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ABE is an equilateral tringle, BCDE is a square, and the are

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ABE is an equilateral tringle, BCDE is a square, and the are [#permalink] New post 05 Mar 2013, 13:07
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In the figure above, ABE is an equilateral triangle, BCDE is a square, and the area of the circle is 36π. What is the area of polygon ABCDE?

A. 108 + 27 \sqrt{3}
B. 108 + 36 \sqrt{3}
C. 189
D. 216 + 81 \sqrt{3}
E. 324 + 27 \sqrt{3}
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 Feb 2014, 06:19, edited 4 times in total.
Edited the question.
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Re: ABE is an equilateral tringle, BCDE is a square, and the are [#permalink] New post 05 Mar 2013, 16:21
Given that ABE is an equilateral triangle, the Radius is equal to 2/3 of the triangle's height:


R = (2/3).L\sqrt{3} /2 = 6



Hence, L = 6\sqrt{3}

The area of the region ABCDE is therefore:

Area = L^2 + (L^2).\sqrt{3}/4

= 108 + 27\sqrt{3}


ANSWER: No option available (maybe a typo on answer choice A)
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Re: ABE is an equilateral tringle, BCDE is a square, and the are [#permalink] New post 05 Mar 2013, 18:49
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Image

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

We know R = 6 because area of circle is 36π
Plug this into the formula above and you will see that a =6\sqrt{3}

Let Area of Equilateral Triangle be represented by "K"
Image
(sub in a =6\sqrt{3})
K = 27\sqrt{3}

Let Area of Square be represented by "S"
S= a^2
S=(6\sqrt{3})^2
S=108

Total Area of Polygon ABCDE is
108 + 27\sqrt{3}
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Re: ABE is an equilateral tringle, BCDE is a square, and the are [#permalink] New post 07 Mar 2013, 02:39
holidayhero wrote:
Image

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.
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Re: ABE is an equilateral tringle, BCDE is a square, and the are [#permalink] New post 07 Mar 2013, 15:48
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pancakeFR wrote:
holidayhero wrote:
Image

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.


Image

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.

R = (2/3) (a\sqrt{3}/2)
R = a\sqrt{3}/3
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Re: ABE is an equilateral tringle, BCDE is a square, and the are [#permalink] New post 08 Mar 2013, 11:27
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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:
Image

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.


Image

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.

R = (2/3) (a\sqrt{3}/2)
R = a\sqrt{3}/3
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Re: ABE is an equilateral tringle, BCDE is a square, and the are [#permalink] New post 08 Mar 2013, 11:29
Thank you all for your answers. I get it now ! :)
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Re: ABE is an equilateral tringle, BCDE is a square, and the are [#permalink] New post 24 Feb 2014, 03:08
holidayhero wrote:
Image

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

We know R = 6 because area of circle is 36π
Plug this into the formula above and you will see that a =6\sqrt{3}

Let Area of Equilateral Triangle be represented by "K"
Image
(sub in a =6\sqrt{3})
K = 27\sqrt{3}

Let Area of Square be represented by "S"
S= a^2
S=(6\sqrt{3})^2
S=108

Total Area of Polygon ABCDE is
108 + 27\sqrt{3}

Hi Bunuel,

Will you please explain this ? I am not even close to understanding !

Thanks.
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Re: ABE is an equilateral tringle, BCDE is a square, and the are [#permalink] New post 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!
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Re: ABE is an equilateral tringle, BCDE is a square, and the are [#permalink] New post 24 Feb 2014, 06:26
Expert's post
Image
In the figure above, ABE is an equilateral triangle, BCDE is a square, and the area of the circle is 36π. What is the area of polygon ABCDE?

A. 108 + 27 \sqrt{3}
B. 108 + 36 \sqrt{3}
C. 189
D. 216 + 81 \sqrt{3}
E. 324 + 27 \sqrt{3}

The area of the circle is 36\pi --> \pi{R^2}=36\pi --> R=6.

The radius of the circumscribed circle in equilateral triangle is R=side\frac{\sqrt{3}}{3}=6 --> side=6\sqrt{3}.

The area of equilateral triangle = side^2*\frac{\sqrt{3}}{4}=27\sqrt{3}.

The area of square = side^2=108.

The area of polygon ABCDE = 108+27\sqrt{3}.

Answer: A.

For more check Triangles chapter of Math Book here: math-triangles-87197.html

Hope it helps.
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Re: ABE is an equilateral tringle, BCDE is a square, and the are   [#permalink] 24 Feb 2014, 06:26
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