done thank you

Hi,

Can somebody please post a better solution to this?

Thanks!

The key to this problem is finding the length of Rhombus diagonal EF, which is the base of both upper and lower triangles of the Rhombus.

The diagonal of the square is root(2). This length of root(2) is comprised of 2 segments of 1 (radius of the circle) "merged partially". So consider this as a Venn Diagram problem.

Full length = Sum of overlapping lengths - common length.
Diagonal = (Radius1 + Radius2) - EF.
root(2) = (1 + 1) - EF.
EF = 2 - root(2).

Now we have the base of the 2 triangles. The height of the triangles are half the length of the diagonal, which is root(2) / 2 = 1 / root(2).

Area of one triangle = 1/2 * base * height = 1/2 (EF) (half diagonal) = 1/2 (2 - root(2)) (1 / root(2)) = (2-root(2))/2root(2)

Area of rhombus = sum of areas of two triangles = 2 * Area of one triangle = 2 * (2 - root(2)) / 2root(2) = (1 - root(2)).

While this solution is very simple, it took me long enough to come up with it, sadly. The main point of this solution is using the Venn Diagram principle to find the length of the overlapping section of 2 line segments.

There is a lot going on in this picture and given information. Let's focus on the main objectives

First Question: What are we trying to find?
The area of the rhombus

Second Question: What do we need to get the area of the rhombus?
We need the areas of triangles AEF and CEF so we can then add them together
Since triangles AEF and CEF are congruent, it is enough to find the area of triangle AEF and multiply it by 2.


Third Question: What do we need to calculate area of traingle AEF?
We need the base EF and height from A to EF.


Our work now focuses on just getting base EF and height of A to EF

Since we are told that arc AEC belongs to a circle centered at D then ED is a radius of the circle
AD is also a radius of this cricle
AD=1
Therefore ED=1

Also since the square ABCD has sides of 1. Then diagonal BD is (2^.5)

Therefore
EF=(2^.5)-2(2^.5-1)............where 2(2^.5-1) is the length of BE and FD
EF=2-(2^.5)


Getting the height

The diagonal of the square is (2^.5)
The height from A to EF is half the diagonal

height from A to EF= (2^.5)/2


Area of triangle AEF
=(1/2)*(EF)*(height from A to EF)
=(1/2)*(2-(2^.5))*(2^.5)/2
=(2(2^.5)-2)/4
=((2^.5)-1)/2

Area of Rhombus= 2*Area of triangle AEF = 2*((2^.5)-1)/2 = 2^.5-1

Final Answer: B

Another method

options D and E are out since these are more than 1 which is not possible. (Area of the full square is 1)

A = 0.6 --> area of rhombus looks like less then or equal to half not more therefore out
B = 0.4 --> Marked this
C = 0.84 --> out since the remaining area outside rhombus is not around 15%.

I understand that the figure is not drawn to scale.....I drew my own to scale figure.

P.S. : methods suggested above are better methods mathematically.

Cheers

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.