Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 26 May 2016, 19:57

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# An equilateral triangle that has an area of 9*3^1/2

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 08 Aug 2005
Posts: 251
Followers: 1

Kudos [?]: 34 [1] , given: 0

An equilateral triangle that has an area of 9*3^1/2 [#permalink]

### Show Tags

06 May 2006, 00:28
1
KUDOS
4
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

61% (02:53) correct 39% (02:03) wrong based on 110 sessions

### HideShow timer Statistics

An equilateral triangle that has an area of $$9\sqrt{3}$$ is inscribed in a circle. What is the area of the circle?

A. $$6\pi$$
B. $$9\pi$$
C. $$12\pi$$
D. $$9\pi \sqrt{3}$$
E. $$18\pi \sqrt{3}$$

OPEN DISCUSSION OF THIS QUESTION IS HERE: an-equilateral-triangle-that-has-an-area-of-9-root-105356.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 21 Nov 2013, 01:46, edited 2 times in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
VP
Joined: 29 Dec 2005
Posts: 1348
Followers: 8

Kudos [?]: 47 [0], given: 0

Re: The area of the circle [#permalink]

### Show Tags

16 May 2006, 12:11
getzgetzu wrote:
An equilateral triangle is inscribed in a circle and has the area of 9*sqrt3. What the area of the circle?

area of an equilateral triangle = [sqrt(3)/4] a^2
9 sqrt(3) = [sqrt(3)/4] a^2
a = 6

if an equilateral triangle is inscribed in a circle, the radius (r) of the circle = a/sqrt(3)

r = 6/sqrt(3) = (3x2)/sqrt(3) = 2 sqrt(3)
the area of the circle = (pi) r^2 = (pi) [2 sqrt(3)]^2 = 12 pi.
Senior Manager
Joined: 09 Mar 2006
Posts: 445
Followers: 1

Kudos [?]: 8 [0], given: 0

### Show Tags

16 May 2006, 13:15
I got 24pi, but my approach is slightly different.

What is the OA?
Manager
Joined: 14 Mar 2006
Posts: 208
Followers: 2

Kudos [?]: 7 [0], given: 0

Re: The area of the circle [#permalink]

### Show Tags

16 May 2006, 13:23
Professor wrote:
getzgetzu wrote:
An equilateral triangle is inscribed in a circle and has the area of 9*sqrt3. What the area of the circle?

area of an equilateral triangle = [sqrt(3)/4] a^2
9 sqrt(3) = [sqrt(3)/4] a^2
a = 6

if an equilateral triangle is inscribed in a circle, the radius (r) of the circle = a/sqrt(3)

r = 6/sqrt(3) = (3x2)/sqrt(3) = 2 sqrt(3)
the area of the circle = (pi) r^2 = (pi) [2 sqrt(3)]^2 = 12 pi.

Prof, what is 'a'? I have never seen these equations for equilateral triangles before.
VP
Joined: 29 Dec 2005
Posts: 1348
Followers: 8

Kudos [?]: 47 [0], given: 0

Re: The area of the circle [#permalink]

### Show Tags

16 May 2006, 13:49
shampoo wrote:
Professor wrote:
getzgetzu wrote:
An equilateral triangle is inscribed in a circle and has the area of 9*sqrt3. What the area of the circle?

area of an equilateral triangle = [sqrt(3)/4] a^2
9 sqrt(3) = [sqrt(3)/4] a^2
a = 6

if an equilateral triangle is inscribed in a circle, the radius (r) of the circle = a/sqrt(3)

r = 6/sqrt(3) = (3x2)/sqrt(3) = 2 sqrt(3)
the area of the circle = (pi) r^2 = (pi) [2 sqrt(3)]^2 = 12 pi.

Prof, what is 'a'? I have never seen these equations for equilateral triangles before.

a = a side of an equilateral triangle..
VP
Joined: 28 Mar 2006
Posts: 1381
Followers: 2

Kudos [?]: 26 [0], given: 0

Re: The area of the circle [#permalink]

### Show Tags

16 May 2006, 16:59
Professor wrote:
getzgetzu wrote:
An equilateral triangle is inscribed in a circle and has the area of 9*sqrt3. What the area of the circle?

area of an equilateral triangle = [sqrt(3)/4] a^2
9 sqrt(3) = [sqrt(3)/4] a^2
a = 6

if an equilateral triangle is inscribed in a circle, the radius (r) of the circle = a/sqrt(3)

r = 6/sqrt(3) = (3x2)/sqrt(3) = 2 sqrt(3)
the area of the circle = (pi) r^2 = (pi) [2 sqrt(3)]^2 = 12 pi.

Prof is correct

Also you may take a longer route

After you find a=6(whihc is the side of eq triangle)

You may get the ht = sqrt(3)/2*a = sqrt(3)*3

now circum radius = 2/3*ht = 2*sqrt(3)

Area = 12pi
Manager
Joined: 23 Jan 2006
Posts: 192
Followers: 1

Kudos [?]: 14 [0], given: 0

Re: The area of the circle [#permalink]

### Show Tags

17 May 2006, 06:50
Professor wrote:
getzgetzu wrote:
An equilateral triangle is inscribed in a circle and has the area of 9*sqrt3. What the area of the circle?

area of an equilateral triangle = [sqrt(3)/4] a^2
9 sqrt(3) = [sqrt(3)/4] a^2
a = 6

if an equilateral triangle is inscribed in a circle, the radius (r) of the circle = a/sqrt(3)

r = 6/sqrt(3) = (3x2)/sqrt(3) = 2 sqrt(3)
the area of the circle = (pi) r^2 = (pi) [2 sqrt(3)]^2 = 12 pi.

