# Difference between $((a+b)/2)^2$ and $ab$?

I was thinking about the difference between the area of a rectangle that is not a square, and a square with sides whose lengths are at the midpoint between the lengths of $a$ and $b$. I did some algebraic manipulation and it seems that the difference between the area of the square, $((a+b)/2)^2$, and the area of the rectangle, $ab$, is $(a^2+b^2)/4 - (ab)/2$.

Now if you had a right triangle with sides $a$ and $b$, it's hypotenuse would be the square root of $a^2 + b^2$.

So what I'm wondering is why the difference between the area of the square and the rectangle is the same as the difference between one fourth the square of the hypotenuse of the right triangle with sides $a$ and $b$, and $1/2$ the rectangle $ab$? If you were trying to find this difference purely with geometry, what steps could you take to reach this conclusion, starting from the original square and rectangle?

• This is a wonderful question. Sep 9, 2016 at 22:10
• I have 2 points of confusion. The square whose lengths are at the mid-point of $a$ and $b$? And $a, b$ are the sides of a rectangle. The lenght of the segment between these two points is half the diagonal of the rectangle. And the square of that is $\frac 14 (a^2 + b^2)$ not $\big(\frac {(a+b)}{2}\big)^2$. Sep 9, 2016 at 22:19
• Actually it's more than half the diagonal of the rectangle. ((a+b)/2)^2 is 1/4(a^2+2ab+b^2) . I combined ab/2 and -ab to get -ab/2, aka half the area of the rectangle ab. Sep 9, 2016 at 22:24
• Also ty @Juan Sebastian Lozano Sep 9, 2016 at 22:25
• second point. This question seems to be asking a circular question about the proof of the Pythagorean theorem. i.e. $4$ right triangles with legs $( a, b)$ can be arranged such that the form a square with side $(a+b)$ and the 4 hypoteni forming a square. The $4$ triangles have combinded area $2ab.$ And the big square has area $(a+b)^2$, meaning that the square formed by the hypoteni has area $a^2 + b^2$ Sep 9, 2016 at 22:26

Here is my proposition building heavily on a well known proof of the Pythagorean Theorem:

The red rectangle is $ab$. You see how the four blue triangles with legs that are half of $a$ and $b$ respectively fit nicely in the left half of the rectangle. The white square is $1/4$ of the square on the hypotenuse.

So by subtracting the rectangle from the large square, we are essentially subtracting the right half of rectangle from the white square.

• This is the sort of thing I'm looking for, but I'm not seeing the geometric representation of the square with sides (a+b)/2. I'm looking for the difference of it and the rectangle ab, and it looks like you have the rectangle ab in your top left (times 4?). Maybe I'm just not understanding your proposition. If you multiplied the difference by 8, you would have 2 of those open squares surrounded by purple, minus that big red rectangle (4ab) and that would give 8 times the difference I'm looking for. Sep 9, 2016 at 22:55
• @Hockeyfan19: the entire red rectangle is $ab$. The blue triangles have half of $a$ and $b$ as sides. Sep 9, 2016 at 23:03
• The square of $(a+b)/2$ is enclosed by the legs of the blue triangles ... Sep 9, 2016 at 23:10
• The big blue square is (a+b)/2 squared. Sep 9, 2016 at 23:10
• That makes a lot of sense String, I just wasn't reading it right. Then when you get rid of the blue squares and one half the red rectangle we have the difference I was looking for. Perfect! Sep 9, 2016 at 23:14

One more note:

$\frac {(a^2 + b^2)}{4} - \frac {ab}2 = \big(\frac {(a-b)}{2}\big)^2$

Maybe a picture will help:

First figure: the larger square has dimensions $(A+B)/2 \times (A+B)/2$ Which has can be cut into smaller rectangles and squares of size $A/2 \times B/2,A/2 \times (B-A)/2, A/2 \times A/2$ leaving the green square that is $(B-A)/2 \times (B-A)/2$

The white rectangles can be re-arranged into an $A\times B$ rectangle.

Part 2. $C$ is the hypotenuse of the $A\times B$ rectangle.

$(C/2)^2 - (AB)/2 = ((B-A)/2)^2$

• This is a great answer, thank you for the information! You demonstrated the difference between the rectangle and square very well geometrically. Sep 9, 2016 at 23:35