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TAGS: GRE Instructor Joined: 10 Apr 2015
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In the above diagram, the circle inscribes [#permalink]
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Question Stats: 63% (03:27) correct 36% (03:43) wrong based on 22 sessions
Attachment: #GREpracticequestion In the above diagram, the circle inscribes the larger.png [ 9.89 KiB | Viewed 592 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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Brent Hanneson – Creator of greenlighttestprep.com  GRE Instructor Joined: 10 Apr 2015
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Re: In the above diagram, the circle inscribes [#permalink]
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Expert's post
GreenlightTestPrep wrote: 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) We can now compare this blue 30-60-90 triangle with the BASE 30-60-90 triangle 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 At this point, we can focus our attention on the GREEN 30-60-90 triangle 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

Cheers
Brent
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Re: In the above diagram, the circle inscribes [#permalink]
Expert's post
This was really tough.

Thank you.

Regards
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GRE Prep Club Members of the Month: Each member of the month will get three months free access of GRE Prep Club tests. GRE Instructor Joined: 10 Apr 2015
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Re: In the above diagram, the circle inscribes [#permalink]
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Carcass wrote:
This was really tough.

Thank you.

Regards

Agreed - very tough!!
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Brent Hanneson – Creator of greenlighttestprep.com Director Joined: 09 Nov 2018
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Re: In the above diagram, the circle inscribes [#permalink]
Tricky Manager  Joined: 19 Nov 2018
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Re: In the above diagram, the circle inscribes [#permalink]
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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.
Attachments rotate inner triangle.jpg [ 319.02 KiB | Viewed 522 times ] Re: In the above diagram, the circle inscribes   [#permalink] 06 Sep 2019, 05:18
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