In the figure shown, if the side of the square is 40, what is the radi

IMO C
AS the diameter of large circle is 40 as the side of square is also 40
then diagonal of square is \(40\sqrt{2}\)
and this diagonal will include 2 small circles and 1 large circle so radius of small circle is approx
\(\frac{(40\sqrt{2} - 40)}{2}\)( 40root2-40/2)
so \(20(\sqrt{2}-1)\) (this will be a bit bigger than diameter of smaller circle)
we need radius then divide it by 2 which make bit bigger than radius as \(10(\sqrt{2}-1)\)

but as you can see both small circles are not touching in the corner of the square then the answer will be little less than the above answer
above answer in terms of value is 10*0.3=3 correct answer will be little less
A.10*0.3= 3 our answer were bith less than 3
B.6 way greater than 3 incorrect
C.6/2.3= 2.3 (bit less correct)
D.40*0.3= 12 (very large)
E.40*0.3/2.3= 5.3(greater than 3 incorrect)

-------------------------------------------------------------------------------------------------------------------------------

If you like my explanation than please give me KUDOS

IMO C.

This is a nice tricky question and I took over 4 mins to solve it. Everyone who have tried a solution have got a different answer and I am excited to find out the OA.

Please refer to the attached figure. We have to find t ( radius of the small circle ) as in the figure.

The figure is zoomed into the 1/4th part of the larger square.

The distance between one corner of the square and the circle is diagonal of the square of side 20 minus the radius of inscribed large circle of radius 20.
I have called the distance x. And x = \(20\sqrt{2} - 20\)

This is the trap option!

However this distance x is split between the radius of the circle t and and distance between the center of small circle and corner of large square t\sqrt{2}.

Hence by multiplying the x in the correct ratio we can find radius t.

t =\(x * 1/{\sqrt{2}+1}\)

t = \(\frac{20(\sqrt{2} - 1)}{\sqrt{2}+1}\)

Hence C.

Please give my answer a Kudos! if you liked the explanation...

Shouldn't it 4r????

Sorry my bad got it!!

Sent from my ONEPLUS A5000 using GMAT Club Forum mobile app

Hi Dexter78424,

Not quite clear what you are asking exactly. But the trick part of this question is - the ability to zoom in onto 1/4th of the larger square and focus on that.

You can see the detailed explanation in my post above & also in many other posts. The OA is correct -> Option (C)

Please go through those and ask specific doubts if you have any..

Best,
Gladi

taking the approach of setting the (Diagonal of the Square) - (Diameter of the Circle) = 2 of the Diameters of the Smaller Circles will end up overvaluing the actual radius because the smaller circles do NOT touch the Vertices of the Square.


Rule 1: and Radius drawn to a Line Tangent to the Circle will be Perpendicular to the Tangent Line.


If you call the Radius of the Small Circle = little "r"

Drawing 2 radii = r ---- from the Center of the Smaller Circles to the Tangent Side of the Square will create a Square of side r in the Corner of the Larger Square.

Thus, the Diagonal from the Center of the Smaller Circle to the Vertex of the Square = r * sqrt(2) = Diagonal of Small created Square


the Distance from the Center of the Smaller Circle to the Point of Tangency at the Larger Circle = r


the Distance across the Center of the Inscribed Circle = Diameter = Side of Square = 40


thus:


Entire Diagonal of Large Square = r * sqrt(2) + r + 40 + r * sqrt(2)

40 * sqrt(2) = 2 * r * sqrt(2) + 2*r + 40

----DIVIDE both sides by 2----

20 * sqrt(2) - 20 = r * sqrt(2) + r

---take r Common---

20 * sqrt(2) - 20 = r * [ sqrt(2) + 1 ]

r = [20 * sqrt(2) - 20] / [1 + sqrt(2)]


-C-

Hi Bunuel
GMATGuruNY

Why doesn't it work to take the centroid formula 1/3 * 20(sqrt{2}-1)

I mean if you imagine the smaller circle as an incircle of an imaginary triangle with base as common tangent to both circles.

Then diameter of the smaller circle is perpendicular tangent and also angular bisector of square (coincides with the diagonal of the square)

Diameter of the large circle = side of the square = 40
Diagonal of the square \(= 40\sqrt{2}\) ≈ 40*1.4 = 56
The following figure is yielded:


Since each red line segment = 8, the diameter of each small circle = LESS THAN 8.
Thus, the radius of each small circle = LESS THAN 4.
Only C yields a value less than 4:
\(\frac{20(\sqrt{2} - 1)}{\sqrt{2}+1}\) ≈ \(\frac{20(1.4-1)}{1.4+1} = \frac{8}{2.4} =\) less than 4

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.