Brent@GMATPrepNow 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

Brent@GMATPrepNow 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

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

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.


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