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?
+129907 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
