If the area of the outer square is 4x2, then the area of the inner squ

One side of the larger square is 2x, making the radius x, and the diameter 2x. Since it is a 45 45 90 triangle, that makes a side xsqrt(2). Squaring that gives us 2x^2
I supposed we could also use the formula Area of a square = Diameter^2 / 2, which would also give us 2x^2

Answer: D

Hi All,

I am getting the Answer as A. x2/2

The side of the bigger square= 2x

This is nothing but, the diameter of the circle=2x.

The diameter of the circle= The diagnoal of the smaller square=2x

Diagnal is \sqrt{2} time the side=> The side of the smaller square= 2x/\sqrt{2}.

= The area of the inner square= x2/2

Bunuel, please correct me if wrong.

Hi EmpowerGmat,

I am getting the Answer as A. Please correct me if wrong.

shriramvelamuri,

The diagonal of the inner square is 2x. This forms a 45 45 90 triangle with hypotenuse of 2x. Since it is as 45 45 90 triangle, the sides are in the ratio of side, side, side*sqrt(2).
In this case 2x is the hypotenuse.

side*sqrt(2)=2x
squaring both sides we get
(side^2)*2 = 4*(x^2)
side^2 =2*x^2 -------> This is the area of the square.

hi shriramvelamuri,
you are perfectly fine till you come to the diagonal of inner square..
ans is D 2x^2... pl look at the coloured portion down below

Hi All,

While this question does involve some specific Geometry rules, it can be solved by TESTing VALUES.

We're told that the area of the outer square is 4(X^2). We're asked for the area of the inner square.

IF....
X = 3
Area of the outer square = 4(3^2) = 36
Side of the outer square = 6

Since the inner square is inscribed in the circle, we can 'rotate' the inner square so that IT'S diagonal is parallel with the side of the outer square.

Diagonal of inner square = 6

A square can be 'cut in half' across the diagonal and will form two 45/45/90 triangles.

Diagonal of inner square = 6
Side of inner square =
6 / (Root2) =
6(Root2) / 2 =
3(Root2)

Area of the small square =
(Side)^2 =
[3(Root2)]^2 =
18

So we're looking for an answer that equals 18 when X = 3. Only one answer matches....

Final Answer:
GMAT assassins aren't born, they're made,
Rich

Very nice question

The height of the big square equals the diagonal of the small square. So the inner square has half the area.

Let's say D. \(2x^2\)

I suppose this is the simplest explanation

---
edit

To give a more visual explanation: rotate the small square until its diagonals are perpendicular to the sides of the big square and draw the diagonals. Just ignore the circle: the figure is now made up of many right isosceles triangles and the big square has twice as many small triangles as the small square.

Side of big square =sqrt(4x^2)=2x

it is diameter of circle and diagonal of the little square

Area of little square=2x*2x/2=2x^2 (rhombus property)


D

Just refer to the modified diagram below:


Rotate the inner square (Pink shaded) by 45 degrees; notice that inner square is inscribed in the outer square

In such cases, area of the inscribed square is 1/2 area of the subscribed square

So, required area \(= \frac{4x^2}{2} = 2x^2\)

Answer = D

MAGOOSH OFFICIAL SOLUTION:

Easier solution would be:
The side of the outer square (2x) equals to the diameter of the circle, which on the other hand equals to the diagonal of the inner square.

The area of a square = \(\frac{diagonal^2}{2}\) = (2x)^2/2 = 2x^2.

Answer: D.

Following 2 formulas come handy in any question involving a square and a circle
Area of circle inscribed in a square = (pi/4) area of square
Area of circle circumscribed around a square = (pi/2) area of square

Area of circle = (pi/4) * 4x^2

Are of inner square = (2/pi) * (pi/4) * 4x^2 = 2x^2

Ans D

Thank you, I love this kind of explanations! Good job You have received a kudo! ^^

Very easy. Area of Outer Square = 4x^2 so side will be 2x and 2x is diameter of inner circle and that is also diagonal of of inner square.
As per formula A=d^2/2 and we know that d = 2x so (2x)^2/2 that is 2x^2
Answer is D

Hello from the GMAT Club BumpBot!

Thanks to another GMAT 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.