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!!
+23246 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.
