The solution of this question is as per the sketch:




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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

#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

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

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.

Choice A is the answer.

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

Please expand on: R + y = 2a

How can that be true if R^2 = a^2 + y^2?

gmatbusters jhavyom

By symmetry, it lies on diameter. It can be proved but better to use the principle of symmetry in GMAT.
Also, since the dimensioning line has 2 arrows, it means the max distance between the circle and circle, which is when E lies on diameter (otherwise the exact point on circle must have been specified).


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