A square with sides of length 10 units is shown in the diagram - Junior Cycle Mathematics - Question 7 - 2015

Let the side of the square to the left of point A be denoted as $A$, and the side of the square to the right be $(10 - A)$. Then, the total area of the shaded squares can be expressed as:

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

Expanding this gives:

$= A^2 + (100 - 20A + A^2)$ $= 2A^2 - 20A + 100$

To find the minimum area, we take the derivative of the area function and set it to zero:

rac{d( ext{Area})}{dA} = 4A - 20 = 0

Solving for A yields:

$A = 5$

Substituting $A = 5$ back into the area equation provides:

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

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