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The figure above shows an equilateral triangle with three circles. Eac

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The figure above shows an equilateral triangle with three circles. Eac  [#permalink]

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New post 08 Mar 2018, 20:26
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The figure above shows an equilateral triangle with three circles. Each point of contact between two circles or between a circle and the triangle is a point of tangency. If the triangle has a height of 18, what is the combined area of the two smaller circles?

A. 4π
B. 8π
C. 16π
D. 36π
E. 44π

Attachment:
Equilateral_triangle_circle_packing_1 (1).png
Equilateral_triangle_circle_packing_1 (1).png [ 13.97 KiB | Viewed 1572 times ]

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Re: The figure above shows an equilateral triangle with three circles. Eac  [#permalink]

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New post 08 Mar 2018, 20:27
Bunuel wrote:
Image
The figure above shows an equilateral triangle with three circles. Each point of contact between two circles or between a circle and the triangle is a point of tangency. If the triangle has a height of 18, what is the combined area of the two smaller circles?

A. 4π
B. 8π
C. 16π
D. 36π
E. 44π

Attachment:
Equilateral_triangle_circle_packing_1 (1).png


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The figure above shows an equilateral triangle with three circles. Eac  [#permalink]

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New post 08 Mar 2018, 23:31
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Bunuel wrote:
Image
The figure above shows an equilateral triangle with three circles. Each point of contact between two circles or between a circle and the triangle is a point of tangency. If the triangle has a height of 18, what is the combined area of the two smaller circles?

A. 4π
B. 8π
C. 16π
D. 36π
E. 44π

Attachment:
The attachment Equilateral_triangle_circle_packing_1 (1).png is no longer available


The radius of an In-Circle in an equilateral triangle = (1/3)* Height of equilateral triangle(DERIVATION IS ATTACHED HERE)

therefore Radius of the bigger circle = (1/3)*18 = 6

Now The Height of Triangle ADE = 18-(2*radius of bigger circle) = 18 - 12 = 6

Now the radius of the smaller circle = (1/3)* Height of equilateral triangle ADE = (1/3)*6 = 2

i.e. Area of Both the smaller circle = 2* πr^2 = 2*π*2^2 = 8π

Answer: option B
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Re: The figure above shows an equilateral triangle with three circles. Eac  [#permalink]

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New post 09 Mar 2018, 01:11
Bunuel wrote:
The figure above shows an equilateral triangle with three circles. Each point of contact between two circles or between a circle and the triangle is a point of tangency. If the triangle has a height of 18, what is the combined area of the two smaller circles?

A. 4π
B. 8π
C. 16π
D. 36π
E. 44π



Since there is only one way to draw an equilateral triangle with an inscribed circle, we can trust the drawing.
That is, we'll visually estimate the answer and look for the closest answer choice.
This is an Alternative approach.

The height of the triangle is 18 so its base is 18\(\sqrt{2}\) (because this is a 30-60-90 triangle) which is about 20*1.4 = 28
Then the area is about 18*28/2 = 18*14 = 140+80+32=252
Dividing our equilateral triangle into 9 identical smaller triangles, we can SEE that the two small circles are about half the area of 2 of the small triangles.
That is, about (1/2)*(2/9) of the total area or 252/9 which is a bit more than 25, say 30.
Option (B) is the only relevant choice.

** Note that with regular polygons and area-related questions it is almost always easier to estimate the answer instead of looking for an extremely complicated Precise geometrical solution.
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Re: The figure above shows an equilateral triangle with three circles. Eac  [#permalink]

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New post 09 Mar 2018, 10:19
GMATinsight wrote:
Bunuel wrote:
Image
The figure above shows an equilateral triangle with three circles. Each point of contact between two circles or between a circle and the triangle is a point of tangency. If the triangle has a height of 18, what is the combined area of the two smaller circles?

A. 4π
B. 8π
C. 16π
D. 36π
E. 44π

Attachment:
Equilateral_triangle_circle_packing_1 (1).png


The radius of an In-Circle in an equilateral triangle = (1/3)* Height of equilateral triangle(DERIVATION IS ATTACHED HERE)

therefore Radius of the bigger circle = (1/3)*18 = 6

Now The Height of Triangle ADE = 18-(2*radius of bigger circle) = 18 - 12 = 6

Now the radius of the smaller circle = (1/3)* Height of equilateral triangle ADE = (1/3)*6 = 2

i.e. Area of Both the smaller circle = 2* πr^2 = 2*π*2^2 = 8π

Answer: option B



Radius =1/3 of the height is obvious since circumcenter,incenter of equilateral triangle lie on the same point. Since circumcenter divides the triangle in 2:1 ratio, the radius should be 1/3 of the height
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Re: The figure above shows an equilateral triangle with three circles. Eac  [#permalink]

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New post 23 Oct 2018, 08:23
GMATinsight wrote:
Bunuel wrote:
Image
The figure above shows an equilateral triangle with three circles. Each point of contact between two circles or between a circle and the triangle is a point of tangency. If the triangle has a height of 18, what is the combined area of the two smaller circles?

A. 4π
B. 8π
C. 16π
D. 36π
E. 44π

Attachment:
Equilateral_triangle_circle_packing_1 (1).png


The radius of an In-Circle in an equilateral triangle = (1/3)* Height of equilateral triangle(DERIVATION IS ATTACHED HERE)

therefore Radius of the bigger circle = (1/3)*18 = 6

Now The Height of Triangle ADE = 18-(2*radius of bigger circle) = 18 - 12 = 6

Now the radius of the smaller circle = (1/3)* Height of equilateral triangle ADE = (1/3)*6 = 2

i.e. Area of Both the smaller circle = 2* πr^2 = 2*π*2^2 = 8π

Answer: option B


Hi GMATinsight,
Please can you tell me, how the smaller triangle ADE is an equilateral triangle. Thank you.
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Re: The figure above shows an equilateral triangle with three circles. Eac  [#permalink]

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New post 23 Oct 2018, 20:07
1
stne wrote:
GMATinsight wrote:
Bunuel wrote:
Image
The figure above shows an equilateral triangle with three circles. Each point of contact between two circles or between a circle and the triangle is a point of tangency. If the triangle has a height of 18, what is the combined area of the two smaller circles?

A. 4π
B. 8π
C. 16π
D. 36π
E. 44π

Attachment:
Equilateral_triangle_circle_packing_1 (1).png


The radius of an In-Circle in an equilateral triangle = (1/3)* Height of equilateral triangle(DERIVATION IS ATTACHED HERE)

therefore Radius of the bigger circle = (1/3)*18 = 6

Now The Height of Triangle ADE = 18-(2*radius of bigger circle) = 18 - 12 = 6

Now the radius of the smaller circle = (1/3)* Height of equilateral triangle ADE = (1/3)*6 = 2

i.e. Area of Both the smaller circle = 2* πr^2 = 2*π*2^2 = 8π

Answer: option B


Hi GMATinsight,
Please can you tell me, how the smaller triangle ADE is an equilateral triangle. Thank you.


stne

DE is parallel to BC i.e. triangle ABC and ADE are similar triangles

Since ABC is equilateral therefore ADE also will be similar triangle... I hope that helps!!!
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Re: The figure above shows an equilateral triangle with three circles. Eac &nbs [#permalink] 23 Oct 2018, 20:07
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