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avatar+545 

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
avatar+106533 
+1

 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

 

 

 

cool cool cool

 Nov 5, 2019

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