3 identical circles of radius ‘r’ are placed in a plane such that all

Question

3 identical circles of radius ‘r’ are placed in a plane such that all three of them are touching the other two circles. A new circle is then made passing through the centers of the three circles. By what percent is the radius of new circle more than the radius of each of the three identical circles?

A. 15.5%
B. 17%
C. 20.5%
D. 25%
E. 30.5%

A. 15.5%
B. 17%
C. 20.5%
D. 25%
E. 30.5%

GMATinsight · Answer

By joining the centers of three circles we get an equilateral triangle of side r+r = 2r length

The height of this equilateral triangle = √3/2(2r) = √3r

The Radius of circle passing through vertex of this new equilateral triangle = (2/3)*Height of triangle = (2/3)*√3r = (2/√3)r = 1.155 r

i.e. The radius of new circle = 15.5% greater than the radius of previous circles

Answer: Option A

Shrey9 · Answer

In your figure, the circle is outside 3 circles, but in the question, it's given that the new circle passes through the centre of the given circles..
Does that make a difference to the answer you've posted?

Posted from my mobile device

GMATinsight · Answer

No ... It doesn;t make any difference on the answer and solution... The answer is correct... I didn't draw the figure instead I searched online for an appropriate figure and this was the closest figure. only the 4th circle was missing.

The answer is correct

Archit3110 · Answer

GMATinsight 
hi sir , i did not understand the highlighted part of how will the 2/3rd of the height of the ∆ be the radius?

Mo2men · Answer

Dear  GMATinsight

In right triangle OBD, OB must be √3r and should be equal to the height of the triangle OA as both are radii of the circle passing through the centers of 3 identical circles. But actually both do not yield same result. What is wrong here?

Can you help please?

firas92 · Answer

The centre of the new circle will be at the centroid of the equilateral triangle of side 2r since it is equdistant from the centres of all three smaller circles.

The centroid divides the median (or h) in the ratio 2:1 or in the fractions 2/3 from vertex to centroid and 1/3 from centroid to opposite edge

Once we have the distance between vertrex and centre of the triangle, we have the radius of the new larger circle

Hope its clear!

Posted from my mobile device

Archit3110 · Answer

firas92 
Okay understood now

GMATinsight · Answer

The highlighted part represents that the center of equilateral triangle always divides the height of equilateral triangle in 2:1 ratio

i.e. Radius of the circle passing through vertices of an equilateral triangle will have Radius = (2/3) height of equilateral triangle

YOu can always prove it with 30º-60º-90º rule as well...

check attached file

Archit3110

Archit3110 · Answer

The highlighted part represents that the center of equilateral triangle always divides the height of equilateral triangle in 2:1 ratio i.e. Radius of the circle passing through vertices of an equilateral triangle will have Radius = (2/3) height of equilateral triangle YOu can always prove it with 30º-60º-90º rule as well... check attached file Archit3110 GMATinsight thank you sir, I have now understood the logical reason behind of how the radius of circle was determined ..

Sreeragc · Answer

************************************************************************************

Hi Sir,

i've reached till (2/sqrt3) r,

but then i took sqrt 3 = 1.7 just like we take 1.4 for sqrt 2 for the ease of calculation and got the result as 1.176r and a 17.6 % increase and apparently choose option B.

So how should we know up to which decimal place should we consider provided we cant use a calculator.

Thanks.

GMATinsight · Answer

I think you are in the right direction when you choose the value of √3 as 1.7 instead of 1.732 GMAT expects us to use √2= 1.4 instead of 1.414 and √=1.7 instead of 1.732 unless clearly mentioned. So the options will certainly not as close as you saw in this question. RELAX... Let me admit that I just used the calculator and broke the law of GMAT while calculating my answer

Sreeragc

bumpbot · Answer

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3 identical circles of radius ‘r’ are placed in a plane such that all

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