This is a much more interesting problem than what we usually see here.
Let's just start by assuming a=1 so we have two unit squares.
PMC and PTC are two congruent triangles, each right triangles with one leg of length 1 and a shared hypotenuse PC.
We'll put it on the grid at P(0,0), T(1,0).
Angle TPM is 45 degrees and PM=1 so
M(cos 45, sin 45) = M(1/√2, 1/√2)
MK has slope -1, thru M, so is
y = -x + √2
C's on that line at x=1, so C's x coordinate is 1, so its y coordinate is √2 - 1
C(1, √2 - 1)
OK, that's enough to compute the area.
P(0,0), T(1,0), C(1, √2 - 1), M(1/√2, 1/√2)
area(PMCT) = area(PTC) + area(PCM)= 2 area(PTC)
The area of a right triangle is half the product of its legs.
area(PMCT) = 2 (1/2) (1) ( √2 - 1)
area(PMCT) = √2 - 1
When we have sides a, the area scales by a²,
Answer: a²(√2 - 1)
