![(1-\sqrt{2})a^2](/tpl/images/0348/4556/adabb.png)
Step-by-step explanation:
Consider irght triangle PRS. By the Pythagorean theorem,
![PS^2=PR^2+RS^2\\ \\PS^2=a^2+a^2\\ \\PS^2=2a^2\\ \\PS=\sqrt{2}a](/tpl/images/0348/4556/7eb8c.png)
Thus,
![MS=PS-PM=\sqrt{2}a-a=(\sqrt{2}-1)a](/tpl/images/0348/4556/d66b8.png)
Consider isosceles triangle MSC. In this triangle
![MS=MC=(\sqrt{2}-1)a.](/tpl/images/0348/4556/634dc.png)
The area of this triangle is
![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}](/tpl/images/0348/4556/ea141.png)
Consider right triangle PTS. The area of this triangle is
![A_{PTS}=\dfrac{1}{2}PT\cdot TS=\dfrac{1}{2}a\cdot a=\dfrac{a^2}{2}](/tpl/images/0348/4556/91042.png)
The area of the quadrilateral PMCT is the difference in area of triangles PTS and MSC:
![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](/tpl/images/0348/4556/23f7b.png)