It is currently 21 Aug 2017, 08:54

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# S98-20

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 40989

Kudos [?]: 119070 [0], given: 12014

### Show Tags

16 Sep 2014, 01:53
Expert's post
3
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

31% (03:05) correct 69% (03:55) wrong based on 62 sessions

### HideShow timer Statistics

Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?

A. 24%
B. 30%
C. 36%
D. 42%
E. 48%
[Reveal] Spoiler: OA

_________________

Kudos [?]: 119070 [0], given: 12014

Math Expert
Joined: 02 Sep 2009
Posts: 40989

Kudos [?]: 119070 [1] , given: 12014

### Show Tags

16 Sep 2014, 01:53
1
KUDOS
Expert's post
Official Solution:

Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?

A. 24%
B. 30%
C. 36%
D. 42%
E. 48%

If polygon $$ABCDEF$$ is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.

We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to $$180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ$$, and there are 6 interior, equal angles, then each of those angles must measure $$\frac{720^\circ}{6} = 120^\circ$$.

Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center $$O$$:

Consider small triangle $$AXB$$. The hypotenuse of this right triangle, $$AB$$, has length $$r$$. Angle $$ABO$$ is $$60^\circ$$, so $$AXB$$ is a 30-60-90 triangle. This means that $$XB$$ is $$\frac{r}{2}$$ in length, and $$AX$$ is $$\frac{\sqrt{3}r}{4}$$ in length. Since $$AX$$ and $$XB$$ are the same length (by symmetry), $$AC$$ is $$\sqrt{3}r$$ in length.

This means that the area of triangle $$ABC$$ is $$\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}$$. Since there are three shaded triangles in all (including $$ABC$$), the total shaded area is $$\frac{3 \sqrt{3} r^2}{4}$$. The area of the circle is , so the ratio of the shaded area to the area of the circle is given by
$$\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43$$

_________________

Kudos [?]: 119070 [1] , given: 12014

Intern
Joined: 07 Sep 2014
Posts: 24

Kudos [?]: 17 [0], given: 6

Location: United States (MA)
Concentration: Finance, Economics

### Show Tags

10 Oct 2014, 16:25
Bunuel wrote:
Official Solution:

Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?

A. 24%
B. 30%
C. 36%
D. 42%
E. 48%

If polygon $$ABCDEF$$ is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.

We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to $$180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ$$, and there are 6 interior, equal angles, then each of those angles must measure $$\frac{720^\circ}{6} = 120^\circ$$.

Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center $$O$$:

Consider small triangle $$AXB$$. The hypotenuse of this right triangle, $$AB$$, has length $$r$$. Angle $$ABO$$ is $$60^\circ$$, so $$AXB$$ is a 30-60-90 triangle. This means that $$XB$$ is $$\frac{r}{2}$$ in length, and $$AX$$ is $$\frac{\sqrt{3}r}{4}$$ in length. Since $$AX$$ and $$XB$$ are the same length (by symmetry), $$AC$$ is $$\sqrt{3}r$$ in length.

This means that the area of triangle $$ABC$$ is $$\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}$$. Since there are three shaded triangles in all (including $$ABC$$), the total shaded area is $$\frac{3 \sqrt{3} r^2}{4}$$. The area of the circle is , so the ratio of the shaded area to the area of the circle is given by
$$\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43$$

Bunuel,

I am confused at the step in which you derive the height side of the first 30-60-90 triangle. the "x" side is r/2, because the "2x" side is r, but why is "x sqrt 3" side not = to (r sqrt 3)/2? instead you have it divided by 4? I must be missing something. You also say that side AX = XB? That can't be because we just found it to be a 30-60-90 right triangle. You must've meant to say that AX = AX, thus we know AC?

Kudos [?]: 17 [0], given: 6

Math Expert
Joined: 02 Sep 2009
Posts: 40989

Kudos [?]: 119070 [0], given: 12014

### Show Tags

11 Oct 2014, 13:28
bsmith37 wrote:
Bunuel wrote:
Official Solution:

Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?

A. 24%
B. 30%
C. 36%
D. 42%
E. 48%

If polygon $$ABCDEF$$ is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.

We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to $$180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ$$, and there are 6 interior, equal angles, then each of those angles must measure $$\frac{720^\circ}{6} = 120^\circ$$.

Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center $$O$$:

Consider small triangle $$AXB$$. The hypotenuse of this right triangle, $$AB$$, has length $$r$$. Angle $$ABO$$ is $$60^\circ$$, so $$AXB$$ is a 30-60-90 triangle. This means that $$XB$$ is $$\frac{r}{2}$$ in length, and $$AX$$ is $$\frac{\sqrt{3}r}{4}$$ in length. Since $$AX$$ and $$XB$$ are the same length (by symmetry), $$AC$$ is $$\sqrt{3}r$$ in length.

This means that the area of triangle $$ABC$$ is $$\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}$$. Since there are three shaded triangles in all (including $$ABC$$), the total shaded area is $$\frac{3 \sqrt{3} r^2}{4}$$. The area of the circle is , so the ratio of the shaded area to the area of the circle is given by
$$\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43$$

Bunuel,

I am confused at the step in which you derive the height side of the first 30-60-90 triangle. the "x" side is r/2, because the "2x" side is r, but why is "x sqrt 3" side not = to (r sqrt 3)/2? instead you have it divided by 4? I must be missing something. You also say that side AX = XB? That can't be because we just found it to be a 30-60-90 right triangle. You must've meant to say that AX = AX, thus we know AC?

