+586 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?
+129850 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
