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In the above diagram, the circle inscribes

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In the above diagram, the circle inscribes [#permalink] New post 08 Jun 2018, 05:30
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#GREpracticequestion In the above diagram, the circle inscribes the larger.png
#GREpracticequestion In the above diagram, the circle inscribes the larger.png [ 9.89 KiB | Viewed 173 times ]


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 \(\sqrt{3}\), what is the area of the larger triangle?

A) \(9\pi - 16 \sqrt{3}\)

B) \(4 \sqrt{3}\)

C) \(8 \sqrt{3}\)

D) \(16 \sqrt{3}\)

E) \(16\pi - 2 \sqrt{3}\)
[Reveal] Spoiler: OA

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Re: In the above diagram, the circle inscribes [#permalink] New post 10 Jun 2018, 06:33
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GreenlightTestPrep 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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Re: In the above diagram, the circle inscribes [#permalink] New post 10 Jun 2018, 08:09
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This was really tough.

Thank you.

Regards
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Re: In the above diagram, the circle inscribes [#permalink] New post 10 Jun 2018, 09:13
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Carcass wrote:
This was really tough.

Thank you.

Regards


Agreed - very tough!!
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Re: In the above diagram, the circle inscribes [#permalink] New post 20 Jan 2019, 19:52
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Re: In the above diagram, the circle inscribes [#permalink] New post 06 Sep 2019, 05:18
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Another way it could be done, if I'm not mistaken, is to rotate the inner triangle 180 degrees. Then you have created four equilateral triangles within a larger equilateral triangle. I imagine there is some rule saying then that you can label all the other small triangles with the same lengths as the inner one.

Not being familiar with the name of any such rule, I tried to see if there was some way to justify doing that. What I came up with was saying that after rotating the small triangle we can say we have created 3 parallelograms and can then infer the missing sides of the large triangle using the property of parallelograms that says that opposite sides are equal.

I've attached how I did it.
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rotate inner triangle.jpg
rotate inner triangle.jpg [ 319.02 KiB | Viewed 104 times ]

Re: In the above diagram, the circle inscribes   [#permalink] 06 Sep 2019, 05:18
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