anonymous
  • anonymous
Find the area of the smaller segment whose chord is 8" long in a circle with an 8" radius.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Directrix
  • Directrix
|dw:1332857218491:dw|
Directrix
  • Directrix
Area of sector ----------- 60/360 = S/ Area of Circle 1/6 = S/(64π) Sector Area = 32π / 3 Area of triangle = 8^2 (√3) / 4 = 16√3 Area of Segment = 32π / 3 - 16√3 Area of Segment = ( 32π - 48√3) / 3 --> written as single fraction
anonymous
  • anonymous
\[area.of.segment = (\frac{\theta}{360} * \pi * r^2) - ( \frac{1}{2}*r^2*\sin \theta)\] \[(\frac{60}{360} * \pi * 8^2) - (\frac{1}{2}*8^2*\sin60)\] \[(\frac{1}{6}*\pi*64) - (\frac{1}{2}*64*\frac{\sqrt{3}}{2})\] \[\frac{32 \pi }{3} - 16\sqrt{3}\] \[\frac{32 \pi - 48\sqrt{3}}{3}\]

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anonymous
  • anonymous
refer the pic
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