anonymous
  • anonymous
Find the area of a regular pentagon with a side of 6 ft. Give the answer to the nearest tenth.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
i believe it would be the area created by 5 isotriangles that have a base of 6
amistre64
  • amistre64
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amistre64
  • amistre64
the length of a side would be s= 6sin(56)/sin(72) \[5*\frac{18sin(56)}{sin(72)}\to\ \frac{90sin(56)}{sin(72)}\] IF i see it right

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amistre64
  • amistre64
my 56 is bad :)
amistre64
  • amistre64
180-70-2 110-2 = 108 108/2 = 54 change 56 to 54
anonymous
  • anonymous
54 doesnt match any of the options i have. Here are my choices: 123.9 ft2 49.5 ft2 61.9 ft2 12.4 ft2
amistre64
  • amistre64
pentagon = 360/5 = 72 for a central angle 180 - 72 = 108; 108/2 = 54 for the base angles law of sine gives us: 6/sin(72) = s/sin(54); therefore; s=6sin(54)/sin(72) Area of a tri = base * height * 1/2 base = 6 height = s sin(54) ... i forgot to include that extra sin(54) part :)
amistre64
  • amistre64
yeah, thats better :) \[Area_P=90sin(54)\frac{sin(54)}{sin(72)}\]
anonymous
  • anonymous
area of 5 equlateral trinagles
anonymous
  • anonymous
(5-2) 180 = 3 X 180 = 540 540 / 5 = 108
anonymous
  • anonymous
108/2 = 54

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