tenth)[/tex]
Therefore, the area of sector JWK, rounded to the nearest tenth, is 299.7 square inches.
