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 = 2So, each side of the smaller equilateral triangle has length
2Using 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 triangleBy the property of similar triangles, we know that the ratios of corresponding sides will be equal.
That is:
1/
√3 =
r/
2Cross 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
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Brent Hanneson – Creator of greenlighttestprep.comSign up for GRE Question of the Day emails