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The figure shows the intersection of a Square ABCD and a Circle. What

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The figure shows the intersection of a Square ABCD and a Circle. What  [#permalink]

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07 Dec 2019, 19:00
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GMATBusters’ Quant Quiz Question -2

The figure shows the intersection of a Square ABCD and a Circle. What is the Perimeter of the Square ABCD?
1) X = 1
2) AC and BD are perpendicular to each other

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123.JPG [ 38.21 KiB | Viewed 843 times ]

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Re: The figure shows the intersection of a Square ABCD and a Circle. What  [#permalink]

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07 Dec 2019, 19:39

I'm assuming in the given diagram the diameter passes through the intersection of the diagonals:

a= side of the square
d=diameter of the circle
Requirement is to find 4a ?
= 4(d-x)

1) x=1
for x=1 perimeters =4(d-1)
multiple answers as d changes hence insufficient.

2) This basically gives product information. In a square diagonal are perpendicular. Not sufficient.
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Re: The figure shows the intersection of a Square ABCD and a Circle. What  [#permalink]

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08 Dec 2019, 00:30
Here lets start by considering radius of circle as r. Let O be the center of circle and let M be the midpoint of side BC where Perpendicular from O meets BC.
By applying right angle property
(OB)^2 = (BM)^2 + (OM)^2
r^2 = (BM)^2 + (r-x)^2
on solving, we get (BM)^2 = x(2r-x)

since from 1 we know the value of x, we get value of (BM)^2 as 2r-1

so we cannot find the perimeter of square just from 1

(2) since its told that abcd is a square, we can already infer what is given in 2

Taking both 1 and 2, still we donot have enough information to find the value of r so answer is E
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Re: The figure shows the intersection of a Square ABCD and a Circle. What  [#permalink]

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08 Dec 2019, 00:52
Ans: A
(1): from the mid point of the circle O draw radius to point B and C .Since the triangle OBC is an isosceles triangle with angle BOC = 90 degree. We can say sin 45 = OB/OX.
Now we can find radius and hence the perimeter.
(2). This is a redundant info . Its given that the figure is square which implies that diagonals will intersect at 90 degree.
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Re: The figure shows the intersection of a Square ABCD and a Circle. What  [#permalink]

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08 Dec 2019, 01:01
#1
X=1 ;
insufficient as we dont know the center of circle also X=1 is the highest point of circle from square
#2
AC and BD are perpendicular to each other
the lines will intersect at center at 90* , but center of circle is not clear ; insufficient
from 1 &2
we can determine the sides of the digonals using 45:45:90 ; but will have unknow value of distance from center to circle side X=1
radius ; a+1 ; side of square ; 2a and perimeter ; 8a ; but value of a is not know
insufficient
IMO E

The figure shows the intersection of a Square ABCD and a Circle. What is the Perimeter of the Square ABCD?
1) X = 1
2) AC and BD are perpendicular to each other
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Re: The figure shows the intersection of a Square ABCD and a Circle. What  [#permalink]

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08 Dec 2019, 01:05
Let us look at each statements one by one

1) X = 1
This statement does not provides us any information which will provide us the dimensions of the square, we cannot arrive at any side dimension length of the square hence statement 1 alone is not sufficient

2) AC and BD are perpendicular to each other

This statement does not provides us any information which will provide us the dimensions of the square, hence statement 1 alone is not sufficient

Even when we combine information in both the statements together, it won't give us any information about square dimensions or of the circle, hence both statements together are also not sufficient.
Answer is Option E, even both the statements together are also not sufficient.
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Re: The figure shows the intersection of a Square ABCD and a Circle. What  [#permalink]

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08 Dec 2019, 03:05
The figure shows the intersection of a Square ABCD and a Circle. What is the Perimeter of the Square ABCD?

Tricky one for me, still i'll try.

1) X = 1,
Insufficient as we can't find the radius of the circle, which can give us diameter of the circle, which in turn could help to find side or diagonal of square.