Is there any way to solve these without knowingn these?
just using the standard properties of triangles & circles?
Senior Manager
Joined: 09 Mar 2006
Posts: 445
Followers: 1

Kudos [?]: 8 [0], given: 0

Re: The area of the circle [#permalink]

### Show Tags

17 May 2006, 07:35
kook44 wrote:
Is there any way to solve these without knowingn these?
just using the standard properties of triangles & circles?

Yes, I used another approach

Assume the triangle is ABC and O is the center of the circle.
AO, BO and CO divide the triangle ABC into 3 equal isosceles triangles,
when each has an area of 9sqrt(3) / 3 = 3sqrt(3) .

Extending the AO up to BC , we get a height AD , so OD is the height of triangle BOC.

Now assume OD = x , hence DC = x*sqrt(3) and OC=2 *x = R ( ODC is a right triangle 30-60-90 )

the area of ODC = 3sqrt(3) / 2

x*x*sqrt(3) / 2 = 3sqrt(3)/2
x = sqrt(3)
OC = R = 2*sqrt(3) , hence the area of the circle is 12pi

Last edited by deowl on 17 May 2006, 07:55, edited 1 time in total.
CEO
Joined: 20 Nov 2005
Posts: 2911
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 21

Kudos [?]: 194 [0], given: 0

Re: The area of the circle [#permalink]

### Show Tags

17 May 2006, 07:38
kook44 wrote:
Professor wrote:
getzgetzu wrote:
An equilateral triangle is inscribed in a circle and has the area of 9*sqrt3. What the area of the circle?

area of an equilateral triangle = [sqrt(3)/4] a^2
9 sqrt(3) = [sqrt(3)/4] a^2
a = 6

if an equilateral triangle is inscribed in a circle, the radius (r) of the circle = a/sqrt(3)

r = 6/sqrt(3) = (3x2)/sqrt(3) = 2 sqrt(3)
the area of the circle = (pi) r^2 = (pi) [2 sqrt(3)]^2 = 12 pi.

Is there any way to solve these without knowingn these?
just using the standard properties of triangles & circles?

Yes there are ways. One of them is here

Side of triangle = a
Height of triangle using pythagoras = SQRT(a^2 - (a/2)^2 ) = a * SQRT(3)/2

Area of triangle = 1/2 * Base * Height = 1/2 * a * a * SQRT(3)/2 = a^2 * SQRT(3)/4 = 9 * SQRT(3)

So a = 6
Height = 6 * SQRT(3)/2 = 3 * SQRT(3)

Now if you drop lines from all three vertices of triangle to the centre of sides then these three lines will meet at the center of the circle. The ratio of the lines from centre to the vertex and centre to the side is 2:1 (This you have to remember or you may waste time to prove this using similar triangles). The line from centre to the vertex is the radius of circle.
So r = 2/3 * Height = 2/3 * 3 * SQRT(3) = 2 * SQRT3

Area of circle = PI * r^2 = PI * (2 * SQRT(3)) ^2 = 12 PI

Hope this helps.
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Manager
Joined: 07 May 2013
Posts: 109
Followers: 0

Kudos [?]: 17 [0], given: 1

Re: An equilateral triangle is inscribed in a circle and has the [#permalink]

### Show Tags

20 Nov 2013, 18:48
$$\sqrt{3}$$/4*$$a^2$$=9$$\sqrt{3}$$. a=6. Now shortcut formula for radius of circle circumsubcribed over an equilateral is $$\sqrt{3}$$a/3= 2$$\sqrt{3}$$. $$area=pi*r^2$$=12pi.
Math Expert
Joined: 02 Sep 2009
Posts: 33037
Followers: 5759

Kudos [?]: 70578 [2] , given: 9849

Re: An equilateral triangle that has an area of 9*3^1/2 [#permalink]

### Show Tags

21 Nov 2013, 01:49
2
KUDOS
Expert's post
An equilateral triangle that has an area of $$9\sqrt{3}$$ is inscribed in a circle. What is the area of the circle?

A. $$6\pi$$
B. $$9\pi$$
C. $$12\pi$$
D. $$9\pi \sqrt{3}$$
E. $$18\pi \sqrt{3}$$

$$area_{equilateral}=a^2*\frac{\sqrt{3}}{4}$$, where $$a$$ is the length of a side --> as given that $$area_{equilateral}=a^2*\frac{\sqrt{3}}{4}=9\sqrt{3}$$ then $$a=6$$;

We are given that this triangle is inscribed in circle. The radius of the circumscribed circle is $$R=a*\frac{\sqrt{3}}{3}=2\sqrt{3}$$ (the radius of the inscribed circle $$r=a*\frac{\sqrt{3}}{6}$$) --> $$area_{circle}=\pi{R^2}=12\pi$$.

Check Triangles chapter of Math Book for more: math-triangles-87197.html

OPEN DISCUSSION OF THIS QUESTION IS HERE: an-equilateral-triangle-that-has-an-area-of-9-root-105356.html
_________________
Re: An equilateral triangle that has an area of 9*3^1/2   [#permalink] 21 Nov 2013, 01:49
Similar topics Replies Last post
Similar
Topics:
7 Both triangles ABC and ADE are equilateral. The shaded area 12 22 Jul 2014, 03:50
7 In the given figure, the area of the equilateral triangle is 8 15 Sep 2013, 10:08
5 If the area of an equilateral triangle is x square meters 21 29 Dec 2010, 18:25
11 An equilateral triangle that has an area of 9*root(3) 15 25 Nov 2010, 02:11
1 Area of a circumcirlce of equilateral triangle 3 28 Jul 2010, 21:13
Display posts from previous: Sort by