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Bunuel
If a circle is inscribed in an equilateral triangle, what is the area of the triangle NOT taken up by the circle?

(1) The area of the circle is 12π
(2) The length of a side of the triangle is 12

Following is the formula that tells the relation between the radius of the circle and the side of the triangle:

r = a * (\(\sqrt{3}\)/ 6)
So if we have either of the side or the radius, we can find the other thing.
The relation can be found by using the figure below:

Attachment:
circle in triangle.JPG

Statement 1: The area of the circle is 12π
We can find the radius of the circle and hence the side of the triangle and the corresponding area
Therefore we can find the difference between the areas
SUFFICIENT

Statement 2: The length of a side of the triangle is 12
We can find the radius of the circle and hence the area of the circle
Therefore we can find the difference between the areas
SUFFICIENT

Option D
NOTE: We do not need to find the area. There is not need to do the calculations.



HI TeamGMATIFY,

Actually I already learnt this formula but haven't tried to figure it out the logic behind this.
The figure provided by you makes a triangle of 30-60-90 = 1x : Root 3* x : 2x

So the opposite length of side 30 degree angle --> R
=> 1x= R or x=R

And opposite length of side 60 degree angle--> A/2
=> root 3 * x = A/2

By this I am getting relation as --
R= A / (root 3 * 2) which is not the same

Can you please assist..??
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HI TeamGMATIFY,

Actually I already learnt this formula but haven't tried to figure it out the logic behind this.
The figure provided by you makes a triangle of 30-60-90 = 1x : Root 3* x : 2x

So the opposite length of side 30 degree angle --> R
=> 1x= R or x=R

And opposite length of side 60 degree angle--> A/2
=> root 3 * x = A/2

By this I am getting relation as --
R= A / (root 3 * 2) which is not the same

Can you please assist..??
Once you have got R= A / (root 3 * 2)
You need to rationalize the denominator, or simply remove the under root from the denominator
Multiply \(\sqrt{3}\) on both numerator and denominator and you will get

R= A \(\sqrt{3}\)/ 6

Does this help?
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Hello,

Can someone elaborate on this question without the use of a formula?

Thanks
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Answer: (D) – each statement on its own is sufficient. The area within the triangle that is not taken up by the circle can be found by subtracting the area of the circle from the area of the full triangle. Therefore, if we can determine both the area of the circle and the area of the triangle, we will be able to answer the question. Watch out, though! If you glance at statements (1) and (2), you will note that it would be relatively simple to do so using both: statement (1) is flat out giving you the area of the circle, and statement (2) is giving you the length of the side of what you know to be an equilateral triangle, which is basically like giving you the area straight up (remember, there’s a formula for this:

A(equilateral triangle) =
(side)234
, so it would just entail plugging in and solving… and even if you don’t remember the formula, you might know that there is one, and since you don’t need to solve for an exact answer anyway, it would theoretically be enough). The fact that the statements lean so heavily towards an easy-C should indicate that you might be missing something. Upon further investigation, you would conclude that these statements are actually giving you the exact same information, which means that each leads to both areas, and each is sufficient on its own. (Here, we will go over how to deduce the length of the triangle side from the radius; going the other way is just a matter of reversing the mathematical logic.)

Consider statement (1): The area of the circle is
12π
. From this we can calculate the radius to be
23
. Sketch the figure and draw in the three radii that hit the circle at the points of intersection with the triangle, and you will see that the circle is divided into three seemingly equal parts. Since the angles of the triangle are all the same (equilateral) and therefore cut off three equivalent arcs around the circle, we can conclude that the three segments created are, in fact, equal, and the central angles must all be 120°. Furthermore, bisecting the central angles with a line from the center of the circle out to the corner of the triangle creates a 30-60-90 right triangle, with the side across from the 30° angle as radius of the circle and the side across from the 60° angle as half the length of the triangle. From here, you could calculate using the 30-60-90 side ratio the length of the base, and from there the area of the triangle.

Geometry_Q20SolutionImage.pg.jpg
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