282. Cross of chips. Arrange 20 chips in the shape of a cross, as shown in the figure. How many different cases can you count where four chips form a perfect square?

For example, squares are formed by the chips at the ends of the cross, the chips located in the center, as well as the chips marked with the letters $A$ and $B$.

![](https://cdn.mathpix.com/cropped/2024_05_21_56f37ca27ac3d928838dg-084.jpg?height=485&width=488&top_left_y=1431&top_left_x=790)

Which 6 chips should be removed so that no four of the remaining chips are located at the vertices of any square?
282. There are 19 such squares in total. Of these, 9 are the same size as the square marked with the letters \(a\), 4 are the same size as the square marked with the letters \(b\), 4 are the size of \(c\), and 2 are the size of \(d\). If 6 chips marked with the letter \(e\) are removed, it will be impossible to form any square from the remaining chips.

![](https://cdn.mathpix.com/cropped/2024_05_21_56f37ca27ac3d928838dg-241.jpg?height=574&width=580&top_left_y=1072&top_left_x=744)

[In fact, there are 21 squares. Will the reader be able to find the two squares that Dudeney missed? The answer to the second part of the problem remains correct, nevertheless. - M. G.]