Inactive Tutor answered • 11/18/21
The area of a square is simply the length of any of its sides squared (since all of the sides are the same length). Because of that, the two squares in question have areas as follows:
Square 1: Area1 = (m+n) * (m+n)
Square 2: Area2 = (m-n) * (m-n)
To multiply correctly in these cases we have to remember FOIL (first, outside, inside, last).
So looked at Square 1, we have first: m*m = m^2, outside: m*n = mn, inside: n * m = nm (or mn because the order does not matter, and phrasing it as mn allows us to combine it with the result of "outside"), last: n*n=n^2.
Altogether, Square 1 should now be represented as follows: m^2 + 2mn + n^2.
Following the same process for square 2 should yield: m^2 - 2mn + n^2 (note that the n^2 is positive because the double negative multiplication of -n * -n makes a positive).
Now, because we're looking at sums, we have to add the two together.
m^2 + 2mn + n^2 + m^2 - 2mn + n^2
Combing like terms the positive 2mn and the negative 2mn cancel out, and we get 2m^2 + 2n^2, the answer.
To verify this works we can pick an arbitrary value for m and n, let's say 5 for m and 2 for n. This means the two squares have sides of 7 (m+n) and 3 (m-n) respectively, meaning their respective areas would be 49 (7*7) and 9 (3*3). 49 + 9 = 58, the sum of their areas. Now the question is, did our formula predict that?
Well, let's plug the values in. 2 * 5^2 + 2 * 2^2 simplifies to 2 * 25 + 2*4 which simplifies to 50 + 8, which is 58. Perfect, our formula works, and we can confidently conclude that 2m^2 + 2n^2 is the answer.
