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Approximately what percent of the area of the circle shown i
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03 Apr 2014, 12:27
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55% (02:26) correct 45% (02:35) wrong based on 193 sessions
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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%
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Re: Approximately what percent of the area of the circle shown i
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05 Apr 2014, 04:55
I checked this with my brother, who is a high school student. He did it in 1 min  Here is his solution (though his method relies on memorising formulae, still i think thats the basic necessity) Circle radius = R So, Area of circle\(A_1 = \pi R^2\) Radius of circumcircle = Length of Sides (for a regular hexagon) So, Area of regular hexagon \(A_2 = \frac{3 \sqrt{3} R^2}{2}\) (6 times area of a equilateral triangle with side= R) ACE is an equilateral triangle (you are supposed to know that too) and Circumradius of equilateral triangle with side 'a' , \(R = \frac{a}{sqrt{3}}\) Hence, \(a = \sqrt{3}R\) Area of equilateral triangle ACE \(A_3 = \frac{sqrt{3} ( sqrt{3} R )^2}{4} = \frac{3 \sqrt{3} R^2}{4}\) Now % of shaded region = \(\frac{A_2  A_3}{A_1}\) in two steps, the fraction will reduce to :\(\frac{3sqrt{3}}{4 \pi}\), then you can approximate as posted by mike or samir.
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Re: Approximately what percent of the area of the circle shown i
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03 Apr 2014, 13:15
ConnectTheDots wrote: 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%
Let's look for a shortcut way. Dear ConnectTheDots, I'm happy to help. The sides of the regular hexagon inscribed in a circle are equal to the radius of the circle. You can easily see this if you divide the hexagon into six congruent equilateral triangles  then each triangle length equals a hexagon side, and two of those triangle lengths are the diameter of the circle. Look at triangle ABC. Sides AB = BC = r. The angle at B is 120 degrees, so the smaller angles are 30 degrees. Consider the midpoint of AC  call that point P. Then segment BP divides triangle ABC into two congruent 306090 triangles, each with hypotenuse = r. In one of those 306090 triangles: hypotenuse = r short leg = r/2 long leg = (r*sqrt(3))/2 We could put two of those triangles together to make a rectangle with sides of (short leg) and (long leg), so the area of this rectangle would equal the area of triangle ABC. That area is A = [(r^2)*sqrt(3)]/4 The entire diagram has three triangles that side in the shaded region, so the area of the shaded region is: A = [(r^2)*3sqrt(3)]/4 That's the easy part. Now, we divide this by Archimedes' magic formula, A = (pi)(r^2), and we get a ratio of [3sqrt(3)]/4 to (pi), and we want to approximate that as a percent without a calculator. Now, MGMAT approximates (pi) as 3, in which case this ratio becomes sqrt(3)/4. It helps to know that sqrt(3) is approx. 1.73, so sqrt(3)/4 is approx 43%. I think this is at the far end of the kind of estimating that the GMAT would expect you to do. I hope this helps. Mike
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Re: Approximately what percent of the area of the circle shown i
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04 Apr 2014, 04:35
Hopefully a shorter approach . Draw all the three diagonals of the hexagon, The diagonals intersect at 'O' , the center of the circle. And this divides hexagon into six equilateral triangles each with side = radius of the circle. Now the triangle AEF and AEO are congruent (using SSS) <one side is common and other two sides of both the triangles are equal to the radius of the circle>. Similarly the other two pairs of triangles are congruent. Sum of the area of all the shaded triangles is now equal to area of the triangle AEC, which is an equilateral triangle <easy to deduce>. Now all we need is the ratio of the area of AEC to the area of circle. Now, we need the length of side of the triangle AEC. Here we can use the property of centroid (point of intersection of the medians of a triangle) of a triangle to calculate the 'height' of the triangle AEC. Property : The centroid of a triangle divides the median in the ratio 2:1. <distance of centroid from the vertex is twice the distance of centroid from side>. Using this property, height of the triangle = 1.5 r , where r is the radius of the circle. We can use the formula : height of an equilateral triangle = \(\frac{sqrt3*a}{2}\) to calculate the side(a) of the equilateral triangle. \(\frac{sqrt3*a}{2} = 1.5 r\) \(a = sqrt3 * r\) We now use formula for area of equilateral triangle : \(Area = \frac{sqrt3*a^2}{4}\) Required ratio = \(\frac{sqrt3 * a^2}{4} / \pi * r^2\) Substitute \(a = sqrt3 * r\), the ratio reduces to \(\frac{sqrt3 * 3 * r^2}{4} / \pi * r^2\) Cancel \(r^2\) from numerator as well as denominator and approximate \(\pi\) as 3 Hence the required ratio now becomes \(\frac{sqrt3}{4}\). We know the value of \(sqrt3=1.71\)(approx). At this point , we can safely mark 42% as answer. Please excuse for formatting errors , I am correcting it.
