Question

show that in case of a square and rectangle having equal perimeter square has more area.​

Answer 1

Step-by-step explanation:

The areas are not equal. The area of the square is greater. But how to prove it?

Let, the length and width of the rectangle is x and y, respectively. The length of the side of the square is a.

Hence,

2(x + y) = 4a

Or, x + y = 2a

Or, x² + 2xy + y² = 4a²

Or, x²/4 + xy/2 + y²/4 = a²

Or, (x/2)² + (y/2)² + xy/2 – xy = a² – xy

Or, (x/2)² – xy/2 + (y/2)² = a² – xy

Or, (x/2)² – 2×(x/2)×(y/2) + (y/2)² = a² – xy

Or, a² – xy = {(x/2) – (y/2)}²

Or, a² = {(x/2) – (y/2)}² + xy

So, we can say that a² > xy as the term in bold letters is a perfect square.

Hence, the square has greater area.

Don't hesitate in case of any doubts or queries. Comment below.

Let’s work this one out together!

So we have a square, side length A.

And a rectangle, length X, width Y.

The question (rephrased in the language of mathematicians) is:

Is it the case if 4A = 2X + 2Y, then A^2 = XY

(You see where this is coming from, right? perimeter equal implies area equal?)

Well, if we have a situation where 4A = 2X + 2Y BUT A^2 not= XY, then we have proven that equal perimeter doesn’t mean equal area.

Observe that A = (X + Y)/2

A^2 = (X^2 + Y^2)/4 + XY/2

But, if perimeter were the same as area, A^2 would’ve been the same as XY.

Since A^2 is clearly different from XY it is not the case the the perimeter being equal implies the areas of the two are equal.

Wow, that proof looks messy. Sorry, can’t use the Maths text formatting yet.