Circles are drawn with four vertices as the center and radius equal to

Question

Circles are drawn with four vertices as the center and radius equal to the side of the square. If the square is formed by joining the mid points of another square of side  2*\sqrt{6}) , what is the area common to all the four circles

A.  \frac{\sqrt{3} -1}{2}*6\pi

B.  4\pi -3*\sqrt{3}

C.  \frac{\pi - 3*\sqrt{3}}{2}

D.  4\pi-12(\sqrt{3}-1)

E.  4\pi-8(\sqrt{3}-1)

EvaJager · Accepted Answer

Circles are drawn with four vertices as the center and radius equal to the side of the square. If the square is formed by joining the mid points of another square of side \(2*\sqrt{6})\), what is the area common to all the four circles?

A. \(\frac{\sqrt{3} -1}{2}*6\pi\)

B. \(4\pi -3*\sqrt{3}\)

C. \(\frac{\pi - 3*\sqrt{3}}{2}\)

D. \(4\pi-12(\sqrt{3}-1)\)

E. \(4\pi-8(\sqrt{3}-1)\)

Please, refer to the attached drawings

From the first figure, we can deduce that the side of the smaller square is half of the diagonal of the larger square, so it is equal to \(\frac{1}{2}\cdot{2}\sqrt{6}\cdot{sqrt{2}=2\sqrt{3}}\).

We have to calculate 4 times the area of the shaded region in the first figure.
Now, please refer to the second drawing.
Since \(OA=2\sqrt{3}\) is twice \(AB=\sqrt{3}\), triangle \(OAB\) is a 30-60-90 triangle. We can see that the quarter of the circle with radius \(OA\) is divided into three equal sectors with a central angle of 30 degrees. The area of each such sector is\(\frac{\pi(2\sqrt{3})^2}{4\cdot{3}}=\pi\).

To find the area of the shaded region, we should subtract from the area of the 30 degrees sector twice the area of the triangle \(ACO\).
\(OB=\sqrt{3}AB=\sqrt{3}\sqrt{3}=3\), therefore \(x=3-\sqrt{3}\).
The area of the shaded region is \(\pi-2\cdot\frac{x\sqrt{3}}{2}=\pi-(3-\sqrt{3})\sqrt{3}=\pi-(3\sqrt{3}-3)=\pi-3(\sqrt{3}-1)\).
Thus the requested area is \(4(\pi-3(\sqrt{3}-1))=4\pi-12(\sqrt{3}-1)\).

Answer D.

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Attachment:

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mikemcgarry · Answer

Circles are drawn with four vertices as the center and radius equal to the side of the square. If the square is formed by joining the mid points of another square of side \(2*\sqrt{6})\), what is the area common to all the four circles

A. \(\frac{\sqrt{3} -1}{2}*6\pi\)

B. \(4\pi -3*\sqrt{3}\)

C. \(\frac{\pi - 3*\sqrt{3}}{2}\)

D. \(4\pi-12(\sqrt{3}-1)\)

E. \(4\pi-8(\sqrt{3}-1)\)

Show Answer

Official Answer

D

First of all, let me say, this is a HARD problem, much hard than anything you could expect to see on the GMAT. This is getting to the level of some of the more elementary formal mathematical competitions (e.g. the AIME) You do not need to know how to do problems like this for the GMAT. Ms. EvaJager showed an elegant solution. Just in case you want to see some more detail, I have posted a pdf of my solution. I hope this helps. Mike

avaneeshvyas · Answer

I have trouble understanding how you can conclusively say that the triangle OAB is a 30-60-90 triangle. I want to understand this particular logic as I feel I lack these nuances which makes geometry a mess for me.....

avaneeshvyas · Answer

In the solution PDF how can we conclusively say that MQL or KPL or JRM are equilateral triangles......

EvaJager · Answer

In a right triangle with angles 30-60-90, the shorter leg is half of the hypotenuse. The reciprocal is also true: if in a right triangle one of the legs is half of the hypotenuse, then the angle opposing that leg is 30 (the other acute angle must necessarily be 60).

These are proven facts. You should go over the Geometry requisites, and understand and memorize the most important properties/facts. Although this is not a typical GMAT question, the above properties are a must on the GMAT.

avaneeshvyas · Answer

What i intended to ask was ...you concluded about the size of the sides assuming that the triangle is a 30-60-90 triangle. How did u conclude this? And just in case it is the other way round ....i.e you concluded that it is 30-60-90 triangle based on the sides then my question would be how do we know that AB = 3^0.5? Hope I am making my question clearer now...

EvaJager · Answer

From the sizes I deduced the angles. 
AB is perpendicular to OB. Call D the point above O on the extension of AC. ODAB is a rectangle, in which opposite sides are equal, so  AB=\sqrt{3} .
Also, compare with the initial larger drawing, in which the interior square is divided into four congruent squares. So, the smallest squares, all have a side of  \sqrt{3} .

mikemcgarry · Answer

Let's take MQL as an example. ML is the length of a radius, because the side of the square JKLM equals the radius of each circle. Point Q is on the circle centered at L, so QL is the length of a radius. Point Q is also on the circle centered at M, so QM is a radius. Thus ML = QL = QM, and the triangle is equilateral. Whenever you draw a segment, and construct circles at each endpoint with radii equal to the length of the segment, that produces equilateral triangles. This fundamental idea is actually the very first theorem in Euclid's Elements . Does this make sense? Mike :-)

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Circles are drawn with four vertices as the center and radius equal to

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