A regular polygon is inscribed in a circle. How many sides does the

Question

A regular polygon is inscribed in a circle. How many sides does the polygon have?

(1) The length of the diagonal of the polygon is equal to the length of the diameter of the circle.
(2) The ratio of area of the polygon to the area of the circle is less than 2:3.

kvazar · Answer

We now have a polygon with equal sides and angles.

(1) The length of the diagonal of the polygon is equal to the length of the diameter of the circle. 
As regular polygons are always convex, any even sided regular polygon will have this property (they will have a diagonal = diameter), see the pic below.

Hence, we have at least two solutions - square (4 sides), hexagon (6 sides).  Insufficient.

(2) The ratio of area of the polygon to the area of the circle is less than 2:3. 
If a square with a diagonal of  \sqrt{2}  is inscribed, the area of such square is equal to  1^2=1 , while the Area of a circle with a diameter of  \sqrt{2}  equals to ( \sqrt{2}/2)^2*\pi=\frac{2}{4}\pi= \frac{\pi}{2} , hence the ratio is  1 : \frac{\pi}{2}  or  2 : 3.14 .

Less than 2:3, but at least one another regular polygon equilateral triangle also meets this requirement.  Insufficient.

Statements 1 and 2 together

The first statement eliminates triangles and other odd-sided polygons, and the second statement eliminates any even-sided polygons with more than 4 sides (as the 6 sided hexagon will have higher ratio than 2:3:14, I'm not sure if we need to actually calculate this?).  Sufficient.

Answer: C

GMATinsight · Answer

Question : How many sides does the Regular polygon have?

Statement 1: The length of the diagonal of the polygon is equal to the length of the diameter of the circle.

In a polygon of n sides, (n-3) diagonals can be drawn from every vertex of the polygon and all the diagonals will not be of the same length

However, Since the length of the diagonal of the polygon is equal to the length of the diameter of the circle therefore we understand that the longest diagonal is passing through the centre and hence the number of sides of the polygon will be even but the no. of sides of polygon may be 4 or 6 or 8

Hence,   NOT SUFFICIENT

Statement 2: The ratio of area of the polygon to the area of the circle is less than 2:3

Case 1: Square inside a Circle or radius=1 
Area of Circle = (Pi)r^2 = 3.14 x 1 = 3.14
Area of Square = Side^2 =   ^2 =   ^2  =  (\sqrt{2})^2  = 2

Area of Square / Area of Circle = 2/3.14 i.e. Less than 2/3

Case 1: Equilateral Triangle inside a Circle or radius=1 
Area of Circle = (Pi)r^2 = 3.14 x 1 = 3.14
Area of Equilateral Triangle inside that circle will be less than the Area of Square inside the circle as calculated in case 1

Area of Equilateral Triangle / Area of Circle will be Less than 2/3

So the Number of Sides of the Regular Polygon may be either 3 or 4

Hence,   NOT SUFFICIENT

Combining the Two statements:

The Number of sides of the Polygon must be even i.e. n= 4 or 6 or 8 etc.

The case of Equilateral Triangle can be ruled out as there is no Diagonal and it has odd number of sides
and
The case of Square is acceptable as per both the Statements
and
The case of Hexagon will result in the ratio of area of Hexagon and Area of Circle as Greater than 2/3 hence not acceptable

Similarly The case of Polygon with more than 4 sides will result in the ratio of area of Polygon and Area of Circle as Greater than 2/3 hence not acceptable

Thus only SQUARE is the possible Polygon hence
  SUFFICIENT

Answer: Option  C

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION:

In this question, we know that the polygon is a regular polygon i.e. all sides are equal in length. As the number of sides keeps increasing, the area of the circle enclosed in the regular polygon keeps increasing till the number of sides is infinite (i.e. we get a circle) and it overlaps with the original circle. The diagram given below will make this clearer.
 https://gmatclub.com/forum/download/file.php?id=26808 
Let’s look at each statement:

Statement I:  The length of one of the diagonals of the polygon is equal to the length of the diameter of the circle.
Do we get the number of sides of the polygon using this statement? No. The diagram below tells you why.
 https://gmatclub.com/forum/download/file.php?id=26809 
Regular polygons with even number of sides will be symmetrical around their middle diagonal and hence the diagonal will be the diameter. Hence the polygon could have 4/6/8/10 etc sides. Hence this statement alone is not sufficient.

Statement II:  The ratio of area of the polygon to the area of the circle is less than 2:3.

Let’s find the fraction of area enclosed by a square.

In the previous post we saw that
Side of the square =  \sqrt{2}  * Radius of the circle
Area of the square =  Side^2 = 2*Radius^2 
Area of the circle =  \pi*Radius^2 = 3.14*Radius^2 
Ratio of area of the square to area of the circle is 2/3.14 i.e. slightly less than 2/3.

So a square encloses less than 2/3 of the area of the circle. This means a triangle will enclose even less area. Hence, we see that already the number of sides of the regular polygon could be 3 or 4. Hence this statement alone is not sufficient.

Using both statements together , we see that the polygon has 4/6/8 etc sides but the area enclosed should be less than 2/3 of the area of the circle. Hence the regular polygon must have 4 sides. Since the area of a square is a little less than 2/3rd the area of the circle, we can say with fair amount of certainty that the area of a regular hexagon will be more than 2/3rd the area of the circle. But just to be sure, you can do this:

Side of the regular hexagon = Radius of the circle

Area of a regular hexagon = 6*Area of each of the 6 equilateral triangles =  6*(\frac{\sqrt{3}}{4})*Radius^2 = 2.6*Radius^2

2.6/3.14 is certainly more than 2/3 so the regular polygon cannot be a hexagon. The regular polygon must have 4 sides only.

Answer (C).

bumpbot · Answer

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A regular polygon is inscribed in a circle. How many sides does the

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