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705-805 (Hard)|   Geometry|      
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Bunuel
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Six congruent circles are packed into an equilateral triangle so that 08 Mar 2018, 21:17
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

75% (hard)

Question Stats:

47% (01:58) correct 53% (01:50) wrong based on 310 sessions

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Six congruent circles are packed into an equilateral triangle so that no circle is overlapping and such that circles are tangent to one another or the triangle at any point of contact, as shown above. What is the area of the part of the triangle that is NOT covered by circles?

(1) The radius of each circle is 2
(2) The area of the triangle is \(48+28\sqrt{3}\)


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six_circles_packed_1.png
six_circles_packed_1.png [ 16.87 KiB | Viewed 12454 times ]
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D
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
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Six congruent circles are packed into an equilateral triangle so that 08 Mar 2018, 21:22
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Bunuel


Six congruent circles are packed into an equilateral triangle so that no circle is overlapping and such that circles are tangent to one another or the triangle at any point of contact, as shown above. What is the area of the part of the triangle that is NOT covered by circles?

(1) The radius of each circle is 2
(2) The area of the triangle is \(48+28\sqrt{3}\)


Show Spoiler
Attachment:
six_circles_packed_1.png
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23. Geometry




24. Coordinate Geometry




25. Triangles




26. Polygons




27. Circles




28. Rectangular Solids and Cylinders




29. Graphs and Illustrations



For other subjects:
ALL YOU NEED FOR QUANT ! ! !
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Six congruent circles are packed into an equilateral triangle so that 09 Mar 2018, 01:07
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Bunuel


Six congruent circles are packed into an equilateral triangle so that no circle is overlapping and such that circles are tangent to one another or the triangle at any point of contact, as shown above. What is the area of the part of the triangle that is NOT covered by circles?

(1) The radius of each circle is 2
(2) The area of the triangle is \(48+28\sqrt{3}\)


Show Spoiler
Attachment:
The attachment six_circles_packed_1.png is no longer available
Show more

Question: Area of uncovered part of triangle = Area of triangle - Area of six circles = ?

Statement 1: The radius of each circle is 2
Using this information the side of the equilateral triangle can be calculated (using 30-60-90 property) as mentioned in attachment, Hence
SUFFICIENT

Statement 2: he area of the triangle is \(48+28\sqrt{3}\)
sing this information the side of the equilateral triangle can be calculated as mentioned in attachment, Hence
SUFFICIENT

Answer: option D
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Screen Shot 2018-03-09 at 1.40.56 PM.png
Screen Shot 2018-03-09 at 1.40.56 PM.png [ 204.67 KiB | Viewed 10876 times ]

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Six congruent circles are packed into an equilateral triangle so that 10 Mar 2018, 20:42
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There is a simpler approach to it, by trying to draw it on paper.
see the Sketch attached.
Attachment:
WhatsApp Image 2018-03-11 at 09.09.28.jpeg
WhatsApp Image 2018-03-11 at 09.09.28.jpeg [ 99.73 KiB | Viewed 10758 times ]

GMATinsight
Bunuel


Six congruent circles are packed into an equilateral triangle so that no circle is overlapping and such that circles are tangent to one another or the triangle at any point of contact, as shown above. What is the area of the part of the triangle that is NOT covered by circles?

(1) The radius of each circle is 2
(2) The area of the triangle is \(48+28\sqrt{3}\)


Show Spoiler
Attachment:
The attachment six_circles_packed_1.png is no longer available

Question: Area of uncovered part of triangle = Area of triangle - Area of six circles = ?

Statement 1: The radius of each circle is 2
Using this information the side of the equilateral triangle can be calculated (using 30-60-90 property) as mentioned in attachment, Hence
SUFFICIENT

Statement 2: he area of the triangle is \(48+28\sqrt{3}\)
sing this information the side of the equilateral triangle can be calculated as mentioned in attachment, Hence
SUFFICIENT

Answer: option D
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Six congruent circles are packed into an equilateral triangle so that 26 Oct 2018, 08:44
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Bunuel


Six congruent circles are packed into an equilateral triangle so that no circle is overlapping and such that circles are tangent to one another or the triangle at any point of contact, as shown above. What is the area of the part of the triangle that is NOT covered by circles?

