+312 A square with sides 6 inches is shown. If P is a point such that the segment PA, PB, PC are equal in length, and segment PC is perpendicular to segment FD, what is the area, in square inches, of triangle APB?
+129852 Let P = (3, b)
Let A = (0, 0)
Let C = ( 3, 6)
And we can draw a circle with center (3, b) and radius PA that will pass through points A, B and C
So....since radius PA = radius PC, we have that
(0 - 3)^2 + ( 0 - b)^2 = (3 - 3)^2 + ( 6 - b)^2
9 + b^2 = b^2 - 12b + 36
12b = 27
b = 27 / 12 = 9 / 4 = height of triangle APB
So P = (3, 9/4)
And the area of triangle APB = (1/2) BC * (9/4) = (1/2) (6) (9/4) = 27/4 = 6.75 units^2
See the following image :
