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# ABC is an equilateral triangle of area 3, and arc DE is cent

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ABC is an equilateral triangle of area 3, and arc DE is cent  [#permalink]

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Updated on: 22 Sep 2013, 03:34
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Question Stats:

69% (02:43) correct 31% (03:03) wrong based on 333 sessions

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ABC is an equilateral triangle of area 3, and arc DE is centered at C. If E is the midpoint of AC, what is the area of the shaded region?

A. $$3-\frac{\sqrt{3}\pi}{2}$$

B. $$3-\frac{\pi}{\sqrt{3}}$$

C. $$3-\frac{\pi}{2}$$

D. $$3-\frac{\pi}{2\sqrt{3}}$$

E. $$3-\frac{\pi}{6}$$

Originally posted by akashb106 on 22 Sep 2013, 00:48.
Last edited by Bunuel on 22 Sep 2013, 03:34, edited 1 time in total.
Edited the question and added the OA
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Re: ABC is an equilateral triangle of area 3, and arc DE is cent  [#permalink]

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22 Sep 2013, 03:46
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akashb106 wrote:

ABC is an equilateral triangle of area 3, and arc DE is centered at C. If E is the midpoint of AC, what is the area of the shaded region?

A. $$3-\frac{\sqrt{3}\pi}{2}$$

B. $$3-\frac{\pi}{\sqrt{3}}$$

C. $$3-\frac{\pi}{2}$$

D. $$3-\frac{\pi}{2\sqrt{3}}$$

E. $$3-\frac{\pi}{6}$$

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

Given that $$side^2*\frac{\sqrt{3}}{4}=3$$ --> $$side=2\sqrt[4]{3}$$.

Since E is the midpoint of AC, then CE = radius = half of the side = $$\sqrt[4]{3}$$.

Next, as ABC is an equilateral triangle, then angle C is 60 degrees, thus sector CED is 1/6 of the area of the whole circle --> $$area=\pi*(\sqrt[4]{3})^2*\frac{1}{6}=\frac{\pi\sqrt{3}}{6}=\frac{\pi}{2\sqrt{3}}$$.

The area of the shaded region = $$3-\frac{\pi}{2\sqrt{3}}$$.

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Re: ABC is an equilateral triangle of area 3, and arc DE is cent  [#permalink]

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22 Sep 2013, 04:48
Bunuel wrote:
akashb106 wrote:

ABC is an equilateral triangle of area 3, and arc DE is centered at C. If E is the midpoint of AC, what is the area of the shaded region?

A. $$3-\frac{\sqrt{3}\pi}{2}$$

B. $$3-\frac{\pi}{\sqrt{3}}$$

C. $$3-\frac{\pi}{2}$$

D. $$3-\frac{\pi}{2\sqrt{3}}$$

E. $$3-\frac{\pi}{6}$$

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

Given that $$side^2*\frac{\sqrt{3}}{4}=3$$ --> $$side=2\sqrt[4]{3}$$.

Since E is the midpoint of AC, then CE = radius = half of the side = $$\sqrt[4]{3}$$.

Next, as ABC is an equilateral triangle, then angle C is 60 degrees, thus sector CED is 1/6 of the area of the whole circle --> $$area=\pi*(\sqrt[4]{3})^2*\frac{1}{6}=\frac{\pi\sqrt{3}}{6}=\frac{\pi}{2\sqrt{3}}$$.

The area of the shaded region = $$3-\frac{\pi}{2\sqrt{3}}$$.

Good call, I missed the $$\frac{1}{6}$$ part...
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Re: ABC is an equilateral triangle of area 3, and arc DE is cent  [#permalink]

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23 Sep 2013, 10:04
I don't know why I'm not good with surds. Whenever sqr(3) I mess up with my calculations. Do we have some notes on basics of surds
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Re: ABC is an equilateral triangle of area 3, and arc DE is cent  [#permalink]

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24 Sep 2013, 00:29
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b2bt wrote:
I don't know why I'm not good with surds. Whenever sqr(3) I mess up with my calculations. Do we have some notes on basics of surds

Theory on roots problems: math-number-theory-88376.html

All DS roots problems to practice: search.php?search_id=tag&tag_id=49
All PS roots problems to practice: search.php?search_id=tag&tag_id=113

Tough and tricky exponents and roots questions (DS): tough-and-tricky-exponents-and-roots-questions-125967.html
Tough and tricky exponents and roots questions (PS): new-tough-and-tricky-exponents-and-roots-questions-125956.html

Hope it helps.
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Re: ABC is an equilateral triangle of area 3, and arc DE is cent  [#permalink]

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30 Jan 2015, 12:47
Good post ! Catch here was the 1/6th part of the circle, which I caught and very well explained by our expert.

