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  1. #1
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    Cool geometry homework

    A hexagonal right prism has a volume of 500 cubic inches. If the base is a regular hexagon with a side 4 inches, what is the altitude of the prism ? Round off your answer to two decimal places.
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  2. #2
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    Hello,

    You have to calculate the hexagon area A, because the volume of the prism is altitude*A. As the hexagon is regular, you know that it's composed of 6 identical triangles, isocele in O :



    As the angle at O is 360/6 = 60°, we can state that all triangles are equilateral, because each side is already equal.
    So each side of a triangle will be 4 inches.

    Now let's calculate the length of the altitude from O to [AB], OH, in the triangle ABO, so as to get the area of the hexagon.
    The altitude will also be the bisection of the segment ([AB] for instance) -> H is the midpoint of [AB]
    So by the Pythagor theorem, OH˛=OA˛-AH˛=4˛-2˛=12 -> OH=\sqrt{12}=2 \sqrt{3}
    The area of the triangle ABO is \frac{OH.AB}{2} = \frac{2 \sqrt{3} . 4}{2} = 4 \sqrt{3}

    A, the area of the hexagon will be 6 times ABO's area. So A=24 \sqrt{3} square inches


    Hence the altitude of the prism, h, is : 500=24 \sqrt{3} * h
    -> h=\frac{500}{24 \sqrt{3}}

    I hope there is no mistake :s
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