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In a certain playground, a square sand box rests in a  [#permalink]

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Difficulty:   15% (low)

Question Stats: 83% (01:29) correct 17% (00:55) wrong based on 141 sessions

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In a certain playground, a square sand box rests in a circular plot of grass so that the corners of the sandbox just touch the edge of the plot of grass at points W, X, Y and Z, as shown. What is the distance from point W to point Y?

(1) The area of the circular plot is 49π
(2) The ratio of the area of the sand box to the area of the circular plot is 2/π.
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Re: In a certain playground, a square sand box rests in a circul  [#permalink]

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macjas wrote:
Attachment:
Q19.gif

In a certain playground, a square sand box rests in a circular plot of grass so that the corners of the sandbox just touch the edge of the plot of grass at points W, X, Y and Z, as shown. What is the distance from point W to point Y?

(1) The area of the circular plot is 49π
(2) The ratio of the area of the sand box to the area of the circular plot is 2/π.

we need to fond radius of circle as WY=2r
1) we can get radius => πr^2=49π=> r=7

2) We cannot get radius as acc to statement=>
2/π= area of square/area of circle= (diagonal\sqrt{2})^2/πr^2
diagonal = 2r hence r^2 cancels out=> we cannot find r

Hence A is the answer
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Re: In a certain playground, a square sand box rests in a  [#permalink]

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In a certain playground, a square sand box rests in a circular plot of grass so that the corners of the sandbox just touch the edge of the plot of grass at points W, X, Y and Z, as shown. What is the distance from point W to point Y?

The distance from W to Y equals to the diameter so to 2r.

(1) The area of the circular plot is 49π --> $$\pi{r^2}=49\pi$$ --> $$r=7$$. Sufficient.

(2) The ratio of the area of the sand box to the area of the circular plot is 2/π --> the area of a square is $$\frac{diagonal^2}{2}=\frac{4r^2}{2}=2r^2$$ and the area of a circle is $$\pi{r^2}$$ --> the ratio thus is $$\frac{2}{\pi}$$. So, we already knew that ratio from the stem. Not sufficient.

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Re: In a certain playground, a square sand box rests in a  [#permalink]

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Bunuel wrote:
In a certain playground, a square sand box rests in a circular plot of grass so that the corners of the sandbox just touch the edge of the plot of grass at points W, X, Y and Z, as shown. What is the distance from point W to point Y?

The distance from W to Y equals to the diameter so to 2r.

(1) The area of the circular plot is 49π --> $$\pi{r^2}=49\pi$$ --> $$r=7$$. Sufficient.

(2) The ratio of the area of the sand box to the area of the circular plot is 2/π --> the area of a square is $$\frac{diagonal^2}{2}=\frac{4r^2}{2}=2r^2$$ and the area of a circle is $$\pi{r^2}$$ --> the ratio thus is $$\frac{2}{\pi}$$. So, we already knew that ratio from the stem. Not sufficient.

Hi Bunnel,

I have assume that distance from W to Y will not pass from centre of the circle. So i have chosen E as the answer.

Is their any rule which specifies that when a square is placed inside a circle (as described in the question), the diameter of the circle will be equal to the diagnoal of a square .
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Re: In a certain playground, a square sand box rests in a  [#permalink]

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riteshgmat wrote:
Bunuel wrote:
In a certain playground, a square sand box rests in a circular plot of grass so that the corners of the sandbox just touch the edge of the plot of grass at points W, X, Y and Z, as shown. What is the distance from point W to point Y?

The distance from W to Y equals to the diameter so to 2r.

(1) The area of the circular plot is 49π --> $$\pi{r^2}=49\pi$$ --> $$r=7$$. Sufficient.

(2) The ratio of the area of the sand box to the area of the circular plot is 2/π --> the area of a square is $$\frac{diagonal^2}{2}=\frac{4r^2}{2}=2r^2$$ and the area of a circle is $$\pi{r^2}$$ --> the ratio thus is $$\frac{2}{\pi}$$. So, we already knew that ratio from the stem. Not sufficient.

Hi Bunnel,

I have assume that distance from W to Y will not pass from centre of the circle. So i have chosen E as the answer.

Is their any rule which specifies that when a square is placed inside a circle (as described in the question), the diameter of the circle will be equal to the diagnoal of a square .

The property is that if you draw a triangle inside a circle such that the diameter is one of the sides of the triangle, then the angle made by triangle on the circumference is a right angle (=90 degrees). In other words, a triangle inside a circle such that the diameter is one of the sides of the triangle will be a right angled triangle.

In a square, the diagonal divides it into 2 right triangles and as such any square drawn inside a circle (such that all the 4 points of the square lie on the circle) will have the diagonal as the diameter of the circle.

Hope this helps.
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Re: In a certain playground, a square sand box rests in a  [#permalink]

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_________________ Re: In a certain playground, a square sand box rests in a   [#permalink] 04 Sep 2018, 02:44
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