Comments on: Exercise dependence?

By: Keith

Keith — Wed, 10 Mar 2010 07:22:59 +0000

Let the diagonal of a square = X
Let Y = the length of one side

Formula for the diagonal
Y^2 + Y^2 = X^2
Y^2 = X^2 / 2

Y^2 also happens to be the area, so the area of the square is:

X^2 / 2

(1.2 * X)^2 / 2 = (1.44 * X^2) / 2
so
Increasing X by 20% increases the area by 44%
References :

By: Mark

Mark — Wed, 10 Mar 2010 07:20:59 +0000

Since the side of a square with diagonal d is d/√2, the Area A = d²/2.

If you increase the diagonal by 20%, the new diagonal is 1.20d and so the

new Area = (1.2d)² / 2 = 1.44(d²/2) = 1.44 A, which is 44% more thatn the original area A.

The general principle is that a change in a linear dimension results in the square of that factor in a quadratic dimension. For cubes, you woulod cube to find change in volume from a change in a linear measurement.

I have outlined the details for the square so that you can see how that happens in more detail.
References :

By: railbuff

railbuff — Wed, 10 Mar 2010 07:18:59 +0000

If the linear dimension is increased by 20% length of lines are increased to 1.2 times the previous length.
Area changes in proportion to the square of the change in linear length.
A square of length 1.2 would have an area of (1.2)^2 = 1.44, reflecting that a change of 20% in length causes a change of 44% in area.
References :
retired math teacher

By: Amiras

Amiras — Wed, 10 Mar 2010 02:16:02 +0000

Square area dependence on diagonal exercise?
Square diagonal increased by 20%, in how many % increased his area?
I know the answer, and it is 44%, but have no idea how to begin.