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In the figure, circle O has center O, diameter AB and a [#permalink]
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11 Feb 2012, 14:21
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In the figure, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region? A. (5/3)pi + 5√3 B. (5/3) pi + 10√3 C. (10/3) pi + 5√3 D. (10/3) pi + 10√3 E. (10/3) pi + 20√3 I can think of few pointers such as The value of x will be 30 degrees because CD is parallel to diameter. Angle COA and AOE will each be 60 degrees as the central angle is double the inscribed angle. But I am still struggling to get through the question.
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enigma123 wrote: In the figure, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region? A. (5/3)pi + 5√3 B. (5/3) pi + 10√3 C. (10/3) pi + 5√3 D. (10/3) pi + 10√3 E. (10/3) pi + 20√3 Attachment:
Perimeter.PNG [ 7.32 KiB  Viewed 60696 times ]
Since CD is parallel to AB then <CBA=<BCD=30 > <CBE=60 > <COE=2*60=120 (according to the central angle theorem) > length of minor arc CE is \(\frac{120}{360}*2\pi{r}=\frac{10}{3}*\pi\); Now, we should find the lengths of BC and BE ( notice that they are equal). Since <CBA=30 then triangle ACB is 306090 right triangle (AB=diameter means that <C=90), thus \(\frac{BC}{AB}=\frac{\sqrt{3}}{2}\), (BC is opposite to 60 degrees so corresponds to \(\sqrt{3}\)) > \(\frac{BC}{10}=\frac{\sqrt{3}}{2}\) (AB=diameter=2r=10) > \(BC=5\sqrt{3}\); Thus the perimeter of the shaded region is (minor arc CE)+BC+BE: \(\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}\). Answer: D.
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Re: In the figure, circle O has center O, diameter AB and a [#permalink]
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01 Mar 2012, 19:40
I got a doubt. In 306090 rule, side opp to 60 is sqrt3* side opp 30 isn't it ? So with that rule if angle CAB is 306090 triangle, whr AB=5 as angle ACB is 90, then angle CBA is 30, so its length is AC= 5/2, and angle is CAB is 60 degree, CB= sqrt3* 5/2. So since CB & BE are of same length (acc to your explnt) then CB+BE = Sqrt3*5/2 + Sqrt3*5/2 = sqrt3*5 . hence, I get ans as C.
Pls explain.



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Re: In the figure, circle O has center O, diameter AB and a [#permalink]
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01 Mar 2012, 20:18
priyalr wrote: I got a doubt. In 306090 rule, side opp to 60 is sqrt3* side opp 30 isn't it ? So with that rule if angle CAB is 306090 triangle, whr AB=5 as angle ACB is 90, then angle CBA is 30, so its length is AC= 5/2, and angle is CAB is 60 degree, CB= sqrt3* 5/2. So since CB & BE are of same length (acc to your explnt) then CB+BE = Sqrt3*5/2 + Sqrt3*5/2 = sqrt3*5 . hence, I get ans as C.
Pls explain. The red parts are not correct. The point is that \(AB=diameter=2*radius=2*5=10\), not 5,. You've done everything right after this step but got different numerical values because of this one mistake. It should be: \(AC=\frac{10}{2}=5\) > \(CB=5\sqrt{3}\) > \(CB+BE=5\sqrt{3}+5\sqrt{3}=10\sqrt{3}\). Hope it's clear.
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Re: In the figure, circle O has center O, diameter AB and a [#permalink]
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01 Mar 2012, 23:29
priyalr wrote: I got a doubt. In 306090 rule, side opp to 60 is sqrt3* side opp 30 isn't it ? So with that rule if angle CAB is 306090 triangle, whr AB=5 as angle ACB is 90, then angle CBA is 30, so its length is AC= 5/2, and angle is CAB is 60 degree, CB= sqrt3* 5/2. So since CB & BE are of same length (acc to your explnt) then CB+BE = Sqrt3*5/2 + Sqrt3*5/2 = sqrt3*5 . hence, I get ans as C.
Pls explain. thnx a ton. My bad ! I want to give my exam by APril end. I have studied the quant notes for this specific topic and leter solve the PS q, for the given topic. The problem is I get stuck at the start or in middle of my solving. I seem to not apply the concepts learned on to the q. What shld i do ?



