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Re S9820
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16 Sep 2014, 00:53
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 306090 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 \(\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43\) The closest answer is 42%. Answer: D
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Re: S9820
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10 Oct 2014, 15:25
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 306090 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 \(\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43\) The closest answer is 42%.
Answer: D Bunuel, I am confused at the step in which you derive the height side of the first 306090 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 306090 right triangle. You must've meant to say that AX = AX, thus we know AC?



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Re: S9820
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11 Oct 2014, 12:28
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 306090 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 \(\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43\) The closest answer is 42%.
Answer: D Bunuel, I am confused at the step in which you derive the height side of the first 306090 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 306090 right triangle. You must've meant to say that AX = AX, thus we know AC? The following discussion might help: approximatelywhatpercentoftheareaofthecircleshowni169744.html
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New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



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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 306090 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 \(\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43\) The closest answer is 42%.
Answer: D Bunuel, I am confused at the step in which you derive the height side of the first 306090 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 306090 right triangle. You must've meant to say that AX = AX, thus we know AC? The following discussion might help: approximatelywhatpercentoftheareaofthecircleshowni169744.htmlthat 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.



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Re: S9820
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23 Dec 2014, 23:42
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 306090 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 \(\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43\) The closest answer is 42%. Answer: D Hey Bunuel, I think the highlighted part above should be ax=xc. and that is why it is creating a bit of confusion.



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Re S9820
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19 Feb 2016, 05:00
I think this is a highquality 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.



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Re: S9820
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17 Feb 2017, 01:18
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
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Re: S9820
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04 Jan 2018, 01:59
I think we don't have to actually calculate and waste time in exam. Rather, we can simply see that if the area of triangle is "x" the highlighted area, when wrapped inside, is also "x". Hence, the total area becomes (2*Triangle area+Remaining area). Now we see clearly that the remaining area is a smaller area as compared to the area of triangle. Hence, Area of triangle can not be less than 33% (Option A&B goes out). At the same time, the area of Triangle can not be 50% or very close to 50%, since that would mean that remaining area is hardly 2% (Option E goes out). Now, If we choose 36% that would mean that remaining area is nearly equal to area of triangle, which is clearly not the case (Option C Goes out). The only remaining and sensible choice that remains is Option D.










