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Inspired from a recent post: How does one obtain the radius

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Current Student
Joined: 31 Aug 2007
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Inspired from a recent post: How does one obtain the radius [#permalink]

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17 Jan 2008, 09:18
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Inspired from a recent post:

How does one obtain the radius of a circle that circumscribes an equilateral triangle with side length x?

Kudos [?]: 163 [0], given: 1

Current Student
Joined: 31 Aug 2007
Posts: 368

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17 Jan 2008, 09:40
Jack.Zhou wrote:
young_gun wrote:
Inspired from a recent post:

How does one obtain the radius of a circle that circumscribes an equilateral triangle with side length x?

I get (√3) * x / 6

actual solutions, as opposed to answers, would be greatly appreciated [walker, what's taking you so long to respond??].

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CEO
Joined: 17 Nov 2007
Posts: 3584

Kudos [?]: 4585 [1], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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17 Jan 2008, 09:45
1
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Expert's post

for me:
$$\frac{x}{2}=r*cos(\frac{60}{2})=r*\frac{sqrt{3}}{2}$$

$$r=\frac{1}{sqrt{3}}x$$
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Kudos [?]: 4585 [1], given: 360

Current Student
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17 Jan 2008, 10:16
walker wrote:
:)

for me:
$$\frac{x}{2}=r*cos(\frac{60}{2})=r*\frac{sqrt{3}}{2}$$

$$r=\frac{1}{sqrt{3}}x$$

any non-trig way to do this?

Kudos [?]: 163 [0], given: 1

CEO
Joined: 17 Nov 2007
Posts: 3584

Kudos [?]: 4585 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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17 Jan 2008, 10:34
young_gun wrote:
walker wrote:
:)

for me:
$$\frac{x}{2}=r*cos(\frac{60}{2})=r*\frac{sqrt{3}}{2}$$

$$r=\frac{1}{sqrt{3}}x$$

any non-trig way to do this?

$$r^2=(\frac{x}{2})^2+(\frac{r}{2})^2$$

$$r=\frac{1}{sqrt{3}}x$$
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Current Student
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17 Jan 2008, 10:44
walker wrote:
young_gun wrote:
walker wrote:
:)

for me:
$$\frac{x}{2}=r*cos(\frac{60}{2})=r*\frac{sqrt{3}}{2}$$

$$r=\frac{1}{sqrt{3}}x$$

any non-trig way to do this?

$$r^2=(\frac{x}{2})^2+(\frac{r}{2})^2$$

$$r=\frac{1}{sqrt{3}}x$$

thanks, but do you think you could post a diagram or explain more...i'm not seeing it. seems like you used pythag? where does (r/2)^2 come from? is it known that in an equil. triangle, the mid point to the middle of the sides is r/2?

Kudos [?]: 163 [0], given: 1

CEO
Joined: 17 Nov 2007
Posts: 3584

Kudos [?]: 4585 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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17 Jan 2008, 11:12
hope this help
Attachments

t58615.gif [ 10.82 KiB | Viewed 740 times ]

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17 Jan 2008, 13:56
young_gun wrote:
Inspired from a recent post:

How does one obtain the radius of a circle that circumscribes an equilateral triangle with side length x?

knew that question would spark some interest.

Walker has a nice picture, but essentially the long side of the smaller triangle is x/2 then small side is xsqrt3/6

So the hypt. is xsqrt3/3 Which is the radius. or as walker puts it x/sqrt3

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Intern
Joined: 15 Jan 2008
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17 Jan 2008, 15:51
Great analysis of the question. You guys deserve some kudos.

GMATBLACKBELT wrote:
young_gun wrote:
Inspired from a recent post:

How does one obtain the radius of a circle that circumscribes an equilateral triangle with side length x?

knew that question would spark some interest.

Walker has a nice picture, but essentially the long side of the smaller triangle is x/2 then small side is xsqrt3/6

So the hypt. is xsqrt3/3 Which is the radius. or as walker puts it x/sqrt3

Kudos [?]: 2 [0], given: 0

Re: Equilateral Tri Question   [#permalink] 17 Jan 2008, 15:51
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