Geometric probability question, find the probability that the area of the triagle is more than twelve. | A B | = 8 , | C D | = 6. We choose the point P by chance on AB and again a point Q by chance on CD find the probability so the the triangle whose height is AP, and base is CQ is larger than 12. The answer is in the picture in it's condesed form, can anyone provide an explanation? y = 24 x , p = ∫ 4 8 ( 6 − 24 x ) d x 48

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Geometric probability question, find the probability that the area of the triagle is more than twelve.      |    A  B      |    =  8  ,      |    C  D      |    =  6. We choose the point P by chance on AB and again a point Q by chance on CD find the probability so the the triangle whose height is AP, and base is CQ is larger than 12. The answer is in the picture in it&#039;s condesed form, can anyone provide an explanation?  y  =      24    x    ,  p  =      ∫          4              8                  (      6      −              24        x            )      d      x        48

Emmalee Reilly · Accepted Answer

Step 1First note that the area of triangle is half the area of rectangle with base CQ and height AP.So we need to find prob   A  P  ×  C  Q  ≥  24.  A  P  =  X and   C  Q  =  Y are independently chosen random variables from the interval [0,8] and [0,6].So   [  0  ,  8  ]  ×  [  0  ,  6  ] is the sample space.Step 2Out of this sample space, the area that satisfies the question is   X  ⋅  Y  &amp;gt;=  24.That&#039;s the area   A  =      ∫          x      =      4        8    (  6  −  24      /    x  )    d  x  =  12  −  24  ln  ⁡      3    2  So probability =   A      /    (  6  ×  8  )  =  1      /    4  −  1      /    2  ln  ⁡  (  3      /    2  )  ≈  4.73  %

|AB|=8, |CD|=6 We choose the point P by chance on AB and again a point Q by chance on CD find the probability so the the triangle whose height is AP, and base is CQ is larger than 12.

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