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 define the area in terms of A. The square on the left has sides of length A, and the square on the right has sides of length (10 - A).

Thus, the total area can be expressed as: $\text{Area} = A^2 + (10 - A)^2$

Expanding this expression: $\text{Area} = A^2 + (100 - 20A + A^2) = 2A^2 - 20A + 100$

Next, we find the minimum possible value of this quadratic function. The vertex of a parabola represented by (ax^2 + bx + c) occurs at (A = -\frac{b}{2a}). Here, (a = 2) and (b = -20).

Calculating the vertex position: $A = -\frac{-20}{2 \times 2} = 5$

Substituting A = 5 back into the area function to find the minimum area: $\text{Area} = 2(5^2) - 20(5) + 100 = 50\n\text{(minimum area)}$

Thus, the minimum possible value of the total area is 50 square units.