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.

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:

If you have an equilateral triangle. That triangle is [#permalink]
18 Aug 2008, 07:52

If you have an equilateral triangle. That triangle is enclosed perfectly in a circle such that each corner is exactly touching the edge of the circle. What is the radius of the circle with relation to one side of the triangle?

Re: Equilateral Triangle enclosed in a circle [#permalink]
18 Aug 2008, 07:56

1

This post received KUDOS

\frac{1}{3}\sqrt{3} explanation in a minute...

I think I remember reading somewhere that the center of the circle will be 2/3 from any of the 3 vertices of triangle. So if an equilateral triangle creates 2 30:60:90 triangles back-to-back, then the height of the triangle will be \sqrt{3}, and the radius should be 2/3 of that length. But the question asks for the relation of the radius to any side of the equilateral triangle. The relationship of the "height" of the equilateral triangle to a side is 2:\sqrt{3}. So this would be \frac{2}{3}\sqrt{3}:2 because the radius of the circle to one side of the triangle. This would be the same as dividing \frac{2}{3}\sqrt{3} by 2. so 2/3 * 1/2 = 1/3...or \frac{1}{3}\sqrt{3}. I'm not sure if this is correct, but it seems logical to me.

Re: Equilateral Triangle enclosed in a circle [#permalink]
18 Aug 2008, 08:47

This makes sense to me too. The link you attached, the images don't appear for me, but I vaguely remember reading that 1/3, 2/3 center point as well. Thanks for the help and quick reply +1

Re: Equilateral Triangle enclosed in a circle [#permalink]
18 Aug 2008, 08:56

i think its very useful to memorize some common sin cos and tan values.

sin 30 = 1/2, sin 60 = sqrt(3)/2, and 45 = 1/sqrt(2) cos 30 = sqrt(3)/2, cos 60 = 1/2, and cos 45 = 1/sqrt(2) tan 30 = 1/sqrt(3), tan 60 = sqrt(3), and tan 45 = 1

just remember these 6 values ... its easy.. and it helps a lot in geamatry questions...

for example all i have to do is hypotneous = r base = a/2 angle = 30

Re: Equilateral Triangle enclosed in a circle [#permalink]
18 Aug 2008, 09:07

3

This post received KUDOS

gmatatouille wrote:

If you have an equilateral triangle. That triangle is enclosed perfectly in a circle such that each corner is exactly touching the edge of the circle. What is the radius of the circle with relation to one side of the triangle?

Thanks.

Attachments

tri-in-circle.gif [ 4.65 KiB | Viewed 1641 times ]

_________________

Your attitude determines your altitude Smiling wins more friends than frowning

Re: Equilateral Triangle enclosed in a circle [#permalink]
18 Aug 2008, 09:43

The triangle can be divided into three equal sectors and their point of intersection will be the center of the circle, which is also the centroid of the triangle. The centroid of a triangle is (2/3)*(height); Height = (side*sqrt[3]/2). This will give us the radius to be side/sqrt[3]. Cheers.

Re: Equilateral Triangle enclosed in a circle [#permalink]
19 Aug 2008, 21:43

an arc on circle makes double the angle on the center to what it makes on the opposite side on the circle.

With this logic, any of the sides of equilateral triangle will make 120 degree angle at the center of the circle.

Now, if I divide the triangle made by two radii and one side of the triangle into two halves, each of the triangles will be 30, 60, 90 and if r is the radius, then the side of triangle will become r multiplied by root 3.

Re: Equilateral Triangle enclosed in a circle [#permalink]
20 Aug 2008, 03:57

Attachment:

CircleAngle.jpg [ 5.01 KiB | Viewed 1555 times ]

Looking at the picture above, does Angle ABC work with this rule? If Angle ABC is 52 degrees, is arc AC 104 degrees even though the angle is not uniform (i.e., isosceles or equilateral).

scthakur wrote:

an arc on circle makes double the angle on the center to what it makes on the opposite side on the circle.

With this logic, any of the sides of equilateral triangle will make 120 degree angle at the center of the circle.

Now, if I divide the triangle made by two radii and one side of the triangle into two halves, each of the triangles will be 30, 60, 90 and if r is the radius, then the side of triangle will become r multiplied by root 3.

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Re: Equilateral Triangle enclosed in a circle [#permalink]
20 Aug 2008, 12:51

gmatatouille wrote:

If you have an equilateral triangle. That triangle is enclosed perfectly in a circle such that each corner is exactly touching the edge of the circle. What is the radius of the circle with relation to one side of the triangle?

Thanks.

Equilatral triangle with side a inscribed in a circle has r equal to a/sqrt (3)
_________________