The vertex of the right isosceles triangle is the center of the square. What is the area of the overlapping region?

Guest May 27, 2021

#3**+1 **

**Geometry help please?**

\(\begin{array}{|rcll|} \hline A &=& \dfrac{5x}{2} + 5*\left(\dfrac{(5-x)+5}{2}\right) \\ A &=& \dfrac{5x}{2} + 5*\left(\dfrac{10-x}{2}\right) \\ A &=& \dfrac{5}{2}\left( x+10-x \right) \\ A &=& \dfrac{5}{2}*10 \\ \mathbf{A} &=& \mathbf{25} \\ \hline \end{array}\)

heureka May 27, 2021