In the figure below four identical circles of radius 19.5 cm have been

Question

https://gmatclub.com/forum/download/file.php?id=55176 In the figure above four identical circles of radius 19.5 cm have been drawn on the sides of a square such that the points of contact of the circles are the vertices of the square as shown below. What is the perimeter of the square? A. 92 cm B. 88 cm C. 100 cm D. 110 cm E. 115 cm 1.png

MayankSingh · Answer

IMO D

Refer attached Image.
Let ABCD be the square & O1, O2, O3, O4 be the center of the circles.
Let join the center of the circles to each other & opposite vertices of the sqaure.

Now, diagonal AC & BD are acting as tangent to the circle at point of contact.
 And, Tangent is perpendicular to the radius at point of contact .

So, AC perpendicular to O1O2 & BD AC perpendicular to O2O3
In Quadrilateral AEBO2, Angle AO2B = 90

so AB = Sqrt (R^2+R^2) = 19.5 Sqrt 2
Perimeter = 4x19.5 Sqrt2 = 110

yashikaaggarwal · Answer

Draw line from centre to the chord of circles.
We will get a square with side 19.5+19.5 = 39
Since a square edge measure = 90°
The triangle form by radius and chord is a right angle triangle, right angled at Centre O.

The radius are the two legs of right angle triangle and chord also the side of short square = hypotenuse.
Therefore, R^2+R^2 = H^2
2(R)^2 = H^2
2*(19.5)^2 = H^2
380.25*2 = H^2
H = √760.5
H = 27.5 = Side of Square 
Perimeter of square = 4*Side
Perimeter = 4*27.5 = 110

Answer is D

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dimri10 · Answer

R AND R creates aright angle triangle with thr ratio R:R: R*ROOT 2
So the Side of the square is about 20 (a bit less)*root 2= 20*1.4=a bit less than 28

a bit less than 28 *4= 110

rajeshforyouu · Answer

Visualization is important here. Let's bring the square into a single Circle.

Diameter of circle: 2*19.5 = 39
Diagonal of Square:  \sqrt{2} *a (a is side of square) 
Here 
Diagonal = Diameter
 \sqrt{2} *a = 39
a =39/ \sqrt{2}

Perimeter of Square : 4*39/ \sqrt{2}  =   110

110

vipulshahi · Answer

Connect the centre of all the circles. 
Lets suppose A, B,C,D be the centre of the circle passses through the vertices of squares M, N, P &Q
M = vertices between the lines AB
N= vertices between the lines BC
P = vertices between the lines CD
Q= vertices between the lines DA
Then,
AB = AM + MB = 19.5+19.5 = 39
BC = BN + NC = 19.5+ 19.5 = 39

In Triangle MBN,

MN^2 = MB^2 + BN^2 = (19.5)^2 + (19.5)^2
MN = sqrt 
MN = sqrt  
MN = 19.5 * 1.4

Perimeter of square = 4* side of square = 4*MN = 4* 19.5*1.4 = 4*195*14/10*10 = 780*14/100
= 110cm

Answer : D

manikmehta95 · Answer

Can you please explain why the diagonal of the square will be a tangent to the circle? the question only mentions that the two circles touch at the vertice of the square

In the figure below four identical circles of radius 19.5 cm have been

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In the figure below four identical circles of radius 19.5 cm have been

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