ABE is an equilateral tringle, BCDE is a square, and the are : Problem Solving (PS)

pancakeFR

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Re: ABE is an equilateral tringle, BCDE is a square, and the are [#permalink]

07 Mar 2013, 03:39

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.

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Re: ABE is an equilateral tringle, BCDE is a square, and the are [#permalink]

07 Mar 2013, 16:48

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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.

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]

08 Mar 2013, 12:27

1

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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:

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.

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]

08 Mar 2013, 12:29

Thank you all for your answers. I get it now !

vishalrastogi

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Re: ABE is an equilateral tringle, BCDE is a square, and the are [#permalink]

24 Feb 2014, 04:08

holidayhero wrote:

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"

(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]

24 Feb 2014, 07:08

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!

L

Bunuel

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Re: ABE is an equilateral tringle, BCDE is a square, and the are [#permalink]

24 Feb 2014, 07:26

Expert Reply

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]

29 Aug 2014, 01:30

Bunuel wrote:

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.

Hi Bunuel,

I have several questions regarding this matter.

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

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Re: ABE is an equilateral tringle, BCDE is a square, and the are [#permalink]

29 Aug 2014, 01:56

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unceldolan wrote:

Bunuel wrote:

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.

Hi Bunuel,

I have several questions regarding this matter.

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

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Re: ABE is an equilateral tringle, BCDE is a square, and the are [#permalink]

14 Sep 2023, 11:32

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Re: ABE is an equilateral tringle, BCDE is a square, and the are [#permalink]

14 Sep 2023, 11:32

GMAT Problem Solving (PS) Questions

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