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10 July, 08:27

What is the approximate area of a regular pentagon with a side lenghtt of 6 feet and a distance from the center to a vertex of 6 feet

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  1. 10 July, 10:11
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    A regular pentagon may be divided into 5 congruent triangles.

    If the side of the pentagon is 6 feet, each base of the triangles will be 6 feet.

    If the distance from the center to a vertex is 6 feet, all the sides of the triangles will 6 feet.

    So there are 5 congruent, equilateral triangles each with side equal to 6 feet.

    The area one of those triangles is calculateb by:

    base * height / 2

    height ^2 = (side length) ^2 - (side length / 2) ^2 = 6^2 - 3^2 = 36 - 9 = 27

    => height = √ (27) ≈ 5.2 feet

    And, using area = base * height / 2,

    area of a triangle = 6 feet * 5.2 feet / 2 = 15.6 feet^2

    And the area of the entire pentagon equals 5 triangles ≈ 5 * 15.6 feet^2 = 78 feet^2

    Answer: 78 feet^2
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