Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

Show Tags

02 Jun 2013, 01:49

In an regular hexagon inscribed in a circle, its side is equal the radius. We can divide the hexagon in 6 triangles each with the base of 4. The heigth will equal \(\sqrt{4^2-2^2}=\sqrt{12}=2\sqrt{3}\). To obtain this just use Pythagoras, the hypotenuse of each triangle it's the radius, and the bases it's \(\frac{4}{2}=2\).

Now you have the height of each triangle, so \(A_t=(4*2\sqrt{3})/2=4\sqrt{3}\).

\(A_h=6*A_t=6*4\sqrt{3}=24\sqrt{3}\)

Attachments

Untitled.png [ 3.1 KiB | Viewed 34397 times ]

_________________

It is beyond a doubt that all our knowledge that begins with experience.

Regular hexagon will have six traingle regions. Central angle will be 60. (360/6) Equilateral triangle will be formed as we have three 60 angles. Length of each side is equal to radius 4. Area of hexagon = area of 6 eq. triangles = 6 * root 3/4 * 4^2 = 24 root3 _________________

A regular hexagon is essentially composed of 6 equilateral trianlges...and the line joining the opposite vertices is the diameter of the circle in which the hexagon is inscribed...So the radius of the circle forms the side of the equilateral triangle... Area is 6* (3^1/2)/4(a^2) where a = radius of the circle. 6*sq. rt3/4 * 4^2=24 root 3

Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

Show Tags

03 Jun 2013, 13:24

fozzzy wrote:

Attachment:

screen_shot_2010_12_23_at_3.41.13_pm.png

If a regular hexagon is inscribed in a circle with a radius of 4, the area of the hexagon is

A. 12 root 3 B. 8 pi C. 18 root 2 D. 24 root 3 E. 48

The Hexagon can be divided into 6 equilateral triangles , each with side = radius of the circle. Since area of equilateral triangle is (root(3)*a^2)4, for 6 such triangles we will get 24root3

Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

Show Tags

05 Jun 2013, 07:15

2

This post received KUDOS

1) Each of the hexagon's angles 120 degrees : formula if you don't know it is [(# of sides - 2) x 180] = (6 - 2) x 180 = 720 ÷ 6 = 120. 2) Next, split up the hexagon into 6 equilateral triangles with 4 for each of its sides. 3) Find the area of one of the triangles: - Base = 4 - find the height by splitting the triangle in half so that it becomes a 30/60/90 triangle and find the height using the pythagorean theorem or knowing the 1,√3,2 triangle. Height = 2√3 - one triangle's area = (1/2)bh = (1/2)(4)(2√3) = 4√3 4) find the area of the hexagon by multiplying the one triangle's area by 6: - 6 x 4√3 = 24√3 Answer is D

Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

Show Tags

12 Jun 2013, 05:22

lchen wrote:

1) Each of the hexagon's angles 120 degrees : formula if you don't know it is [(# of sides - 2) x 180] = (6 - 2) x 180 = 720 ÷ 6 = 120. 2) Next, split up the hexagon into 6 equilateral triangles with 4 for each of its sides. 3) Find the area of one of the triangles: - Base = 4 - find the height by splitting the triangle in half so that it becomes a 30/60/90 triangle and find the height using the pythagorean theorem or knowing the 1,√3,2 triangle. Height = 2√3 - one triangle's area = (1/2)bh = (1/2)(4)(2√3) = 4√3 4) find the area of the hexagon by multiplying the one triangle's area by 6: - 6 x 4√3 = 24√3 Answer is D

Great method we could use the formula for the area of equilateral triangle and save a few steps Root 3/4 a^2 where a = 4 _________________

Re: If a regular hexagon is inscribed in a circle with a radius [#permalink]

Show Tags

04 Feb 2016, 12:21

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Part 2 of the GMAT: How I tackled the GMAT and improved a disappointing score Apologies for the month gap. I went on vacation and had to finish up a...

So the last couple of weeks have seen a flurry of discussion in our MBA class Whatsapp group around Brexit, the referendum and currency exchange. Most of us believed...

This highly influential bestseller was first published over 25 years ago. I had wanted to read this book for a long time and I finally got around to it...