In the figure above, each of the four large circles is tangent to two

Question

https://gmatclub.com/forum/download/file.php?id=58246 In the figure above, each of the four large circles is tangent to two of the other large circles, the small circle, and two sides of the square. If the radius of each of the large circles is 4, what is the diameter of the small circle? A. \sqrt{2} B. 1 C. 8 \sqrt{2} - 8 D. 4 \sqrt{2} - 4 E. \sqrt{2} - 1 1.png

SHoylandBPrep · Answer

You need to draw a triangle by connecting the centers of two large circles with the center of the small circle.

https://i.imgur.com/Oi027U3.png

You can tell that this triangle is a right triangle based on the symmetrical layout of the four large circle. By this same symmetry, the other angles must be 45º each –– you can also show this by noting that the triangle must be isosceles.

Therefore, this is a 45º-45º-90º triangle, so the sides must be in the ratio 1:1:√2. (If you don't remember this ratio, you can also use the Pythagorean theorem.) The hypotenuse of the triangle is made up of two large radii, so its length is 8. Thus, the legs are 8 ÷ √2. You can simplify this as a fraction.

→  \frac{8}{\sqrt{2}} = \frac{8}{\sqrt{2}} × \frac{\sqrt{2}}{\sqrt{2}} = \frac{8\sqrt{2}}{2} = 4\sqrt{2}

Therefore, the legs are  4\sqrt{2}  each. Each leg is made up of a large radius and a small radius.

→ Large Radius + Small Radius =  4\sqrt{2}

→  4  + Small Radius =  4\sqrt{2}

→ Small Radius =  4  –  4\sqrt{2}

The diameter is twice this radius, so the answer is  (C)  8 – 8\sqrt{2}  .

CrackverbalGMAT · Answer

Each side of the square is the diameter of the 2 circles = 8 + 8 = 16

The diagonal of the square =  a \sqrt{2} = 16 \sqrt{2}

This diagonal is made up of 5 parts AB, BC, CD (the diameter of the smaller circle, DE and EF, as shown in the diagram below.

BC = DE = 4

GFEH form a square of side length 4. The diagonal of the square = EF =  4 \sqrt{2}

Therefore  16 \sqrt{2} = AB + BC + d + DE + EF = 4 \sqrt{2} + 4 + d + 4 + 4 \sqrt{2}

d = 16 \sqrt{2} - 8 \sqrt{2} - 8

d = 8 \sqrt{2} - 8

Option C

Arun Kumar

QuantMadeEasy · Answer

Step 1: Understanding the question 
Let A, B and O be the centers of the circles and radius of the smaller circle be x
As the diagonals of the square intersect at right angle, triangle AOB is a right angled at O

AB = 4+4 = 8
AO = (x+4)
BO = (x+4)

Applying Pythagoras
AB^2=AO^2+BO^2
64 = 2* (x+4)^2
32 = (x+4)^2
x + 4 = 4√2
x = 4√2 - 4

Hence diameter of the smaller circle = 8√2 - 8
 C is correct

ramlala · Answer

Diagonal of the square 16√2 = 2 times diameter of larger circle (16) + 2 times diameter of smaller circle (2d)
therefore 2d = 16√2 - 16
d = 8√2 - 8

Answer is c

ScottTargetTestPrep · Answer

Solution: Since the radius of the larger circle is 4, the diameter of the larger circle is 8 and the side length of the square is twice the diameter of the larger circle, so the side length of the square is 16. The diagonal of the square is then 16√2. Notice that we can draw a quarter of the smaller circle at each corner of the square that is tangent to the larger circle. So, if we draw the diagonal of the square, say from the upper left corner to the lower right corner of the square, it will contain, in order, the radius of (a quarter of) the smaller circle, the diameter of the larger circle, the diameter of the smaller circle, the diameter of the larger circle, and finally, the radius of (a quarter of) the smaller circle, Since the diagonal of the square = 16√2 and if we let r = the radius of the smaller circle, we can create the equation: r + 8 + 2r + 8 + r = 16√2 4r + 16 = 16√2 4r = 16√2 - 16 r = 4√2 - 4 Therefore, the diameter of the smaller circle is 2r = 2(4√2 - 4) = 8√2 - 8. Alternate Solution: Connect the centers of the top large circles and form a triangle with the center of the smaller circle, as shown: Circles.png Note that the length of AC = the length of BC. Now, use the Pythagorean theorem to determine the length of side AC: (AC)^2 + (AC)^2 = 8^2 2(AC)^2 = 64 (AC)^2 = 32 AC = 4√2 Note that the length of AC is the sum of the radii of the large and small circles. We know that the radius of the large circle is 4, so the radius of the small circle will be 4√2 - 4. Thus, its diameter is 2(4√2 - 4) = 8√2 - 8. Answer: C

In the figure above, each of the four large circles is tangent to two

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In the figure above, each of the four large circles is tangent to two

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