In which model does removing y^2 from x^2 show the remaining part equal to (x+y)(x-y)?

Correct answer: Difference-of-squares model

The correct answer is A: the difference-of-squares model. A large square with side x has area x^2. Removing a smaller square with side y removes area y^2, leaving x^2 - y^2. The remaining L-shaped region can be cut and rearranged into a rectangle whose sides are x+y and x-y. Its area is therefore (x+y)(x-y), which proves x^2-y^2 = (x+y)(x-y). Option A describes exactly this removal and rearrangement. Option B represents (x+y)^2, whose expansion contains the positive middle term 2xy. Option C represents (x-y)^2, whose expansion contains the negative middle term -2xy; it is not simply x^2-y^2. Option D concerns boundary length, not the area identity being shown. A useful warning is that difference of squares has no middle term, while squares of sums or differences do have one. Memory cue: remove one square from another, then think difference of squares.