2) AC and BD are perpendicular to each other
It implies that diagonals pass through the center of the circle. But still we can't find the radius/diameter of the circle/diagonal of the square/side of square. Hence insufficient.

1) + 2)
Square is inscribed from the circle as AC and BD are perpendicular to each other. Still can not find the side of square. Hence insufficient.

Imo. E
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Re: The figure shows the intersection of a Square ABCD and a Circle. What  [#permalink]

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08 Dec 2019, 06:37

EXPLANATION:

Please refer to the attached figure:
We have assumed a radius is drawn from B and C, meeting at the point which is coolinear to the point x distance is separated from the square. Hence, if the total distance is R, the point till the square edge (where x distance is separated from the circle), is r - x.

We can see, perimeter is:
r + k + 2l + r + k + 2l.

Also, we know, $$l = \sqrt{r^2 - (r-1)^2 }$$
Hence, L equals, $$\sqrt{2r - 1 }$$

Hence, solving the equation, we get perimeter as function of R as:
$$4r - 2 + 4\sqrt{2r-1}$$

Since the value of R can vary, the perimeter can vary and have more than one answer.
Attachments

File comment: Diagram

IMG_20191208_184450__01.jpg [ 1.58 MiB | Viewed 624 times ]

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Re: The figure shows the intersection of a Square ABCD and a Circle. What  [#permalink]

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08 Dec 2019, 11:04
2
In the above figure, let OB = OC = R. Let, OE = y, AB = 2a
Now, R = x + y ...(i)
R + y = 2a ...(ii)
In triangle OEB, R^2 = y^2 + a^2 ...(iii)

Substituting (i) in (iii), we get
(x+y)^2 = y^2 + a^2
x^2 + 2xy + y^2 = y^2 + a^2
x^2 + 2xy = a^2 ...(iv)

Substituting (i) in (ii), we get
x+2y = 2a

Multiplying above equation by x, we get

x^2 + 2xy = 2ax ... (v)

From (iv) and (v), we get

a^2 = 2ax

a=0 or a=2x

Since, a is a length, it can not be non - positive.

Therefore, a = 2x ... (vi)

(1) x = 1
Substituting in (vi), we get
a = 2. Therefore, perimeter = 8a = 16.
Clearly, statement 1 is sufficient.

(2) AC and BD are perpendicular to each other
Since, ABCD is a square AC and BD are perpendicular to each other. We do not get any information about the length of the square.
Therefore statement 2 is insufficient.

Attachments

20191208_231723.jpg [ 1.39 MiB | Viewed 578 times ]

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Re: The figure shows the intersection of a Square ABCD and a Circle. What  [#permalink]

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09 Dec 2019, 05:07
For every square and circle which exist as shown in the figure, distance x will be unique and no two combination of square and circle can have same value of x.

(1) For x=1, we have a unique figure and thus the perimeter of the square can be determined.
Therefore, statement 1 is sufficient.

(2) AC and BD are perpendicular to each other
ABCD is a square AC and BD are always perpendicular to each other.
Not sufficient.

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Re: The figure shows the intersection of a Square ABCD and a Circle. What  [#permalink]

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06 Jan 2020, 00:19
jhavyom can you explain how R+Y =2a in your explanation
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Re: The figure shows the intersection of a Square ABCD and a Circle. What  [#permalink]

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06 Jan 2020, 01:00
The solution of this question is as per the sketch:

Attachment:

WhatsApp Image 2020-01-06 at 1.29.51 PM.jpeg [ 58.55 KiB | Viewed 170 times ]

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Re: The figure shows the intersection of a Square ABCD and a Circle. What  [#permalink]

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06 Jan 2020, 01:02
As per the jhyavyom solution, Side of the square is taken as 2a
Hence R+2a = side of the square = 2a

I hope it is clear, feel free to tag me in case of any doubt.

Happy Learning

saiprasanna wrote:
jhavyom can you explain how R+Y =2a in your explanation

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Re: The figure shows the intersection of a Square ABCD and a Circle. What   [#permalink] 06 Jan 2020, 01:02
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