The following discussion might help: approximately-what-percent-of-the-area-of-the-circle-shown-i-169744.html
_________________

Kudos [?]: 119070 [0], given: 12014

Intern
Joined: 07 Sep 2014
Posts: 24

Kudos [?]: 17 [0], given: 6

Location: United States (MA)
Concentration: Finance, Economics

### Show Tags

11 Oct 2014, 15:57
Bunuel wrote:
bsmith37 wrote:
Bunuel wrote:
Official Solution:

Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?

A. 24%
B. 30%
C. 36%
D. 42%
E. 48%

If polygon $$ABCDEF$$ is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.

We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to $$180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ$$, and there are 6 interior, equal angles, then each of those angles must measure $$\frac{720^\circ}{6} = 120^\circ$$.

Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center $$O$$:

Consider small triangle $$AXB$$. The hypotenuse of this right triangle, $$AB$$, has length $$r$$. Angle $$ABO$$ is $$60^\circ$$, so $$AXB$$ is a 30-60-90 triangle. This means that $$XB$$ is $$\frac{r}{2}$$ in length, and $$AX$$ is $$\frac{\sqrt{3}r}{4}$$ in length. Since $$AX$$ and $$XB$$ are the same length (by symmetry), $$AC$$ is $$\sqrt{3}r$$ in length.

This means that the area of triangle $$ABC$$ is $$\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}$$. Since there are three shaded triangles in all (including $$ABC$$), the total shaded area is $$\frac{3 \sqrt{3} r^2}{4}$$. The area of the circle is , so the ratio of the shaded area to the area of the circle is given by
$$\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43$$

Bunuel,

I am confused at the step in which you derive the height side of the first 30-60-90 triangle. the "x" side is r/2, because the "2x" side is r, but why is "x sqrt 3" side not = to (r sqrt 3)/2? instead you have it divided by 4? I must be missing something. You also say that side AX = XB? That can't be because we just found it to be a 30-60-90 right triangle. You must've meant to say that AX = AX, thus we know AC?

The following discussion might help: approximately-what-percent-of-the-area-of-the-circle-shown-i-169744.html

that link was helpful, but now I'm just trying to tie it back in with your slightly alternative explanation above. Still, i cannot see why you say that AX = $$\frac{\sqrt{3}r}{4}$$ because we have an equilateral triangle (AOB), with sides R, and height AX. The formula states that the height AX would be $$\frac{\sqrt{3}r}{2}$$. Please advise.

Kudos [?]: 17 [0], given: 6

Manager
Joined: 12 May 2013
Posts: 81

Kudos [?]: 38 [0], given: 12

### Show Tags

24 Dec 2014, 00:42
Bunuel wrote:
Official Solution:

Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?

A. 24%
B. 30%
C. 36%
D. 42%
E. 48%

If polygon $$ABCDEF$$ is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.

We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to $$180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ$$, and there are 6 interior, equal angles, then each of those angles must measure $$\frac{720^\circ}{6} = 120^\circ$$.

Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center $$O$$:

Consider small triangle $$AXB$$. The hypotenuse of this right triangle, $$AB$$, has length $$r$$. Angle $$ABO$$ is $$60^\circ$$, so $$AXB$$ is a 30-60-90 triangle. This means that $$XB$$ is $$\frac{r}{2}$$ in length, and $$AX$$ is $$\frac{\sqrt{3}r}{4}$$ in length. Since $$AX$$ and $$XB$$ are the same length (by symmetry), $$AC$$ is $$\sqrt{3}r$$ in length.

This means that the area of triangle $$ABC$$ is $$\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}$$. Since there are three shaded triangles in all (including $$ABC$$), the total shaded area is $$\frac{3 \sqrt{3} r^2}{4}$$. The area of the circle is , so the ratio of the shaded area to the area of the circle is given by
$$\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43$$

Hey Bunuel,
I think the highlighted part above should be ax=xc.
and that is why it is creating a bit of confusion.

Kudos [?]: 38 [0], given: 12

Intern
Joined: 17 Mar 2015
Posts: 1

Kudos [?]: 0 [0], given: 7

### Show Tags

19 Feb 2016, 06:00
I think this is a high-quality question and I agree with explanation. There is a typo. AX equals (√3*r)/2, not (√3*r)/4 as it mentioned in the second line below the picture in the explanation. The rest is correct.

Kudos [?]: 0 [0], given: 7

Manager
Joined: 31 Jan 2017
Posts: 57

Kudos [?]: 28 [0], given: 17

Location: India
Schools: ISB '19, IIMA , IIMB, IIMC , XLRI, IIM
GMAT 1: 680 Q49 V34
GPA: 4
WE: Project Management (Energy and Utilities)

### Show Tags

17 Feb 2017, 02:18
This is a great question.

Typo errors in the solution:

AX = √3 * r/2
AX and XC are same in length
Area of triangle ABC is 1/2 * (3√r) *(r/2) = (√3 * r^2 )/4
_________________

__________________________________
Kindly press "+1 Kudos" if the post helped

Kudos [?]: 28 [0], given: 17

Re: S98-20   [#permalink] 17 Feb 2017, 02:18
Display posts from previous: Sort by

# S98-20

Moderator: Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.