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# A similar question has been asked, but: (CPhill answered it I think)

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A similar question has been asked, but: A point P is chosen at random within right triangle ABC. Find the probability that the area of triangle PBC is less than a QUARTER of the area of triangle ABC.

TY

Jan 5, 2021
edited by Guest  Jan 5, 2021

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Since  we  can use any  convenient right triangle...let's use a 3 - 4 - 5  right triangle  where the legs are AB, AC  and the  hypotenuse is BC

The area  = (1/2) (3)(4)  = 6

(1/4)area  =  6/4   =  3/2

Also....this  area   =      (1/2)  BC  ( height)

3/2  = (1/2) (5)  (height)

(3/2) /( 5/2)  =  6/10   = 6/10  =  3/5   = height  of   PBC

See the following image : Draw a perpendicular  from  A to BC    gives us the  altitude  = 12/5  of the original triangle  from A to BC

But.....we need an altitude of only 3/5

So at D = (1.44,1.92).....construct a circle  of radius (3/5)

E =  (1.08,1.44)   = the point of tangency  to the circle  on a line parallel to BC

The equation of this line is   y = (-3/4) ( x - 1.08)+ 1.44

And any P chosen  on this line will  form a triangle  PBC  that will have an  area of 1/4 of ABC  because the distance  from this line to BC  = 3/5

So any P between  this line and BC  will  form a triangle that has an area < (1/4) [ABC ]  because  the altitude of this triangle   PBC drawn to BC will  be < 3/5

This line will  intercept BA  at F  = (0, 2.25)  and G at (3,0)

So....the area of   GAF =  (1/2) ( 3)(2.25)  ≈ 3.375

So.....the probability  that a "P"  will be chosen  such  that  PBC is  < (1/4) [ABC] is

1 -  area of GAF / area ABC    =

1   -  3.375 / 6    =  .4375  =   7/16

EDIT TO CORRECT A PREVIOUS MISTAKE   Jan 5, 2021
edited by CPhill  Jan 5, 2021
edited by CPhill  Jan 5, 2021
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It was wrong but thanks for trying :)

HelpPls123abc  Jan 5, 2021
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I made a small mistake earlier....see my edt   CPhill  Jan 5, 2021