(1) The radius of each circle is 2
(2) The area of the triangle is \(48+28\sqrt{3}\)
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\(?\,\,\, = \,\,\,{S_\Delta } - 6 \cdot {S_ \circ }\)

The GMATH widget (a.k.a "GW") is defined in the figure below. It is formed combining an equilateral triangle (side length 4r) and three "legs" (length r each), where r is any positive number.



With the GW in mind, the answer is (D) immediately:

(1) The value of r is given, hence the corresponding GW is unique, hence the triangle that circumscribes the GW is unique (dotted triangle in the figure shown in the bottom-left). Our FOCUS is unique.

(2) The equilateral triangle is given, hence the corresponding inscribed GW is unique (dotted GW shown in the bottom-right), hence r is unique. Our FOCUS is unique.


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Six congruent circles are packed into an equilateral triangle so that 11 Mar 2021, 22:52
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Bunuel


Six congruent circles are packed into an equilateral triangle so that no circle is overlapping and such that circles are tangent to one another or the triangle at any point of contact, as shown above. What is the area of the part of the triangle that is NOT covered by circles?

(1) The radius of each circle is 2
(2) The area of the triangle is \(48+28\sqrt{3}\)


Show Spoiler
Attachment:
six_circles_packed_1.png

Question: Area of uncovered part of triangle = Area of triangle - Area of six circles = ?

Statement 1: The radius of each circle is 2
Using this information the side of the equilateral triangle can be calculated (using 30-60-90 property) as mentioned in attachment, Hence
SUFFICIENT

Statement 2: he area of the triangle is \(48+28\sqrt{3}\)
sing this information the side of the equilateral triangle can be calculated as mentioned in attachment, Hence
SUFFICIENT

Answer: option D
Show more

Can you explain statement 2? How does knowing the side length allow us to get the area of the circle?

Re-did this question.

To clarify, both statements are sufficient. The key is knowing that a 30-60-90 triangle forms at each corner of the equilateral triangle. If we can get the radius of the circle, then we can solve.

For statement 2, we can get the side length from knowing the area of the equilateral triangle. That side length in turn is equal to (4 x radius) + (2 x radius√3).

D.
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Six congruent circles are packed into an equilateral triangle so that 01 Nov 2023, 07:43
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CEdward
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Bunuel


Six congruent circles are packed into an equilateral triangle so that no circle is overlapping and such that circles are tangent to one another or the triangle at any point of contact, as shown above. What is the area of the part of the triangle that is NOT covered by circles?

(1) The radius of each circle is 2
(2) The area of the triangle is \(48+28\sqrt{3}\)


Show Spoiler
Attachment:
six_circles_packed_1.png

Question: Area of uncovered part of triangle = Area of triangle - Area of six circles = ?

Statement 1: The radius of each circle is 2
Using this information the side of the equilateral triangle can be calculated (using 30-60-90 property) as mentioned in attachment, Hence
SUFFICIENT

Statement 2: he area of the triangle is \(48+28\sqrt{3}\)
sing this information the side of the equilateral triangle can be calculated as mentioned in attachment, Hence
SUFFICIENT

Answer: option D

Can you explain statement 2? How does knowing the side length allow us to get the area of the circle?

Re-did this question.

To clarify, both statements are sufficient. The key is knowing that a 30-60-90 triangle forms at each corner of the equilateral triangle. If we can get the radius of the circle, then we can solve.

For statement 2, we can get the side length from knowing the area of the equilateral triangle. That side length in turn is equal to (4 x radius) + (2 x radius√3).

D.
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Hi CEdward
How can I know that a 30-60-90 triangle forms at each corner of the equilateral triangle?
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