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Bunuel

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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Bunuel

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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Bunuel

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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stne
GMATinsight
Bunuel

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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The perpendicular height = median = line of symmetry from each vertex in an equilateral triangle.

The radius of an inscribed circle inside an Equilateral triangle = (1/3) * (Height)

In this case r = 6

Then if we draw a radius straight down to the base of the equilateral triangle and then another radius to the vertex (through the center of one of the smaller circles) we will form a 30-60-90 Right Triangle

r = 6 will be across from the 30 degree angle

The Hypotenuse will extend from the center of the triangle/larger circle/overall figure to the vertex and pass through the center of the smaller circle.

This length will be = (2) * (6) = 12

Because of the ratio of 30-60-90 triangles


Lastly, we can draw a line of tangency through the point at which the smaller circle touches the larger center circle.

This line of tangency will create a smaller equilateral triangle with the smaller circle inscribed inside of it.

The length of the radius of the center circle will cover 6 of the hypotenuse’s length of 12 —— the rest of this distance will represent the height of the smaller equilateral triangle we just created


Thus the height of this smaller equilateral triangle will be = 6

If the height = 6, then we can find the radius of the smaller inscribed circle

(1/3) * 6 = 2

Radius of smaller circle = 2

Area of both smaller triangles = (2) * (2)^2 (pi)


(B) 8 (pi)

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