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Math Help - semi circle question

  1. #1
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    semi circle question

    Let the interval [-r,r] be the base of a semicircle. If a point is selected at random from this interval, assign a probability to the event that the length of the perpendicular segment from the point to the semicircle is less than r/2.
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  2. #2
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    Quote Originally Posted by calculuskid1 View Post
    Let the interval [-r,r] be the base of a semicircle. If a point is selected at random from this interval, assign a probability to the event that the length of the perpendicular segment from the point to the semicircle is less than r/2.
    You can use Pythagoras' theorem to calculate the fraction of the diameter for which the vertical line
    to the semicircle is under half the radius.

    \displaystyle\left(\frac{r}{2}\right)^2+x^2=r^2\Ri  ghtarrow\ x^2=r^2-\frac{r^2}{4}=\frac{4r^2-r^2}{4}=\frac{3r^2}{4}

    \displaystyle\Rightarrow\ x=\frac{\sqrt{3}r}{2}

    This is the base length of two right-angled triangles both sides of the y-axis.

    You only need to work with one of them.

    To calculate the length of the line segment for which the perpendicular is under half the radius...

    \displaystyle\ r-\frac{\sqrt{3}r}{2}=r\left(1-\frac{\sqrt{3}}{2}\right)

    The probability is the ratio of that length to the radius.
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  3. #3
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    The probability is the ratio of that length to the radius.
    Do I have to show that.
    The radius is r so how would I make that a ratio?
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  4. #4
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    Quote Originally Posted by calculuskid1 View Post
    The probability is the ratio of that length to the radius.
    Do I have to show that.
    The radius is r so how would I make that a ratio?
    The probability is the ratio of 2(r-x) to 2r,

    because if the point is on either line segment of length r-x (on the diameter,
    touching the circumference), then the perpendicular height to the semicircle
    is less than half the radius.

    You can also use trigonometry to find the ratio.

    \displaystyle\ probability=\frac{2(r-x)}{diameter}=\frac{2(r-x)}{2r}=\frac{r-x}{r}=\frac{r\left(1-\frac{\sqrt{3}}{2}\right)}{r}=\left(1-\frac{\sqrt{3}}{2}\right)
    Attached Thumbnails Attached Thumbnails semi circle question-probability-semicircle.jpg  
    Last edited by Archie Meade; September 23rd 2010 at 12:57 PM. Reason: typo
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