A square with sides of length 10 units is shown in the diagram - Junior Cycle Mathematics - Question 7 - 2011

To find the total area of the two shaded squares, we denote the length of the side of the square at point A as A. The square to the left has sides of length A, and the square to the right has sides of length (10 - A).

The formulas for the areas of the squares are:

• Area of the left square: $A^2$
• Area of the right square: $(10 - A)^2$

Thus, the total area will be:

$ext{Total Area} = A^2 + (10 - A)^2 = A^2 + 100 - 20A + A^2 = 2A^2 - 20A + 100$

To minimize this quadratic function, we can complete the square or take the derivative and set it to zero:

1. Finding the vertex of the parabola: The vertex form is given by: A = -rac{b}{2a} = -rac{-20}{2 imes 2} = 5

2. Substituting A back into the area equation:

   Total Area when A = 5: $ext{Total Area} = 2(5)^2 - 20(5) + 100 = 50$

3. Conclusion: Therefore, the minimum possible value of the total area of the two shaded squares is 50 square units.