excellent question, checks you ability and flexibility working with geometric concepts as well as fractions and ratios, especially the last part when you get square root 2 - 1 divided by square root 2 and you should multiply both numerator and denominator with \sqrt{2} to get a satisfactory A as an answer.

Still don't really understand this, even with the answers. Could anyone illustrate it better/differently?

Here, see the diag. above

Now let the smaller square be ABCD, where it is mentioned the diagonal = 1 foot

Now , if a figure is square and all angles are 90 degree, hence the sides can be divided

as $1: 1: \sqrt2$ (45-45-90 $\triangle$)

Let us divide the square in 2 equal triangles ABC and ADC

For $\triangle$ ABC

we have AC = 1 ( as it is a diagonal)

since the sides are in $1: 1: \sqrt2$

i.e diagonal AC has to be the largest side and should be equal = $\sqrt2$

but how to make this possible?

we can divide$1:1:\sqrt2$ by$\sqrt2$

i.e$\frac{1}{{\sqrt2}} : \frac{1}{{\sqrt2}} : 1$

Now, the largest side (diagonal) is 1 and the other two sides in the $\triangle$ ABC are AB = BC =$\frac{1}{{\sqrt2 }}$

Now we have figure out the side of the smaller square ABCD

Hence, the Area of the smaller square =${side}^2$ = $({\frac{1}{\sqrt2} })^2 = \frac{1}{2}$

Now,

Larger Square = 2 * Area of the smaller square = $2 * \frac{1}{2}= 1$

i.e the side of the larger square = $1$

Ok, now we have the side of the Larger square as well as for the smaller square

Length of the side of the Larger square =$1$

and length of the side of the smaller square = $\frac{1}{{\sqrt2}}$

Then the side of the length of the larger square greater than that of the smaller square = $1 - \frac{1}{\sqrt2} = \frac{{\sqrt2 -1}}{\sqrt2} * \frac{{\sqrt2}}{{\sqrt2}}$ = $\frac{{(2-\sqrt2)}}{2}$

You're great. Thanks so much.

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.