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terricherri

  • 4 years ago

Find the area of an equilateral triangle (regular 3-gon) with the given measurement. 9-inch perimeter A = sq. in Help me my answer needs to be in square root.

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  1. terricherri
    • 4 years ago
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    i have 9/3sqrt4 but thats not right

  2. id21
    • 4 years ago
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    P=9in a=9/3=3in High of triangle (all corners are 60 degrees):\[h=a \times \sin60^o=a \sqrt3 /2\] \[Area=a \times h/2=a^2\sqrt3/2=9\sqrt{3}/2}\]

  3. id21
    • 4 years ago
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    \[Area=a \times h/2=a^2 \sqrt3/2=9\sqrt{3}/2\]

  4. terricherri
    • 4 years ago
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    so its 9sqrt3/2

  5. Harkirat
    • 4 years ago
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    in case of an equilateral triangle, perimeter = 3 * side so side = perimeter/3 so each side = 9 in/3 = 3 in root(3) Area of an equilateral triangle = --------- * s² where s = length of side 4 root(3) 9 root(3) so area = --------- * (3cm)² = ---------- cm² 4 4

  6. id21
    • 4 years ago
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    Oh, yes, I miss one divider 2. Answer is area is \[9\sqrt3/4\] in^2

  7. id21
    • 4 years ago
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    \[Area=a \times h/2=a^2\sqrt3/4=9\sqrt{3}/4\]

  8. terricherri
    • 4 years ago
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    Find the area of an equilateral triangle (regular 3-gon) with the given measurement. 6-inch apothem A = sq. in. for this one i havwe 144sqrt2

  9. terricherri
    • 4 years ago
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    does anyone know that

  10. id21
    • 4 years ago
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    I get \[108\sqrt3\]

  11. id21
    • 4 years ago
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    must check it again

  12. id21
    • 4 years ago
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    Apothem is 1/3 rd from high, right? Then h=3*6=18 in

  13. id21
    • 4 years ago
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    \[Area=a \times h /2=|a=2h/\sqrt3|=2h^2/(\sqrt3 \times 2)=324/\sqrt3=108\sqrt3\]

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