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The △AOB consists of point A=(0,1), point O=(0,0), and point B lying somewhere on the x-axis.  Let P be the point in the first quadrant such that AP = PB and AO parallel  to PB, as shown above.  If the length of OP is sqrt{194}​, what is the area of the quadrilateral AOBP?

 

 Dec 28, 2020
 #1
avatar+114325 
+2

Let OB   =  x   and  BP  = y

 

Triangle   OBP  is right  and we  have that

 

x^2  + y^2  =   194      rearrange  as

 

y  =  sqrt  [ 194  - x^2  ]

 

So  P   =   ( x , sqrt (194 - x^2)

 

And   the distance PA  = PB

 

So....equating  the  squares  of  the   distances we  have that

 

194  - x^2   =   [ (x - 0)^2  +  ( sqrt (194 - x^2)  - 1)^2 ]

 

(194 - x^2 ) =  x^2  +  1 - 2sqrt (194 - x^2)  +  (194 - x^2)

 

x^2  + 1    =   2sqrt (194 - x^2)       square both sides

 

x^4   + 2x^2  + 1  =  4 (194 - x^2)

 

x^4  + 2x^2  + 1   =  776  - 4x^2

 

x^4  + 6x^2 - 775  =  0       factor as

 

(x^2  - 25) ( x^2 + 31)   = 0

 

Only the first  factor  set to  0  gives  us a positive solution  for  x   =  5  =   OB

 

And BP    = sqrt (194 - 5^2) = sqrt (169)  =  13

 

And AB  =  sqrt ( 5^2  +  1^2)  = sqrt (26)

 

So   area of triangle  AOB  =  (1/2) OB * OA  = (1/2) (5) * 1   = 5/2    (1)

 

And  cos ABO  =  OB / AB  =  5/sqrt (26)    =  sin PBA

 

So...area of triangle  PBA  =(1/2) (BP) (AB) ( sin PBA  )  =  (1/2) (13) ( sqrt (26)) ( 5/sqrt (26))  =

 

(1/2)( 13) ( 5)   =  65/2     (2)

 

So....  [ AOBP]  =  (1)  + (2)  =   5/2  +  65/2  =   70 / 2   =   35

 

 

cool cool cool

 Dec 28, 2020
 #2
avatar+859 
+1

Good job, Phill!

jugoslav  Dec 29, 2020
 #3
avatar+114325 
0

THX, jugoslav   !!!

 

cool cool cool

CPhill  Dec 29, 2020

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