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 shaded squares, we set the side lengths as follows: Let the side length of the first shaded square be denoted as $A$ and the side length of the second shaded square will then be $10 - A$. Thus, the total area of the two shaded squares can be expressed as:

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

Expanding this expression: $= A^2 + (100 - 20A + A^2)$ $= 2A^2 - 20A + 100$

This is a quadratic equation in the standard form $ax^2 + bx + c$ where $a = 2$, $b = -20$, and $c = 100$. To find the minimum value of a quadratic function, we can use the vertex formula, which gives the x-coordinate of the vertex as:

$A = -\frac{b}{2a} = -\frac{-20}{2 \cdot 2} = \frac{20}{4} = 5$

Now substituting $A = 5$ back into the area equation: $ext{Total Area} = 2(5)^2 - 20(5) + 100 = 2(25) - 100 + 100 = 50$

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