✒️SEGMENT
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[tex] \large\underline{\mathbb{ANSWER}:} [/tex]
[tex] \qquad \Large \: \rm{\approx 9.48 \: sq. \: cm.} [/tex]
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[tex] \large\underline{\mathbb{SOLUTION}:} [/tex]
» Find the area of the sector that is bounded with 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{\,100\degree}{360\degree} \cdot \pi (5)^2 \: cm^2 \\ [/tex]
- [tex] A_{\,Sector} = \frac{5}{\,18\,} \cdot 25 \pi \: cm^2 \\ [/tex]
- [tex] A_{\,Sector} = \frac{\,125\pi \,}{18} \: cm^2 \\ [/tex]
- [tex] A_{\,Sector} \approx 6.94\pi \: cm^2 \\ [/tex]
» Let 3.14 be the approximate value of pi.
- [tex] A_{\,Sector} \approx 6.94(3.14) \: cm^2 \\ [/tex]
- [tex] A_{\,Sector} \approx 21.79 \: cm^2 \\ [/tex]
» Find the area of the triangle which is 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 (5)^2 \sin(100\degree)\\ [/tex]
- [tex] A_{\,Triangle} = \frac{\,1\,}{2} \cdot 25 \sin(100\degree)\\ [/tex]
- [tex] A_{\,Triangle} \approx \frac{\,1\,}{2} \cdot 24.62\\ [/tex]
- [tex] A_{\,Triangle} \approx 12.31 \\ [/tex]
» Find the area of the segment which is bounded by a chord and a subtended arc.
[tex] \begin{align} & \bold{Formula:} \\ & \quad \boxed{\rm A_{\,Segment} = A_{\,Sector} - A_{\, Triangle} } \end{align} [/tex]
- [tex] A_{\,Segment} \approx 21.79 \: cm^2 - 12.31 \: cm^2 [/tex]
- [tex] A_{\,Segment} \approx 9.48 \: cm^2 [/tex]
[tex] \therefore [/tex] The area of the shaded region is about 9.48 sq. centimeters
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