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In the above diagram, the circle inscribes the larger equilateral,

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In the above diagram, the circle inscribes the larger equilateral,  [#permalink]

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New post 08 Jun 2018, 06:22
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Question Stats:

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In the above diagram, the circle inscribes the larger equilateral, and it circumscribes the smaller equilateral triangle. If the area of the smaller triangle is √3, what is the area of the larger triangle?

A) 9π - 16√3
B) 4√3
C) 8√3
D) 16√3
E) 16π - 2√3

*kudos for all correct solutions

NOTE: There are at least 2 very different approaches we can take to solve this question. How many approaches can you find?

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In the above diagram, the circle inscribes the larger equilateral,  [#permalink]

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New post 08 Jun 2018, 08:35
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GMATPrepNow wrote:
Image
In the above diagram, the circle inscribes the larger equilateral, and it circumscribes the smaller equilateral triangle. If the area of the smaller triangle is √3, what is the area of the larger triangle?

A) 9π - 16√3
B) 4√3
C) 8√3
D) 16√3
E) 16π - 2√3

*kudos for all correct solutions

NOTE: There are at least 2 very different approaches we can take to solve this question. How many approaches can you find?


Excellent approaches have already posted by pushpitkc

To add to the above, the below mentioned approach may be seen:-

Refer to the diagram affixed herewith,

All triangles are equilateral triangles and circles are inscribed in these triangles. If the sides of triangle ABC =a unit , then the side of the triangle DEF=\(\frac{a}{2}\) unit and the side of the triangle GHI=\(\frac{a}{4}\) unit and so on.
<<<<-------Property

We are given, area of smaller triangle DEF=\(\sqrt{3}\) sq. unit
Or,\((\sqrt{3}/4)\)*\((\frac{a}{2})^2\)= \(\sqrt{3}\)
Or, \((\sqrt{3}/4)\)* \(\frac{a^2}{4}\)= \(\sqrt{3}\)
Or, \((\sqrt{3}/4)\)*\(a^2\)=4*\(\sqrt{3}\)
Or, Area of larger triangle ABC=\(4*\sqrt{3}\) sq. unit

Hence option (B)
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In the above diagram, the circle inscribes the larger equilateral,  [#permalink]

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New post 08 Jun 2018, 07:24
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Attachment:
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Given: Area of smaller equilateral triangle is \(\sqrt{3}\)

Formula used:
Area = \(\frac{\sqrt{3}}{4} * a^2\) | Height(h) = \(\frac{\sqrt{3}}{2} * a\) | Radius on incircle(i) = \(\frac{2}{3} * h\)
where a - side of equilateral triangle

Since area of the triangle is \(\sqrt{3}\), \(\frac{\sqrt{3}}{4} * a^2 = \sqrt{3}\) -> a = 2

Height of the triangle(h) = \(\frac{\sqrt{3}}{2}* a = \sqrt{3}\) | Radius of circumcircle = \(\frac{2}{3} * h = \frac{2}{3} * \sqrt{3}\) = \(\frac{2}{\sqrt{3}}\)

OR = OD = \(\frac{2}{\sqrt{3}}\)

In triangle OAD

1. OAD = 30 degrees, ODA = 90 degree (We have a 30-60-90 triangle)
2. Sides of the triangle are in ratio \(1:\sqrt{3}:2\)

Since OD = \(\frac{2}{\sqrt{3}}\), Length of AD = \(2\). AC = 2*AD \(= 2*2 = 4\)
Area of the triangle ABC = \(\frac{\sqrt{3}}{4} * AC^2 = \frac{\sqrt{3}}{4} * 4^2 = 4\sqrt{3}\) (Option B)
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In the above diagram, the circle inscribes the larger equilateral,  [#permalink]

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New post 08 Jun 2018, 07:40
If you imagine and invert the smaller triangle inside the circle, you will note that it divides the larger triangle in 4 equal triangles. The area of the larger triangle is therefore 4 times the area of the smaller triangle. Hence B is the answer.
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Re: In the above diagram, the circle inscribes the larger equilateral,  [#permalink]

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New post 08 Jun 2018, 09:12
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Given: Area of the smaller triangle = √3

To find: Area of the larger triangle

Approach: Let side of the smaller equilateral triangle= a
Area of the smaller triangle =√3 = (√3/4) a^2
=> a= 2

Radius of the circumcircle of an equilateral triangle= Side of the equilateral triangle/√3= 2/√3

For a circle inscribed in an equilateral triangle, Side of the triangle = (2√3)* Radius of the circle

Therefore, Side of the larger triangle = (2√3)*(2/√3) = 4

Area of the larger triangle= (√3/4)* Side of the larger triangle = (√3/4) *4 ^2 = 4√3

Correct Answer= B
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Re: In the above diagram, the circle inscribes the larger equilateral,  [#permalink]

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New post 10 Jun 2018, 07:34
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GMATPrepNow wrote:
Image
In the above diagram, the circle inscribes the larger equilateral, and it circumscribes the smaller equilateral triangle. If the area of the smaller triangle is √3, what is the area of the larger triangle?

