ABCD is a square inscribed in a circle with circumference : Problem Solving (PS)

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ABCD is a square inscribed in a circle with circumference

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ABCD is a square inscribed in a circle with circumference [#permalink]

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08 Jun 2015, 09:04

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ABCD is a square inscribed in a circle with circumference $2\pi{\sqrt{x}}$. What is the area of the shaded region in the diagram above?

A. 2x
B. $\pi{x}-2x$
C. $\pi{x}-x\sqrt{2}$
D. $1-\frac{2}{\pi}$
E. $1-\frac{2}{x}$

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Spoiler: OA

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Re: ABCD is a square inscribed in a circle with circumference [#permalink]

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08 Jun 2015, 09:31

Radius of the circle is equal to √x (2πr = 2π√x).
Diagonal of the square is twice the radius of the circle = 2√x. Therefore side of the square is equal to √2x. (a√2 = 2√x)

Area of the circle is equal to πx (πr^2). Area of the square is equal to 2x. (a^2)

Therefore, area of the shaded part = πx - 2x.

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Re: ABCD is a square inscribed in a circle with circumference [#permalink]

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08 Jun 2015, 09:34

Bunuel wrote:

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circumference = $2\pi{R} = 2\pi{\sqrt{x}}$
So radius = $\sqrt{x}$ and diagonal of square = $2\sqrt{x}$

Let's find side of square y:
$y^2+y^2 = (2\sqrt{x})^2$ --> $y = \sqrt{2x}$
Area of square = 2x and area of circle = $\pi{R^2}$ --> $\pi{x}$

Difference between areas of circle and square = $\pi{x} - 2x$

Answer is B
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Re: ABCD is a square inscribed in a circle with circumference [#permalink]

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08 Jun 2015, 10:39

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

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Re: ABCD is a square inscribed in a circle with circumference [#permalink]

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08 Jun 2015, 10:50

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.

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Re: ABCD is a square inscribed in a circle with circumference [#permalink]

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09 Jun 2015, 01:54

Bunuel wrote:

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
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Re: ABCD is a square inscribed in a circle with circumference [#permalink]

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10 Jun 2015, 02:20

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

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Re: ABCD is a square inscribed in a circle with circumference [#permalink]

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14 Jun 2015, 04:45

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

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Re: ABCD is a square inscribed in a circle with circumference [#permalink]

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15 Jun 2015, 03:41

Bunuel wrote:

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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: $side^2=(\sqrt{2x})^2=2x$

Attachment:

2015-06-15_1439.png [ 35.38 KiB | Viewed 7580 times ]

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

The correct answer is B.
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Re: ABCD is a square inscribed in a circle with circumference [#permalink]

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