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555-605 (Medium)|   Geometry|      
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Bunuel
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The figure below shows a regular hexagon tangent to circle O at six po 14 Dec 2020, 04:45
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35% (medium)

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78% (02:59) correct 22% (02:16) wrong based on 27 sessions

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The figure below shows a regular hexagon tangent to circle O at six points. If the area of the hexagon is \(6 \sqrt{3}\), the circumference of circle O =


(A) \(\frac{3π\sqrt{3}}{2}\)

(B) \(\frac{12\sqrt{3}}{π}\)

(C) \(2π\sqrt{3}\)

(D) 12

(E) 6π

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2020-12-14_14-30-12.png
2020-12-14_14-30-12.png [ 15.76 KiB | Viewed 2939 times ]
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C
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callmeDP
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The figure below shows a regular hexagon tangent to circle O at six po 14 Dec 2020, 05:28
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Bunuel

The figure below shows a regular hexagon tangent to circle O at six points. If the area of the hexagon is \(6 \sqrt{3}\), the circumference of circle O =


(A) \(\frac{3π\sqrt{3}}{2}\)

(B) \(\frac{12\sqrt{3}}{π}\)

(C) \(2π\sqrt{3}\)

(D) 12

(E) 6π

Show Spoiler
Attachment:
2020-12-14_14-30-12.png
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Approach, a hexagon of side "X" is composed of 6 equal equilateral triangle. Thus if the area of the hexagon is \(6\sqrt{3} \)then the area of each equilateral triangle is\( \sqrt{3}\)

The area of equilateral triangle is nothing but \(\frac{\sqrt{3}* X^2 }{4} \) = \( \sqrt{3}\)

Thus, X = 2 >> or >> radius R is : R = \(\frac{\sqrt{3}*X}{2} \)

Thus, R = \(\sqrt{ 3}\)

And cirumference is: 2* pie* R =>> (C) \(2π\sqrt{3}\)

IMO C
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abhinavsodha800
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The figure below shows a regular hexagon tangent to circle O at six po 15 Dec 2020, 12:36
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Bunuel

The figure below shows a regular hexagon tangent to circle O at six points. If the area of the hexagon is \(6 \sqrt{3}\), the circumference of circle O =


(A) \(\frac{3π\sqrt{3}}{2}\)

(B) \(\frac{12\sqrt{3}}{π}\)

(C) \(2π\sqrt{3}\)

(D) 12

(E) 6π

Show Spoiler
Attachment:
2020-12-14_14-30-12.png


Approach, a hexagon of side "X" is composed of 6 equal equilateral triangle. Thus if the area of the hexagon is \(6\sqrt{3} \)then the area of each equilateral triangle is\( \sqrt{3}\)

The area of equilateral triangle is nothing but \(\frac{\sqrt{3}* X^2 }{4} \) = \( \sqrt{3}\)

Thus, X = 2 >> or >> radius R is : R = \(\frac{\sqrt{3}*X}{2} \)

Thus, R = \(\sqrt{ 3}\)

And cirumference is: 2* pie* R =>> (C) \(2π\sqrt{3}\)

IMO C
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Thus, X = 2 >> or >> radius R is : R = 3√∗X2 ( How did you got R ?)
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ayerhs
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The figure below shows a regular hexagon tangent to circle O at six po 20 Dec 2020, 23:32
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abhinavsodha800

The area of equilateral triangle is nothing but 3√∗X243∗X24 = 3√3

Thus, X = 2

AFTER THIS… Pick any one triangle.
Draw the radius of the circle in that triangle.
The radius will now be a perpendicular on the tangent, and a bisector.
So the sides of this new small RIGHT ANGLED triangle formed will be 1, 2 and r(radius).
Use Pythagoras theorem to solve and you will get r = √3

Tip : In equilateral triangle, each angle is 60
When you drop the perpendicular bisector, one angle is 90, one is 60 so remaining one is 30.
A 30-60-90 triangle is a special right triangle where sides are in the ratio 1: √3 : 2 each corresponding to the angles mentioned earlier.
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