Cheers !!!
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Re: ABC is an equilateral triangle of area 3, and arc DE is cent  [#permalink]

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31 Jan 2015, 06:16
The area of equilateral triangle is $$side^2*\frac{\sqrt{3}}{4}$$.

Given that $$side^2*\frac{\sqrt{3}}{4}=3$$ --> $$side=2\sqrt[4]{3}$$.

Can someone explain this step in more detail?
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Re: ABC is an equilateral triangle of area 3, and arc DE is cent  [#permalink]

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31 Jan 2015, 06:23
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LaxAvenger wrote:
The area of equilateral triangle is $$side^2*\frac{\sqrt{3}}{4}$$.

Given that $$side^2*\frac{\sqrt{3}}{4}=3$$ --> $$side=2\sqrt[4]{3}$$.

Can someone explain this step in more detail?

$$side^2*\frac{\sqrt{3}}{4}=3$$;

$$side^2*\frac{\sqrt{3}}{4}=(\sqrt{3})^2$$;

$$side^2*\frac{1}{4}=\sqrt{3}$$;

$$side^2*=4*\sqrt{3}$$;

$$side=2\sqrt[4]{3}$$.

Hope it's clear.
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Re: ABC is an equilateral triangle of area 3, and arc DE is centered at C.  [#permalink]

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28 Jul 2015, 11:08
reto wrote:
Attachment:
The attachment T8888.png is no longer available

ABC is an equilateral triangle of area 3, and arc DE is centered at C. If E is the midpoint of AC, what is the area of the shaded region?

A. $$3 - \frac{\sqrt(3)*\pi}{2}$$
B. $$3 - \frac{\pi}{\sqrt(3)}$$
C. $$3 - \frac{\pi}{2}$$
D. $$3 - \frac{\pi}{2*\sqrt(3)}$$
E. $$3 - \frac{\pi}{6}$$

2 ways to solve this:

Method 1:

Solving by calculating areas of the triangle and the circle.

Let r be the radius of the arc.

From area of the triangle = $$\frac{\sqrt{3}*a^2}{4}$$ , a = side of the triangle.

Thus, $$\frac{\sqrt{3}*a^2}{4}$$ = 3 ---> $$a^2 = 4*\sqrt{3}$$

Now, given that r=a/2 --> area of the sector ( = 60 degree arc) = $$(60/360)* \pi*r^2$$ = $$\frac{\pi*\sqrt{3}}{6}$$

Thus the area of the shaded region = area of the triangle - area of the sector = $$3 - \frac{\pi*\sqrt{3}}{6}$$ = option D.

Method 2: Approximations

*Per the attached picture, the shaded area is 3/4 ths of the triangle area $$\approx{2.25}$$.

Start plugging in $$\pi=3 and \sqrt{3} = 1.7$$ and see which one gives you the closest value.

All values below are approximate

A: 0.75
B: 1.2
C: 1.5
D: 2.2
E: 2.5

Thus option D is the correct one.
Attachments

trangles.jpg [ 10.35 KiB | Viewed 6723 times ]

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ABC is an equilateral triangle of area 3, and arc DE is centered at C.  [#permalink]

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28 Jul 2015, 22:06
reto wrote:
Attachment:
T8888.png

ABC is an equilateral triangle of area 3, and arc DE is centered at C. If E is the midpoint of AC, what is the area of the shaded region?

