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?
If you could figure it out that would be great!
THX in Advance!!
ContinueAP down to line AB. Call this point of intersection "Q".
Let AP = BP = CP = x.
Then PQ = 6 - x.
Triangle(APQ) is a right triangle with AP = x, AQ = 3, and PQ = 6 - x.
By the Pythagorean Theorem: x2 = 32 + (6 - x)2
x2 = 9 + (36 - 12x + x2)
0 = 9 + 36 - 12x
12x = 45
x = 3.75 = AP = BP = CP
PQ = CQ - CP
PQ = 6 - 3.75 = 2.25
Area(triangle(APB) = ½·AB·PQ = ½·6·2.25 = 6.75 sq in.