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:

I am looking for a quicker method ?? If equilateral triangle of area 9Sqrt(3) is inscribed with in a circle what is the area of the circle?

i do not have any quicker methos than this.........
[sqrt(3) L^2]/4= 9sqrt(3), which gives you L=6

draw Prependicular lines from each vertex of the triangle to the other respective sides(base) of the triangle. now we have different triangle. take one triangle and find the H (Hypoteneous) of that triangle and the H is the redius of that triangle.

H of smal triangle = r = 2sqrt(3). then, area of the circle= formula.......

Praveen, if you have any short cut method, pls do share with us.

A triangle if inscribed in a circle is a circumcircle. So lets suppose S is the centre of the circle and A.B and C be the vertices.
Then SA=SB=SC=circumradius

sqrt(3)/4*a^2 = 9*sqrt(3)

a=6 side of triangle

Now angle ASB = 2*Angle C = 2* 60=120(Where S is the Circumcenter . Angle at vertex is half at the circumcenter ,S)

Since AB=6 ===>> AS = 6*Sin120 = 3

So circumradius = 3 and the area of circumcircle = PI*9

PLease ignore the previous posting of mine...

Last edited by gmat2me2 on 27 Mar 2005, 15:48, edited 1 time in total.

Well Given (sqrt3)/4 a^2 = 9sqrt3 => a = 6 length of equilateral triangle

Now The centroid of an equilateral triangle is where the 3 medians cut each other and it should be the centre of the circle and the radius then becomes (2/3)*a*cos 60 deg = 2/3*6*1/2 = 2
Therefore area = pi84 = 4pi

What I would do is to memorize the special right triangle whose three angels are 90, 60, 30. The ratio of its three sides are 1:sqrt(3):2.

The unit area of this special triangle is sqrt(3)/2. Since the area of our triangle (Call it ABC) here is 9sqrt(3), half of the triangle (ABD) is 9sqrt(3)/2. We know that the triangle's three sides very quickly: 3,3sqrt(3),6.

AOD is again this same kind of special triangle and AD=3. We know that AO:AD=2:sqrt(3). So AO=r=6/sqrt(3)=2sqrt(3).

From here it's easy to get the area of the circle. S=pi*r^2=12pi.

Verification: The area of the triangle is 9sqrt(3)=9*1.7=15
Area of the circle =12pi=12*3.14=37

If you draw the graph, you'll see that the area that is formed by AB and the arc AB is a little bit larger than the area of the triangle ABO. It is true for the other two triangles. In other words, the total area of the triangle ABC is a little smaller than the area within the circle and outside of the triangle. In other words it should be a little less than half of the area of the circle. Our results checked out.

If it is a PS question and you have different choices to choose from. You may not even need to calculate if you remember the area of the circle is a little bit larger than twice of the area of the triangle.

Well Given (sqrt3)/4 a^2 = 9sqrt3 => a = 6 length of equilateral triangle

Now The centroid of an equilateral triangle is where the 3 medians cut each other and it should be the centre of the circle and the radius then becomes (2/3)*a*cos 60 deg = 2/3*6*1/2 = 2 Therefore area = pi84 = 4pi

Sorry Anirban you cannot take centroid in all cases... But it is an eq trangle you may probably but it fails for othercases......