A. $$3 - \frac{\sqrt(3)*\pi}{2}$$
B. $$3 - \frac{\pi}{\sqrt(3)}$$
C. $$3 - \frac{\pi}{2}$$
D. $$3 - \frac{\pi}{2*\sqrt(3)}$$
E. $$3 - \frac{\pi}{6}$$

To approximate, you can use the property of similar triangles. If you join DE, you get that triangle DEC is similar to triangle ABC such that side of DEC is half of the side of ABC. So area of DEC will be 1/4 of the area of ABC (which is 3).

The shaded region is a little less than 3 - 3/4. All options are in the form of (3 - Something).
This 'something' would be slightly more than 3/4 but perhaps less than 1.

A. $$3 - \frac{\sqrt(3)*\pi}{2}$$
Something = 1.7*3.14/2 (much greater than 1)
Not possible

B. $$3 - \frac{\pi}{\sqrt(3)}$$
Something = 3.14/1.7 (much greater than 1)
Not possible

C. $$3 - \frac{\pi}{2}$$
Something = 3.14/2 (much greater than 1)
Not possible

D. $$3 - \frac{\pi}{2*\sqrt(3)}$$
Something = 3.14/2*1.7 = 3.14/3.4
Certainly possible

E. $$3 - \frac{\pi}{6}$$
Something = 3.14/6 = Almost 1/2
Not possible since it must be greater than 3/4

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Re: ABC is an equilateral triangle of area 3, and arc DE is cent  [#permalink]

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29 Jan 2017, 02:01
akashb106 wrote:
Attachment:
T8888.png
ABC is an equilateral triangle of area 3, and arc DE is centered at C. If E is the midpoint of AC, what is the area of the shaded region?

A. $$3-\frac{\sqrt{3}\pi}{2}$$

B. $$3-\frac{\pi}{\sqrt{3}}$$

C. $$3-\frac{\pi}{2}$$

D. $$3-\frac{\pi}{2\sqrt{3}}$$

E. $$3-\frac{\pi}{6}$$

Let "x" be the side of the equilateral triangle.
Given: $$x^2$$$$\sqrt{3}/4$$ (Area of the equilateral traiangle) = 3
By simplifying the above eqn we get $$x^2$$ =$$4\sqrt{3}$$
WKT any angle of this triangle is $$60^o$$.
Hence the area of the sector = $$\pi$$$$R^2$$ $$\frac{60}{360}$$

And its given that $$R = x/2$$ (E is the mid point)
Solving we get $$\pi/2\sqrt{3}$$
Hence the area of the shaded portion is 3 - $$\pi/2\sqrt{3}$$
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Re: ABC is an equilateral triangle of area 3, and arc DE is cent  [#permalink]

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08 Oct 2019, 23:04
Bunuel wrote:
akashb106 wrote:

ABC is an equilateral triangle of area 3, and arc DE is centered at C. If E is the midpoint of AC, what is the area of the shaded region?

A. $$3-\frac{\sqrt{3}\pi}{2}$$

B. $$3-\frac{\pi}{\sqrt{3}}$$

C. $$3-\frac{\pi}{2}$$

D. $$3-\frac{\pi}{2\sqrt{3}}$$

E. $$3-\frac{\pi}{6}$$

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

Given that $$side^2*\frac{\sqrt{3}}{4}=3$$ --> $$side=2\sqrt[4]{3}$$.

Since E is the midpoint of AC, then CE = radius = half of the side = $$\sqrt[4]{3}$$.

Next, as ABC is an equilateral triangle, then angle C is 60 degrees, thus sector CED is 1/6 of the area of the whole circle --> $$area=\pi*(\sqrt[4]{3})^2*\frac{1}{6}=\frac{\pi\sqrt{3}}{6}=\frac{\pi}{2\sqrt{3}}$$.

The area of the shaded region = $$3-\frac{\pi}{2\sqrt{3}}$$.

Why do we need to put the radical in the denominator in this example?
I was told that if I have a solution with a radical in the denominator I always should try to reverse it, so Pi*Sqrt(3) / 6 should normally be the right answer option already.

Is this just another way to confuse test takers or what am I missing.
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Re: ABC is an equilateral triangle of area 3, and arc DE is cent   [#permalink] 08 Oct 2019, 23:04
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