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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!

First: is this "The radius of the circumscribed circle in equilateral triangle is R=side\frac{\sqrt{3}}{3} " a general rule? If a equilateral triangle is inscribed in a circle, the formula applies?

And how do you get from the first step to the second step?

R=side\frac{\sqrt{3}}{3}=6 --> side=6\sqrt{3}

The way I see it, 6 = side * sqrt3 / 3 => 18/sqrt3 . What am I mssing?

First: is this "The radius of the circumscribed circle in equilateral triangle is R=side\frac{\sqrt{3}}{3} " a general rule? If a equilateral triangle is inscribed in a circle, the formula applies?

And how do you get from the first step to the second step?

R=side\frac{\sqrt{3}}{3}=6 --> side=6\sqrt{3}

The way I see it, 6 = side * sqrt3 / 3 => 18/sqrt3 . What am I mssing?

Rest is clear.

Thank you

1) Please refer to the Math resources section on geometry for relationships between circumradius, inradius, etc wrt equilateral triangles. It will be useful to learn these concepts so you can save time.

2) 18/sqrt3 Multiple both numerator and denominator with sqrt3, = 6*sqrt3

gmatclubot

Re: ABE is an equilateral tringle, BCDE is a square, and the are
[#permalink]
29 Aug 2014, 00:56

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...