✒️SEGMENTS
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[tex] \large\underline{\mathbb{ANSWER}:} [/tex]
[tex] \qquad \Large \: \rm{\approx 66.27 \: sq. \: cm} [/tex]
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[tex] \large\underline{\mathbb{SOLUTION}:} [/tex]
» Take one segment. Since there are three of them, each would have a 120° angle.
- 3 identical Segments
- Each subtended arcs measures 120°
» Find its area. But first, find the area of the sector that is bounded with the the two radii and an intercepted arc.
[tex] \begin{align} & \bold{Formula:} \\ & \quad \boxed{\rm A_{\,Sector} = \frac{\theta}{\,360\degree} \cdot \pi r^2} \end{align} [/tex]
- [tex] A_{\,Sector} = \frac{120\degree}{\,360\degree} \cdot \pi(6)^2 \: cm^2 \\ [/tex]
- [tex] A_{\,Sector} = \frac{\,1\,}{3} \cdot 36\pi \: cm^2 \\ [/tex]
- [tex] A_{\,Sector} = \frac{\,36\pi\,}{3} \: cm^2 \\ [/tex]
- [tex] A_{\,Sector} = 12\pi \: cm^2 [/tex]
» Let 3.14 be the approximate value of pi.
- [tex] A_{\,Sector} \approx 12(3.14) \: cm^2 [/tex]
- [tex] A_{\,Sector} \approx 37.68 \: cm^2 [/tex]
» Find the area of the triangle that is bounded with the two radii and a chord.
[tex] \begin{align} & \bold{Formula:} \\ & \quad \boxed{\rm A_{\,Triangle} = \frac{\,1\,}{2} \cdot r^2\sin\theta} \end{align} [/tex]
- [tex] A_{\,Triangle} = \frac{\,1\,}{2} \cdot (6)^2 \sin(120\degree) \: cm^2 \\ [/tex]
- [tex] A_{\,Triangle} = \frac{\,1\,}{2} \cdot 36 \sin(120\degree) \: cm^2 \\ [/tex]
- [tex] A_{\,Triangle} \approx \frac{\,1\,}{2} \cdot 31.18 \: cm^2 \\ [/tex]
- [tex] A_{\,Triangle} \approx 15.59\: cm^2 [/tex]
» Find the area of the segment that is bounded with a chord and an arc.
[tex] \begin{align} & \bold{Formula:} \\ & \quad \boxed{\rm A_{\,Segment} = A_{\,Sector} - A_{\,Triangle}} \end{align} [/tex]
- [tex] A_{\,Segment} \approx 37.68 \, cm^2 - 15.59\,cm^2 [/tex]
- [tex] A_{\,Segment} \approx 22.09 \, cm^2 [/tex]
» Multiply the approximate area by three to find the area of all the shaded regions.
- [tex] 22.09(3) \: cm^2 \approx 66.27 \: cm^2 [/tex]
[tex] \therefore [/tex] The area of all the shaded regions are about 66.27 sq. centimeters
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