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Math Help - Need help please

  1. #1
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    Need help please

    1) What is the area of the square circumscribed by the circle with a radius of 10*sqrt(2) units?



    2) If the smaller angles in an obtuse triangle are x and 2x, what is a possible value for x?
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  2. #2
    Member Danshader's Avatar
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    1)
    Well since the square contains the circle of radius 10 sqrt(2) hence the length of one side of the square is 2 times the radius of the circle (diameter) which is 20sqrt(2), from this you can find the area of the square:

    Area of square = (length of one side)^2
    = [20 sqrt(2)]^2
    = 400(2)
    = 800

    2)
    for an obtuse triangle, the obtuse angle must be larger than 90 degree and smaller than 180 degree.

    From an internal angle of a triangle which sums up to 180 degree,

    90 < obtuse angle < 180
    -90 < -obtuse angle < -180
    180 < -obtuse angle <90
    180 - 180 <180 - obtuse angle <180 - 90
    0 < 180 - obtuse angle < 90

    and we know that:
    obtuse angle + x + 2x = 180
    3x = 180 - obtuse angle

    therefore;

    0 < 3x < 90

    where 3x can take any value in the range such as 80
    which gives us x=26.6667 degree ,2x=53.3333 degree and the obtuse angle = 100
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