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Re: Approximately what percent of the area of the circle shown i
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02 Sep 2014, 08:22
ConnectTheDots wrote: I checked this with my brother, who is a high school student. He did it in 1 min  Here is his solution (though his method relies on memorising formulae, still i think thats the basic necessity) Circle radius = R So, Area of circle\(A_1 = \pi R^2\) Radius of circumcircle = Length of Sides (for a regular hexagon) So, Area of regular hexagon \(A_2 = \frac{3 \sqrt{3} R^2}{2}\) (6 times area of a equilateral triangle with side= R) ACE is an equilateral triangle (you are supposed to know that too) and Circumradius of equilateral triangle with side 'a' , \(R = \frac{a}{sqrt{3}}\) Hence, \(a = \sqrt{3}R\) Area of equilateral triangle ACE \(A_3 = \frac{sqrt{3} ( sqrt{3} R )^2}{4} = \frac{3 \sqrt{3} R^2}{4}\) Now % of shaded region = \(\frac{A_2  A_3}{A_1}\) in two steps, the fraction will reduce to :\(\frac{3sqrt{3}}{4 \pi}\), then you can approximate as posted by mike or samir. HI I have one query here. How we can say Triangle ACE is an equilateral triangle. Could you please clarify this. Thanks.



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Re: Approximately what percent of the area of the circle shown i
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02 Sep 2014, 10:16
PathFinder007 wrote: HI
I have one query here.
How we can say Triangle ACE is an equilateral triangle. Could you please clarify this.
Thanks. Dear PathFinder007, I'm happy to help. One absolutely crucial piece of information in the prompt is the following: polygon ABCDEF is a regular hexagon. This word, " regular," is a funny word. In colloquial speech, it means "ordinary, commonplace." In geometry, it means exactly the opposite  it means the most elite, most symmetrical possible shape if its kind. You see, anything with six sides is a hexagon. A hexagon can be completely misshapen and asymmetrical, as long as it has six sides. A regular hexagon has six equal sides, and six equal angles. When we inscribe it into a circle, we know that the six points are equally spaced around the circumference and divide the circumference in to six equal arcs. That means that arc ABC, arc DCE, and arc EFA are also equal, which makes points A & C & E equally spaced around the circle, which means the three sides of triangle ACE are congruent, which makes it, by definition, an equilateral triangle. For more on the geometry of regular shapes, see: http://magoosh.com/gmat/2012/polygonsa ... thegmat/Does all this make sense? Mike
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Re: Approximately what percent of the area of the circle shown i
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18 Apr 2018, 22:12
mikemcgarry wrote: ConnectTheDots wrote: 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%
Let's look for a shortcut way. Dear ConnectTheDots, I'm happy to help. The sides of the regular hexagon inscribed in a circle are equal to the radius of the circle. You can easily see this if you divide the hexagon into six congruent equilateral triangles  then each triangle length equals a hexagon side, and two of those triangle lengths are the diameter of the circle. Look at triangle ABC. Sides AB = BC = r. The angle at B is 120 degrees, so the smaller angles are 30 degrees. Consider the midpoint of AC  call that point P. Then segment BP divides triangle ABC into two congruent 306090 triangles, each with hypotenuse = r. In one of those 306090 triangles: hypotenuse = r short leg = r/2 long leg = (r*sqrt(3))/2 We could put two of those triangles together to make a rectangle with sides of (short leg) and (long leg), so the area of this rectangle would equal the area of triangle ABC. That area is A = [(r^2)*sqrt(3)]/4 The entire diagram has three triangles that side in the shaded region, so the area of the shaded region is: A = [(r^2)*3sqrt(3)]/4 That's the easy part. Now, we divide this by Archimedes' magic formula, A = (pi)(r^2), and we get a ratio of [3sqrt(3)]/4 to (pi), and we want to approximate that as a percent without a calculator. Now, MGMAT approximates (pi) as 3, in which case this ratio becomes sqrt(3)/4. It helps to know that sqrt(3) is approx. 1.73, so sqrt(3)/4 is approx 43%. I think this is at the far end of the kind of estimating that the GMAT would expect you to do. I hope this helps. Mike OA : D I will try to do it in sligthly different way than mikemcgarry. Quote: The sides of the regular hexagon inscribed in a circle are equal to the radius of the circle. You can easily see this if you divide the hexagon into six congruent equilateral triangles  then each triangle length equals a hexagon side, and two of those triangle lengths are the diameter of the circle. Look at triangle ABC. Sides AB = BC = r. The angle at B is 120 degrees, We can find area of triangle ABC by using = \(\frac{1}{2}*AB*BC* sin(Angle at B)\) =\(\frac{{1*r^2*sin120^{\circ}}}{2}\) =\(\frac{{1*r^2*cos30^{\circ}}}{2}\) ( As \(sin(90^{\circ}+A)= cosA\)) =\(\frac{r^2}{2}*\frac{\sqrt{3}}{2}\) (As \(cos30^{\circ}=\frac{\sqrt{3}}{2}\)) Total shaded area =\(3 *\frac{r^2}{2}*\frac{\sqrt{3}}{2}\)(As there are 3 shaded triangle) Area of circle= \(\pi*r^2\) Ratio of total shaded area to circle: \(3*\frac{\sqrt{3}}{4\pi}\) taking \(\pi≈3\), ratio will come out to be \(.433\)(Approx) Ratio multiplied by 100 would be about \(43\)%(Approx) Option D is closest to \(43\)%, Hence OA=D
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