A) 9π - 16√3
B) 4√3
C) 8√3
D) 16√3
E) 16π - 2√3


We're told that the area of the smaller triangle is √3
USEFUL FORMULA: Area of an equilateral triangle = (√3)(side²)/4
So, we can write: (√3)(side²)/4 = √3
Divide both sides by √3 to get: (side²)/4 = 1
Multiply both sides by 4 to get: side² = 4
Solve: side = 2
So, each side of the smaller equilateral triangle has length 2

Using this information, we can create a 30-60-90 triangle (in blue)
Image

We can now compare this blue 30-60-90 triangle with the BASE 30-60-90 triangle
Image
By the property of similar triangles, we know that the ratios of corresponding sides will be equal.
That is: 1/√3 = r/2
Cross multiply to get: (√3)(r) = 2
Solve: r = 2/√3
So, the RADIUS of the circle = 2/√3
We'll add this information to our diragram
Image

At this point, we can focus our attention on the GREEN 30-60-90 triangle
Image
Since we already know that the RADIUS of the circle = 2/√3, we can apply the property of similar triangles again.
The ratios of corresponding sides will be equal.
So, we get: (2/√3)/1 = (x)/√3
Cross multiply to get: (2/√3)(√3) = (x)(1)
Simplify: x = 2
Since x = HALF the length of one side of the larger triangle, we know that the ENTIRE length = 4

What is the area of the larger triangle?
We'll re-use our formula that says: area of an equilateral triangle = (√3)(side²)/4
Area = (√3)(4²)/4
= (√3)(16)/4
= 4√3

Answer: B

Cheers
Brent
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In the above diagram, the circle inscribes the larger equilateral,  [#permalink]

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New post 10 Jun 2018, 07:52
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GMATPrepNow wrote:
Image
In the above diagram, the circle inscribes the larger equilateral, and it circumscribes the smaller equilateral triangle. If the area of the smaller triangle is √3, what is the area of the larger triangle?

A) 9π - 16√3
B) 4√3
C) 8√3
D) 16√3
E) 16π - 2√3


Another approach is to use the following fact:
For GMAT Problem Solving questions (i.e., questions that are NOT data sufficiency questions), all diagrams are DRAWN TO SCALE, unless stated otherwise.
So, we can use this to ESTIMATE the area of the larger triangle.

We know the smaller triangle has area √3
Image
On test day, we must memorize 3 approximations: √2 ≈ 1.4, √3 ≈ 1.7, √5 ≈ 2.2

Given this, what would you estimate the area of LARGE triangle to be?

ASIDE: If you can mentally "move" the smaller triangle to one corner, it might help your estimation.
Image

If the area of the small triangle is 1.7, we might estimate the area of the large triangle to be somewhere in the 5 to 8 range

Now let's check our answer choices....

A) 9π - 16√3 ≈ (9)(3.1) - (16)(1.7) ≈ 1. This is pretty far from our estimate of 5 to 8. ELIMINATE.
B) 4√3 ≈ (4)(1.7) ≈ 6.8. This is within our estimated range 5 to 8. KEEP.
C) 8√3 ≈ (8)(1.7) ≈ 13.6. This is VERY far from our estimate of 5 to 8. ELIMINATE.
D) 16√3 ≈ (16)(1.7) ≈ 27. This is VERY far from our estimate of 5 to 8. ELIMINATE.
E) 16π - 2√3 ≈ (16)(3.1) - (2)(1.7) ≈ 47. This is VERY far from our estimate of 5 to 8. ELIMINATE.

Answer: B

Cheers,
Brent
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Re: In the above diagram, the circle inscribes the larger equilateral,  [#permalink]

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New post 10 Jun 2018, 23:19
GMATPrepNow wrote:
Image
In the above diagram, the circle inscribes the larger equilateral, and it circumscribes the smaller equilateral triangle. If the area of the smaller triangle is √3, what is the area of the larger triangle?

A) 9π - 16√3
B) 4√3
C) 8√3
D) 16√3
E) 16π - 2√3

*kudos for all correct solutions

NOTE: There are at least 2 very different approaches we can take to solve this question. How many approaches can you find?



Area of smaller triangle = \sqrt{3} = 1\frac{2[}{fraction] x b x h

Area of larger triangle = 1[fraction]2} x B x H

Now B=2b & H=2h ( Property of equilateral triangles inscribed in a circle...)

Therefore, Area of larger triangle = 1\frac{2[}{fraction] x 4 x b x h = 4 x 1[fraction]2} x b x h = 4 \sqrt{3}

Answer is B

Kudos if it helped..!
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Re: In the above diagram, the circle inscribes the larger equilateral, &nbs [#permalink] 10 Jun 2018, 23:19
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