$2$-inch squares are cut from the corners of a $10$-inch square. What is the area of the largest square that fits the remaining space?
You'll want something like this:
i.e. where the side of the inscribed square goes through the corner points of the little cut out squares.
Draw some helpful little lines:
And you find that the area is the middle $6\times 6$ square plus exactly half of the area of all the black and blue triangles (since for every blue triangle there is a corresponding black triangle). So, the area is:
$$36 + \frac{4 \cdot 2 \cdot 6}{2}=36+24=60$$
Here's a graphical assist from @Blue:
The blue part has a total area of $2\times 2\times 6 = 24$, Thus, the largest area that can still fit is
$$10^2 - 4\times 2^2 - 24 = 60$$