Find the area of a regular pentagon with a side of 6 ft. Give the answer to the nearest tenth.

ok can't remember how to do this

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  1. A regular pentagon can be drawn as 5 conguent isosceles triangles with a central angle of 72° each, and equal base angles of 54°.
    So just find the area of one of these by first finding the altitude of one of them
    tan 54° = h/3
    h = 3tan54°
    area of one of them = (1/2)base x height
    = (1/2)(6)(3tan54)
    = 9tan 54°
    so the whole pentagon = 45tan 54°
    = appr. 61.9 square ft

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  2. You forgot to give us the lenght of the side of the line that connects from the middle of the pentagon to a side of the pentagon.

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  3. It's there; you just need to recognize it:

    h = 3tan54°

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