ABCD is a square inscribed in a circle with circumference

Ok I tried my best here;

What do we need to find is A(shaded region) = A(circle) - A(square)

A(circle) = r^2*pi
A(square) = s^2

We are given the circumference of the circle, which also operates as the diagonal of the square.
The circumference is 2*pi*sqrt(x)

The circumference normally is diameter * pi
Since the circumference given already has pi in it, the diameter must be 2*sqrt(x)
Since the radius is diameter divided by 2 then the radius would be the sqrt(x)
So the area of the circle would be sqrt(x)^2 * pi which would be x*pi

Okay now the square;
the diameter of the circle is the diagonal of the square, splitting it in two 45-45-90 triangles.
We know that 45-45-90 triangles have a ratio of x : x : x*sqrt(2)
Since the diagonal is already given which is 2*sqrt(x), would make the side the same as 2*x since, sqrt(2) * 2 = 2 and sqrt(x) * x = x

Therefore the answer should be B.

Hope i'm right

Bit tricky question to understand...

Lets take x=4, so area of circumference 2πr=2π. Then r=1 and Diameter=2r=2.

Diameter of circle=Diagonal of square. 2=√(L^2+L^2). Therefore Length of square L=√2.

Area of circle - Area of Square=πr^2 - L^2=π-2.

In the answer choices, only B will fit.

Thanks,

Please give me Kudos.

Ans: B

Solution: given that 2pr=rp * x^1/2
this means r=x^1/2 then diagonal of square= 2* x^1/2
so the side of square is (2x)^1/2
and area is therefor 2x

area of the shaded region = Area of circle- Area of square
px-2x

If the circumference is \(2\pi\sqrt{x}\) then the diameter is \(2\sqrt{x}\). This is also the diagonal of the square. Therefore the area of the square is (using area using the diagonal formula) \(\frac{(2\sqrt{x})^2}{2}=2x\). The area of the circle is \(x\pi\) so the shaded area is \(x\pi-2x\) or B

Radius of the circle is equal to rootx (2πrootx).
Diagonal of the sq= 2rootx. Side= root2x.

Area of the circle = πx.
Area of the sq= 2x

Area of the shaded part = πx - 2x.
Ans B

MANHATTAN GMAT OFFICIAL SOLUTION:

The circumference of the circle = \(2\pi{r}=2\pi{\sqrt{x}}\), so \(r=\sqrt{x}\).

The area of the circle = \(\pi{r^2}=\pi{({\sqrt{x}})^2}=\pi{x}\).

The area of a square is side^2. The diagonal of this square is the diameter of the circle = \(2r=2\sqrt{x}\). The diagonal of a square is always \(\sqrt{2}(side)\), so side = \(\frac{diagonal}{\sqrt{2}}\). Therefore, side = \(\frac{2\sqrt{x}}{\sqrt{2}}=\sqrt{2x}\)
and the area of square ABCD is: \(side^2=(\sqrt{2x})^2=2x\)

The shaded area is the area of the circle minus the area of the square = πx – 2x.

The correct answer is B.

Hi,

I didnt understand how did you get 2x? Why divided by 2?