Three circles with radius 2 cm are tangent to each other as shown abov

Question

Competition Mode Question https://gmatclub.com/forum/download/file.php?id=54360 Three circles with radius 2 cm are tangent to each other as shown above. What is the area of the circle, circumscribing the above figure? A. \frac{π}{4}(4 + 2√3)^2 B. \frac{π}{3}(4 + 2√3)^2 C. 3π(4 + √3)^2 D. 4π(4 + √3)^2 E. 5π(4 + √3)^2 Are You Up For the Challenge: 700 Level Questions mensuration-15.png

nick1816 · Answer

Assume A, B and C are the center of the 3 circles and r is the radius.

ABC is an equilateral triangle having each side equals to 2r and the center of the circle circumscribing the 3 identical circle lies on the centroid, E of the triangle ABC.

AE =\frac{ 2}{3} * \frac{√3}{2} * (2r) = \frac{2r}{√3}

Radius of bigger circle =  r+\frac{2r}{√3} = \frac{(2+√3)r}{√3}

Area of bigger circle =  π * ^2 = \frac{π}{3} * ^2

the27s · Answer

Since all circles are of same size, the centers will form an equivalent triangle.

Make a right angled triangle from center of one of the circle to the tagent point and the third vortex is the common one for all three circles, the angle is 60deg.

Hence, distance from center of the 3 circles to the tangent point is √3/2*2 = √3. Raidus of the circumscribing circle is 2+√3, area is π*(2+√3)^2, which can be re-written as option (A).

Posted from my mobile device

freedom128 · Answer

Find the attached sketch and explanation

The triangle that connects the centre of each smaller circle (r=2) is an equilateral triangle with side of 4.

The big circle that circumscribes the above figure has a radius of  R . This radius R divides each corner angle of the equilateral triangle into 2 equal part (2*30 deg).

cos 30 deg = 2 /(R-2) = √3 /2
--> R = 4/√3 + 2

Area of big circle = π. R^2
= π.(4/√3 + 2)^2
= π/3.(4 + 2√3)^2

FINAL ANSWER IS (B)

Posted from my mobile device

unraveled · Answer

Image
Three circles with radius 2 cm are tangent to each other as shown above. What is the area of the circle, circumscribing the above figure?

A. \(\frac{π}{4}(4+2√3)^2\)

B. \(\frac{π}{3}(4+2√3)^2\)

C. \(3π(4+√3)^2\)

D. \(4π(4+√3)^2\)

E. \(5π(4+√3)^2\)
Refer diagram.

Attachment:

Three circles.png [ 120.01 KiB | Viewed 9555 times ]

Sides of triangle in blue = 4 so it an equilateral triangle.
Here CD is perpendicular to side AB. CD = \(\sqrt{4^2 - 2^2}\) {from triangle CDB}
CD = \(2\sqrt{3}\)

Now, length of CO = \(\frac{2}{3}* any perpendicular = \frac{2}{3}*2\sqrt{3} = \frac{4}{\sqrt{3}}\) {where centroid lies \(\frac{2}{3}\) from any vertex, here C}

Radius of bigger circle = PC + CO = \(2 + \frac{4}{\sqrt{3}}\)
Area of bigger circle = π * r^2 = π * \((2 + \frac{4}{\sqrt{3}})^2\)
= \(\frac{π}{3}\) * \((2\sqrt{3} + 4)^2\)

Answer B.

Note: Since all the circles are of radius 2 and tangent to each other, the center of the figure of the three circles would coincide with centroid of the triangle formed by joining the centers of all the three circles.

ArunSharma12 · Answer

img3.png if we join the center of the three circles we'll get a equilateral triangle. we need a point that is equidistant from the three vertices. this point O is the circumcenter of the equilateral triangle. OA = \frac{side}{\sqrt{3}} = \frac{4}{\sqrt{3}} Radius of the bigger circle circumscribing the figure = OA + AP = \frac{4}{\sqrt{3}} + 2 = \frac{4 + 2\sqrt{3}}{\sqrt{3}} Area = \frac{\pi}{3}(4+ 2\sqrt{3})^2 Ans: B

Ezio23 · Answer

Joining the centers of 3 circles, we get an equilateral triangle with each side 4 units

Circumradius of equilateral triangle = 2/3 * (height of triangle)

height of triangle =  2 * \sqrt{3}

Therefore, Circumradius =  4/\sqrt{3}

Radius of outer circle =  4/\sqrt{3}  + 2 (circumradius + radius)

Area of outermost circle = Pi * ( 4/\sqrt{3}  + 2) * ( 4/\sqrt{3}  + 2)

On simplifying we get choice B

Answer choice: B

GMATGuruNY · Answer

https://i.ibb.co/ChYNFy0/3-inscribed-circles.png

The blue equilateral triangle can be divided into 6 congruent 30-60-90 triangles, as shown in the figure.
In each of the 30-60-90 triangles, the leg opposite the 60-degree angle is equal to the 2-cm radius of each smaller circle, with the result that the hypotenuse  = \frac{4}{\sqrt{3}} .
In the outer circle, the radius to point A  = 2 + \frac{4}{\sqrt{3}} = \frac{2\sqrt{3} + 4}{\sqrt{3}} .
Thus:
Area of the outer circle  = π(\frac{2\sqrt{3} + 4}{\sqrt{3}})^2 = \frac{π}{3}(2\sqrt{3} + 4)^2

B

exc4libur · Answer

find the circle that circumscribes all three smaller circles
find the distance from small circle origin to large circle origin
join the small origins, creating an equilateral triangle side 4
find the distance of perp bisector of the equi's side to large origin 
this forms a 30-60-90 triangle with ratio's x:xV3:2x
the distance we want is the hypotenuse: 2x
60 is the radius of small circle = 2 = xV3, x=2/V3=2V3/3
90 is 2x=4V3/3
radius large circle is radius small circle + hypotenuse 
radius large circle = 4V3/3 + 2 = 4V3+6/3
area large circle = (4V3+6/3)^2*pi

ans (B)

lacktutor · Answer

Please See The Attached File.

We got an equilateral triangle by connecting the centers of three small circles.

The area =  (\frac{√3}{4})4^{2}= 3(\frac{4h}{2} )

4√3= 6h

-->  h = \frac{2}{√3}

r = √((\frac{2}{√3})^{2}+ 4)= √\frac{16}{3}= \frac{4}{√3}

---> The center of outer circle of that equilateral triangle is the center of that big circle too.
And the radius of that big circle  R = r +2

---> R= r +2= \frac{4}{√3}+ 2= \frac{(4+2√3)}{√3}

The area of Big Circle =  πR^{2}= π\frac{(4+2√3)^{2}}{(√3)^{2}}= \frac{π}{3}(4+2√3)^{2}

Answer (B).

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

Three circles with radius 2 cm are tangent to each other as shown abov

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Three circles with radius 2 cm are tangent to each other as shown abov

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards