The Figure Below Shows A Regular Hexagon Tangent To Circle O At Six GMAT Problem Solving

Since the hexagon is regular (all sides are congruent), the area of △AOB in the following figure \(\sqrt3\) is − one sixth the area of the hexagon

AOP is equilateral; hence you can divide it into two 1 :\(\sqrt3\) : 2 triangles, as shown in the figure.
Since the common leg, whose length is\(\sqrt3\), is also the circle's radius, the circle's circumference must be \(2π\sqrt3\)

Correct option: C

A hexagon of side ‘x’ is composed of six equilateral triangles.
Thus, if the area of the equilateral triangle is nothing but\( \sqrt3*x^2\over4 \) = \( \sqrt3\)

Thus, x = 2 >> or >> radius R is : R = \(\sqrt3*x\over 2\)

Thus, R =\(\sqrt3\)

And, circumference = 2πR=\(2π \sqrt3\)

Correct option: C

As this is a regular hexagon, we can divide this hexagon in six right-angles triangles
The area of each right-angled triangle is same i.e., one-sixth the area of the regular hexagon

Area of the equilateral triangle = \(\sqrt3 \over4\)*\((over)^2\)

Area of one triangle = \(Area of hexagon\over 6\)=\(6\sqrt3 \over 6\)=\(\sqrt3\)

Therefore, \( \sqrt3\over 4\)* \((side)^2\) = \( \sqrt3\)

\((side)^2\)= 4

Side = 2

R =\(\sqrt 3*side\over 2\)=\( \sqrt3*2 \over 2\)=\( \sqrt3\)

Circumference = 2πR=\(2π \sqrt3\)

Correct option: C