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anonymous

  • one year ago

WILL MEDAL AND FAN!!! Express answer in exact form. Find the area of the larger segment whose chord is 8" long in a circle with an 8" radius. (Hint: A chord divides a circle into two segments. In problem 1, you found the area of the smaller segment.) and is there any way to put the answer in this format? https://suwannee.owschools.com/media/g_geo_2013/8/q255_x.gif

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  1. anonymous
    • one year ago
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    Thanks Bro

  2. anonymous
    • one year ago
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    I appreciate you helpin'

  3. anonymous
    • one year ago
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    We know that the triangle formed by the two radii and the chord is an equilateral triangle, all angles are 60 degrees, which is 1/6 of 360 degrees : Find area inside the 60 degree arc 1/6*pi8^2 = 35.51 : Find the area of the equilateral triangle 1/2*8*SQRT8^2-4^2 = 27.71 sq/in : Find the area of the shape enclosed by the 60 degree arc and the chord 35.51 - 27.71 = 7.8 sq/in : Find the area of the larger segment pi*8^2 - 7.8 = 193.26 sq/inches NO PROBLEM damn caps

  4. anonymous
    • one year ago
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    Hehehe Lol! I know. It happens to me all the time. It's a real pain

  5. anonymous
    • one year ago
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    ik rite and gtg ill message u when i get back around 1

  6. anonymous
    • one year ago
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    Alright. I appreciate the help

  7. anonymous
    • one year ago
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    i see you later

  8. anonymous
    • one year ago
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    Yeah I still need a little help

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