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A regular triangular pyramid has an altitude of 9m and a volume of 46.8cu. meters. Find the length of the edges and the lateral area of the pyramid.

  • MATH -

    please answer this

  • MATH -

    V = 1/3 base * height
    area of the base = 46.8/3 = 15.6
    the base is a equilateral triangle.
    the area of which is sqrt(3)/4 * edge^2
    edge = sqrt(15.6*4/sqrt 3) = 6.002 m
    The mid-point of the triangle is 2/3 of the way from the angle to the base.
    sqrt(3)/3 * 6 = 2 sqrt 3 ~ 3.46 m
    the length of the other 3 edges then using the Pythagorean theorem.
    12+81 = edge^2
    sqrt 93 ~ 9.64 m
    the slant height...
    3^2 + slant^2 = 93
    slant height = sqrt 84 ~ 9.165 m

    lateral area = 3/2 * 6.002 * 9.165 = 82.5 m^2

  • MATH solid mensuration -

    V = 1/3 base * height
    area of the base = 46.8/3 = 15.6
    the base is a equilateral triangle.
    the area of which is sqrt(3)/4 * edge^2
    edge = sqrt(15.6*4/sqrt 3) = 6.002 m
    The mid-point of the triangle is 2/3 of the way from the angle to the base.
    sqrt(3)/3 * 6 = 2 sqrt 3 ~ 3.46 m
    the length of the other 3 edges then using the Pythagorean theorem.
    12+81 = edge^2
    sqrt 93 ~ 9.64 m
    the slant height...
    3^2 + slant^2 = 93
    slant height = sqrt 84 ~ 9.165 m

    lateral area = 3/2 * 6.002 * 9.165 = 82.5 m^2

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