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Math Expert V
Joined: 02 Sep 2009
Posts: 58402
Six congruent circles are packed into an equilateral triangle so that  [#permalink]

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11 00:00

Difficulty:   85% (hard)

Question Stats: 45% (02:00) correct 55% (01:42) wrong based on 124 sessions

HideShow timer Statistics 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 $$48+28\sqrt{3}$$

Attachment: six_circles_packed_1.png [ 16.87 KiB | Viewed 1999 times ]

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Math Expert V
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Posts: 58402
Re: Six congruent circles are packed into an equilateral triangle so that  [#permalink]

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Bunuel wrote: 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 $$48+28\sqrt{3}$$

Attachment:
six_circles_packed_1.png

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Six congruent circles are packed into an equilateral triangle so that  [#permalink]

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Bunuel wrote: 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 $$48+28\sqrt{3}$$

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

Attachments

File comment: www.GMATinsight.com Screen Shot 2018-03-09 at 1.40.56 PM.png [ 204.67 KiB | Viewed 1598 times ]

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Six congruent circles are packed into an equilateral triangle so that  [#permalink]

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2
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 [ 99.73 KiB | Viewed 1506 times ]

GMATinsight wrote:
Bunuel wrote: 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 $$48+28\sqrt{3}$$

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

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Joined: 12 Oct 2010
Posts: 935
Re: Six congruent circles are packed into an equilateral triangle so that  [#permalink]

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Bunuel wrote: 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 $$48+28\sqrt{3}$$

$$?\,\,\, = \,\,\,{S_\Delta } - 6 \cdot {S_ \circ }$$

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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Our high-level "quant" preparation starts here: https://gmath.net Re: Six congruent circles are packed into an equilateral triangle so that   [#permalink] 26 Oct 2018, 08:44
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