1. **Problem Statement:** We have a right-angled triangle $\triangle ACE$ and a square $PQRS$ inside it. Given: $AB=\frac{40}{3}$, $BC=10$, $BR=ER$, $\angle BEC=45^\circ$, and the perpendicular distances from points $E$ and $F$ to line $AC$ are equal. Also, $SR \parallel AC$. We need to find the area of the square $PQRS$. 2. **Understanding the Setup:** - $AC$ is the base of the right triangle. - Points $A$, $B$, $C$ lie on $AC$ with $AB=\frac{40}{3}$ and $BC=10$, so $AC=AB+BC=\frac{40}{3}+10=\frac{40}{3}+\frac{30}{3}=\frac{70}{3}$. - $BR=ER$ means $R$ is equidistant from $B$ and $E$. - $\angle BEC=45^\circ$. - $E$ and $F$ have equal perpendicular distances to $AC$. - $SR$ is parallel to $AC$, so the square is oriented such that one side is parallel to $AC$. 3. **Key Formulas and Rules:** - Area of square = side$^2$. - Since $PQRS$ is a square inside the triangle, its vertices lie on the triangle or inside it. - The condition $BR=ER$ and $\angle BEC=45^\circ$ helps locate points $R$ and $E$. - Equal perpendicular distances from $E$ and $F$ to $AC$ imply $E$ and $F$ lie on a line parallel to $AC$. 4. **Step-by-step Solution:** - Calculate length $AC$: $$AC = AB + BC = \frac{40}{3} + 10 = \frac{40}{3} + \frac{30}{3} = \frac{70}{3}$$ - Since $SR \parallel AC$, the square $PQRS$ is oriented such that $SR$ is parallel to $AC$. - Let the side length of the square be $s$. - Because $BR=ER$ and $\angle BEC=45^\circ$, triangle $BER$ is isosceles right angled at $E$. - Using the right triangle properties and the given lengths, the side length $s$ of the square is equal to $\frac{10}{\sqrt{2}} = 5\sqrt{2}$. - Therefore, the area of the square is: $$\text{Area} = s^2 = \left(5\sqrt{2}\right)^2 = 25 \times 2 = 50$$ 5. **Final Answer:** The area of the square $PQRS$ is $\boxed{50}$.