To solve the problem (assuming the context of a square with side length (x+y) where points (E) and (F) are on sides (AB) and (AD) such that (AE=x) and (AF=y)), the area of the shaded region (triangle (CEF)) can be derived as follows:

Key Observations:

• The square has side length (s = x + y).
• The area of the square is ((x+y)^2).
• Subtract the areas of the three unshaded triangles ((\triangle AEF), (\triangle BEC), (\triangle DFC)) from the square's area to get the shaded area.

Calculations:

1. Area of (\triangle AEF = \frac{1}{2}xy).
2. Area of (\triangle BEC = \frac{1}{2} \times EB \times BC = \frac{1}{2}y(x+y)).
3. Area of (\triangle DFC = \frac{1}{2} \times FD \times DC = \frac{1}{2}x(x+y)).

Sum of unshaded areas:
[ \frac{1}{2}xy + \frac{1}{2}y(x+y) + \frac{1}{2}x(x+y) = \frac{1}{2}(x^2 + y^2 + 3xy) ]

Shaded area = Area of square - Sum of unshaded areas:
[ (x+y)^2 - \frac{1}{2}(x^2 + y^2 + 3xy) = \frac{x^2 + xy + y^2}{2} ]

Answer: (\boxed{\dfrac{x^2 + xy + y^2}{2}})

作者声明:本文包含人工智能生成内容。