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:

Re: ABE is an equilateral tringle, BCDE is a square, and the are [#permalink]
07 Mar 2013, 15:48

1

This post received KUDOS

pancakeFR wrote:

holidayhero wrote:

R = radius of circle a = side of equilateral triangle (also side of square)

Hi,

I don't understand where you got that from. Could you explain please ?

Thank you.

Here's an explanation. Caioguima also explained it.

Let the side length of the equilateral triangle be a. The circumradius of the equilateral triangle is equal to 2/3 of the altitude. The altitude is equal to a\sqrt{3}/2 (Drawing the altitude creates two 30,60,90 triangles). The circumradius of this triangle is also the radius of the circle.

Re: ABE is an equilateral tringle, BCDE is a square, and the are [#permalink]
08 Mar 2013, 11:27

2

This post received KUDOS

in the equilateral triangle, the circum-centre divides the height in the ratio 2:1 (smaller length towards the base)

and area of circle = pi x r*r hence r= 6

so 2/3 * (3^ 1/2)/2*a = 6 where a is side of triangle hence solving for a we get a = 6* (3^0.5)

the required are = a*a + (3^0.5)/4 * a*a

= {108 + 27(3^0.5)}

holidayhero wrote:

pancakeFR wrote:

holidayhero wrote:

R = radius of circle a = side of equilateral triangle (also side of square)

Hi,

I don't understand where you got that from. Could you explain please ?

Thank you.

Here's an explanation. Caioguima also explained it.

Let the side length of the equilateral triangle be a. The circumradius of the equilateral triangle is equal to 2/3 of the altitude. The altitude is equal to a\sqrt{3}/2 (Drawing the altitude creates two 30,60,90 triangles). The circumradius of this triangle is also the radius of the circle.

Re: ABE is an equilateral tringle, BCDE is a square, and the are [#permalink]
24 Feb 2014, 06:08

This might not be the best approach. But it can do the trick if you don't know how to proceed with the sum Area of triangle = √3/4 a^2 Area of square = a^2

Total area = area of triangle + area of square = √3/4 a^2 + a^2

If you see the options except C all have √3 part. It is very less likely for our answer to not have √3 part. (this is a part where we might be wrong the the answer might have been approximated to near number, unless if GMAT doesn't do so without specifying)

Now the other options. Lets see, if we have a + x√3 in answer choices then it must be true that x = 4a (Why? try playing with the total area) This only true for option A. Voila!