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# In the above diagram, the circle inscribes the larger equilateral,

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In the above diagram, the circle inscribes the larger equilateral, [#permalink]
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If you imagine and invert the smaller triangle inside the circle, you will note that it divides the larger triangle in 4 equal triangles. The area of the larger triangle is therefore 4 times the area of the smaller triangle. Hence B is the answer.
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Re: In the above diagram, the circle inscribes the larger equilateral, [#permalink]
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Given: Area of the smaller triangle = √3

To find: Area of the larger triangle

Approach: Let side of the smaller equilateral triangle= a
Area of the smaller triangle =√3 = (√3/4) a^2
=> a= 2

Radius of the circumcircle of an equilateral triangle= Side of the equilateral triangle/√3= 2/√3

For a circle inscribed in an equilateral triangle, Side of the triangle = (2√3)* Radius of the circle

Therefore, Side of the larger triangle = (2√3)*(2/√3) = 4

Area of the larger triangle= (√3/4)* Side of the larger triangle = (√3/4) *4 ^2 = 4√3

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Re: In the above diagram, the circle inscribes the larger equilateral, [#permalink]
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GMATPrepNow

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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In the above diagram, the circle inscribes the larger equilateral, [#permalink]
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Top Contributor
GMATPrepNow

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.

Cheers,
Brent
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Re: In the above diagram, the circle inscribes the larger equilateral, [#permalink]
GMATPrepNow

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

*kudos for all correct solutions

NOTE: There are at least 2 very different approaches we can take to solve this question. How many approaches can you find?

Area of smaller triangle = \sqrt{3} = 1\frac{2[}{fraction] x b x h

Area of larger triangle = 1[fraction]2} x B x H

Now B=2b & H=2h ( Property of equilateral triangles inscribed in a circle...)

Therefore, Area of larger triangle = 1\frac{2[}{fraction] x 4 x b x h = 4 x 1[fraction]2} x b x h = 4 \sqrt{3}

Kudos if it helped..!
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In the above diagram, the circle inscribes the larger equilateral, [#permalink]
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

Hello GMATPrepNow

I just have a question, why we can solve it like the following to find r?

x√3 = 1

So it gives me x= √3/3

Kind regards!
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Re: In the above diagram, the circle inscribes the larger equilateral, [#permalink]
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jfranciscocuencag
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

Hello GMATPrepNow

I just have a question, why we can solve it like the following to find r?

x√3 = 1

So it gives me x= √3/3

Kind regards!

I believe you're calculating the enlargement factor.
That is x = the enlargement factor

So, to find the radius, we multiply 2 (in the base triangle) by the enlargement factor to get: radius = (2)(√3/3) = 2√3/3
This is equal to the radius I calculated in my solution (2/√3)
To see why, take 2/√3 and multiply top and bottom by √3 to get 2√3/3

Does that help?

Cheers,
Brent
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Re: In the above diagram, the circle inscribes the larger equilateral, [#permalink]
We are given the area of smaller equilateral triangle:--√3
Quote:
Step:1--We find the side of this smaller triangle using equilateral triangle formula
√3/4*a^2=√3
a=2

Quote:
Step :2--The circle that is circumscribed to this triangle has radius
r=√3/3 *a (Standard formula for radius of a circumscribed circle to an equilateral triangle)
From this the radius of smaller circle r=2√3/3
Quote:
Step :3--This circle will now act as inscribed circle for bigger triangle whose area we have to find out.
Now the circle that is inscribed to this triangle has radius
R=√3/6 *A (Standard formula for radius of a inscribed circle to an equilateral triangle)
R:-- Radius of circle inscribed(which is r that we calculated above)
Here A is the side of bigger triangle.

2√3/3=√3/6*A
Solve and get A=4

Finally :-
Area of bigger triangle =√3/4*A*A=(√3/4)*16=4√3

IMO B
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Re: In the above diagram, the circle inscribes the larger equilateral, [#permalink]
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Re: In the above diagram, the circle inscribes the larger equilateral, [#permalink]
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