✒️SEGMENT
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[tex] \large\underline{\mathbb{PROBLEM}:} [/tex]
- Find the area of shaded region given m∠ABC = 90° and mBC = 6cm. Show your complete solution.
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[tex] \large\underline{\mathbb{ANSWER}:} [/tex]
[tex] \qquad \Large \: \rm{\approx 10.26 \: 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{90\degree}{\,360\degree} \cdot \pi(6)^2 \: cm^2 \\ [/tex]
- [tex] A_{\,Sector} = \frac{\,1\,}{4} \cdot 36\pi \: cm^2 \\ [/tex]
- [tex] A_{\,Sector} = \frac{\,36\pi\,}{4} \: cm^2 \\ [/tex]
- [tex] A_{\,Sector} = 9\pi \: cm^2 [/tex]
» Let 3.14 be the approximate value of pi.
- [tex] A_{\,Sector} \approx 9(3.14) \: cm^2 [/tex]
- [tex] A_{\,Sector} \approx 28.26 \: 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(90\degree) \: cm^2 \\ [/tex]
- [tex] A_{\,Triangle} = \frac{\,1\,}{2} \cdot 36 \: cm^2 \\ [/tex]
- [tex] A_{\,Triangle} = 18\: 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 28.26 \, cm^2 - 18\,cm^2 [/tex]
- [tex] A_{\,Segment} \approx 10.26 \, cm^2 [/tex]
[tex] \therefore [/tex] The area of the shaded region is about 10.26 sq. centimeters
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