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# Geometry Problem

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A square with sides 6 inches is shown. If P is a point such that the segment $$\overline{PA}$$$$\overline{PB}$$$$\overline{PC}$$ are equal in length, and segment $$\overline{PC}$$ is perpendicular to segment $$\overline{FD}$$, what is the area, in square inches, of triangle APB? Nov 5, 2019

#1
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Here's one way to do this

By symmetry....FC  =   DC   = 3

Connect  CA

Triangle FCA  is right  and CA  =   sqrt  (3^2  + 6^2)  =  sqrt ( 45)  = 3sqrt (5)

And by the Law  of Sines

sin AFC / CA  =  sin FCA / FA

sin (90)  / [ 3sqrt (5) ]  = sin  FCA  / 6

6 / [ 3sqrt (5) ] =  sin FCA

2 /sqrt (5)  = sin FCA

arcsin (2/sqrt (5))  =  FCA

So  angle  ACP  = angle CAP =  arcsin (2/sqrt(5))

So  angle  APC  =  [180 - 2arcsin (2/sqrt(5))]

Draw PE perpendicular to   AB....so   AE   = 3

And since angles APE  and APC are supplemental sin APC  = sin  APE

So  by the Law of Sines

PA/ sin PEA   =    3 / sin APE

PA  / sin 90   =  sin   [180 - 2arcsin (2/sqrt(5))] /  3

PA  =   3  /  [ sin180 * cos (2 arcsin (2/sqrt(5)))  -  cos 180 * sin (2arcsin (2 /sqrt(5))]

PA  =   3 /  [  sin (2 arcsin (2/sqrt(5))) ]

PA  =   3 / (4/5)

PA  =  15/4

PA  =  3.75

So....PE  =  sqrt  ( PA^2  - AE^2) =  sqrt  (3.75^2 - 3^2)  =  2.25

So....the area  of  APB  =

AE   *  PE   =

3  *  2.25  =

6.75  units^2   Nov 5, 2019