(1) The length of the side of the outermost regular hexagon is 6 cm.

As this hexagon is regular all sides are equal, it is also apparent that only one type of circle can be inscribed in a hexagon of certain area, hence a ratio can be derived. Next, we can calculate the area of hexagon, and using the ratio from above - calculate the area of the circle inscribed, then we now have only one type of hexagon that can be inscribed in this circle.
The area of nth circle/regular hexagon inscribed can be determined if we know the area of any of these figures and its place in the sequence. Sufficient.

(2) The length of a diagonal of the outermost regular hexagon is 12 cm.

Ok, now this seems to be similar to the previous statement, and if not for the "700" flair I would've maid a silly mistake, remember, we are talking about hexagon, it has 6 sides and diagonals are of 2 different lengths! Hence, we can have 2 different hexagons that fully meet the criteria stated in the second statement. Insufficient.


Answer: A
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VERITAS PREP OFFICIAL SOLUTION:

Thankfully, in DS questions, we don’t need to calculate the answer. We just need to establish the sufficiency of the given data. Note that we have found that there is a defined relation between the sides of a regular hexagon and the radius of an inscribed circle and there is also a defined relation between the radius of a circle and the side of an inscribed regular hexagon.

When the circle is inscribed in a regular hexagon,

Radius of the inscribed circle = \(\frac{\sqrt{3}}{2}\)* Side of the hexagon

When a regular hexagon is inscribed in a circle,

Side of the inscribed regular hexagon = Radius of the circle

So all we need is the side of any one regular hexagon or the radius of any one circle and we will know the length of the sides of all hexagons and the radii of all circles.

Statement I: The length of the side of the outermost regular hexagon is 6 cm.
If length of the side of the outermost regular hexagon is 6 cm, the radius of the inscribed circle is \((\sqrt{3}/2)*6 = 3\sqrt{3}\) cm

In that case, the side of the regular hexagon inscribed in this circle is also \(3\sqrt{3}\) cm. Now we can get the radius of the circle inscribed in this second hexagon and go on the same lines till we reach the tenth circle. This statement alone is sufficient.

Statement II: The length of a diagonal of the outermost regular hexagon is 12 cm.

Note that a hexagon has diagonals of two different lengths. The diagonals that connect vertices with one vertex between them are smaller than the diagonals that connect vertices with two vertices between them. Length of AC will be shorter than length of AD. Given the length of a diagonal, we do not know which diagonal it is. Is AC = 12 or is AD = 12? The length of the side will be different in the two cases. So this statement alone is not sufficient.

Answer (A).
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Hello Ergenekon

I think this is GMAT like question because actually this task tests knowledge of pythagorean triangle.
We can split this hexagon on 6 parts and received 6 isosceles triangles and as we know length (6 cm) of one of the sides of these triangles we can find height of this triangle because we know ratios of sides in right triangle with angles 30, 60, 90: \(1\), \(\sqrt{3}\), \(2\) and so on.

So this example is unusual but I think we can meet such tasks on exam.

P.S. IMHO )
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Hello BuggerinOn

1 step: we draw three lines: AD, CF, BE. Each of these lines divide hexagon on two equal parts. So in result we received six equal isosceles triangles with all angles = 60°
we see that they are isosceles visually, but can prove: sum of angles any ...gon equal to (n-2)*180° where n is number of angles. We have 6 angles in this figure
(6-2)*180° = 720°. 720° / 6 = 120° this is degrees of any angle in this hexagon. And as we divide each angle on two equal parts we receive 12 angles by 60° each.
So each of these triangles have two 60° angles and third angle should be 60° too.

2 step: we draw line from center O to L - this is height of isosceles trianlge so we receive two right triangles with angles 30°, 60°, 90°

3 step: we know that sides in triangle with angles 30°, 60°, 90° have a ratio 1, \(\sqrt{3}\), 2. As we know hypotenuse = 6 we can find length of OL = \(3\sqrt{2}\) and this will be radius.

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