Answer:
[tex](1-\sqrt{2})a^2[/tex]
Step-by-step explanation:
Consider irght triangle PRS. By the Pythagorean theorem,
[tex]PS^2=PR^2+RS^2\\ \\PS^2=a^2+a^2\\ \\PS^2=2a^2\\ \\PS=\sqrt{2}a[/tex]
Thus,
[tex]MS=PS-PM=\sqrt{2}a-a=(\sqrt{2}-1)a[/tex]
Consider isosceles triangle MSC. In this triangle
[tex]MS=MC=(\sqrt{2}-1)a.[/tex]
The area of this triangle is
[tex]A_{MSC}=\dfrac{1}{2}MS\cdot MC=\dfrac{1}{2}\cdot (\sqrt{2}-1)a\cdot (\sqrt{2}-1)a=\dfrac{(\sqrt{2}-1)^2a^2}{2}=\dfrac{(3-2\sqrt{2})a^2}{2}[/tex]
Consider right triangle PTS. The area of this triangle is
[tex]A_{PTS}=\dfrac{1}{2}PT\cdot TS=\dfrac{1}{2}a\cdot a=\dfrac{a^2}{2}[/tex]
The area of the quadrilateral PMCT is the difference in area of triangles PTS and MSC:
[tex]A_{PMCT}=\dfrac{(3-2\sqrt{2})a^2}{2}-\dfrac{a^2}{2}=\dfrac{(2-2\sqrt{2})a^2}{2}=(1-\sqrt{2})a^2[/tex]
