ABC is an isosceles right triangle with AB = AC = 6. Given that SDPFSDPF is a square, find the area of the square.
+23246 In triangle(ABC), angle(B) = angle(C) = 45o.
In triangle(AFP), angle(AFP) = angle(APF) = 45o.
Let AP = x ---> PF = x·sqrt(2).
Since AC = 6 and AP = x ---> PC = 6 - x ---> DP = (6 - x) / sqrt(2).
Since FSDP is a square ---> PF = DP ---> x · sqrt(2) = (6 - x) / sqrt(2).
---> x · 2 = 6 - x
---> 2x = 6 - x
---> 3x = 6
---> x = 2
Since PF = x·sqrt(2) ---> PF = 2·sqrt(2)
From here, you can calculate the area of the square.
