GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
In the above diagram, the circle inscribes
[#permalink]
08 Jun 2018, 05:30
08 Jun 2018, 05:30
3
12
Bookmarks
00:00
Question Stats:
70% (03:11) correct
29% (03:24) wrong based on 48 sessions
Hide Show timer Statistics
Attachment:
#GREpracticequestion In the above diagram, the circle inscribes the larger.png [ 9.89 KiB | Viewed 5384 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) \(16\pi - 2 \sqrt{3}\)
Re: In the above diagram, the circle inscribes
[#permalink]
06 Sep 2019, 05:18
06 Sep 2019, 05:18
7
1
Bookmarks
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.
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 5232 times ]
General Discussion
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: In the above diagram, the circle inscribes
[#permalink]
10 Jun 2018, 06:33
10 Jun 2018, 06:33
4
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
Answer: B
Cheers
Brent