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Re: Perimeter of the shaded region. [#permalink]
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Bunuel wrote: Now, we should find the lengths of BC and BE (notice that they are equal). Since <CBA=30 then triangle ACB is 306090 right triangle (AB=diameter means that <C=90), thus \(\frac{BC}{AB}=\frac{\sqrt{3}}{2}\), (BC is opposite to 60 degrees so corresponds to \(\sqrt{3}\)) > \(\frac{BC}{10}=\frac{\sqrt{3}}{2}\) (AB=diameter=2r=10) > \(BC=5\sqrt{3}\);
I didnt get this part. could u please explain sqroot3/2, or give some links where I can read theory? btw, Bunuel, is there another way to solve this question?
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Re: Perimeter of the shaded region. [#permalink]
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02 Mar 2012, 02:42
LalaB wrote: Bunuel wrote: Now, we should find the lengths of BC and BE (notice that they are equal). Since <CBA=30 then triangle ACB is 306090 right triangle (AB=diameter means that <C=90), thus \(\frac{BC}{AB}=\frac{\sqrt{3}}{2}\), (BC is opposite to 60 degrees so corresponds to \(\sqrt{3}\)) > \(\frac{BC}{10}=\frac{\sqrt{3}}{2}\) (AB=diameter=2r=10) > \(BC=5\sqrt{3}\);
I didnt get this part. could u please explain sqroot3/2, or give some links where I can read theory? btw, Bunuel, is there another way to solve this question? Triangle ACB is a 30°60°90° right triangle. AC is opposite the smallest angle 30°, BC is opposite 60° angle, and hypotenuse AB is opposite the largest angle 90°. • A right triangle where the angles are 30°, 60°, and 90°.This is one of the 'standard' triangles you should be able recognize on sight. A fact you should commit to memory is: The sides are always in the ratio \(1 : \sqrt{3}: 2\). Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°). So, for our question BC (the side opposite 60° angle) and AB (the side opposite the largest angle 90°) must be in the ratio \(\sqrt{3}:2\) > \(\frac{BC}{AB}=\frac{\sqrt{3}}{2}\), and since AB=diameter=2r=10 then \(\frac{BC}{10}=\frac{\sqrt{3}}{2}\) > \(BC=5\sqrt{3}\). For more on this subject check Triangles chapter of Math Book: mathtriangles87197.htmlHope it helps.
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Oh, I forgot that ABS ia a right triangle. thats why I didnt get whats sqroot3/2. my fault, sorry.
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Re: Perimeter of the shaded region. [#permalink]
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Bunuel This is the only part that alvvays troubles me right triangle (AB=diameter means that <C=90) Hovv I knovv angle at center subtended by diameter is 90 but cant form it here please explain Bunuel wrote: enigma123 wrote: In the figure, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region? A. (5/3)pi + 5√3 B. (5/3) pi + 10√3 C. (10/3) pi + 5√3 D. (10/3) pi + 10√3 E. (10/3) pi + 20√3 Attachment: Perimeter.PNG Since CD is parallel to AB then <CBA=<BCD=30 > <CBE=60 > <COE=2*60=120 (according to the central angle theorem) > length of minor arc CE is \(\frac{120}{360}*2\pi{r}=\frac{10}{3}*\pi\); Now, we should find the lengths of BC and BE ( notice that they are equal). Since <CBA=30 then triangle ACB is 306090 right triangle (AB=diameter means that <C=90), thus \(\frac{BC}{AB}=\frac{\sqrt{3}}{2}\), (BC is opposite to 60 degrees so corresponds to \(\sqrt{3}\)) > \(\frac{BC}{10}=\frac{\sqrt{3}}{2}\) (AB=diameter=2r=10) > \(BC=5\sqrt{3}\); Thus the perimeter of the shaded region is (minor arc CE)+BC+BE: \(\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}\). Answer: D.



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Re: Perimeter of the shaded region. [#permalink]
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02 Jul 2012, 00:28
venmic wrote: Bunuel This is the only part that alvvays troubles me right triangle (AB=diameter means that <C=90) Hovv I knovv angle at center subtended by diameter is 90 but cant form it here please explain Bunuel wrote: enigma123 wrote: In the figure, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region? A. (5/3)pi + 5√3 B. (5/3) pi + 10√3 C. (10/3) pi + 5√3 D. (10/3) pi + 10√3 E. (10/3) pi + 20√3 Attachment: The attachment Perimeter.PNG is no longer available Since CD is parallel to AB then <CBA=<BCD=30 > <CBE=60 > <COE=2*60=120 (according to the central angle theorem) > length of minor arc CE is \(\frac{120}{360}*2\pi{r}=\frac{10}{3}*\pi\); Now, we should find the lengths of BC and BE ( notice that they are equal). Since <CBA=30 then triangle ACB is 306090 right triangle (AB=diameter means that <C=90), thus \(\frac{BC}{AB}=\frac{\sqrt{3}}{2}\), (BC is opposite to 60 degrees so corresponds to \(\sqrt{3}\)) > \(\frac{BC}{10}=\frac{\sqrt{3}}{2}\) (AB=diameter=2r=10) > \(BC=5\sqrt{3}\); Thus the perimeter of the shaded region is (minor arc CE)+BC+BE: \(\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}\). Answer: D. Look at the diagram below: Attachment:
Circletr.png [ 7.39 KiB  Viewed 47858 times ]
Hope it helps.
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Hi Bunuel, 1 doubt.. when you calculated length of the arc CE, why you used angle as 120 degree, why not 60 degree...??
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Bunuel wrote: ishdeep18 wrote: Hi Bunuel, 1 doubt.. when you calculated length of the arc CE, why you used angle as 120 degree, why not 60 degree...??
Thanks Arc CE is a part of the circumference > central angle COE is 120 degrees > a central angle in a circle determines an arc. For more check here: mathcircles87957.htmlHope this helps. Yes definitely, I was thinking why 60* s not considered , Now I am clear that central angle is considered while calculating the length of the arc.Thank You Bunuel..



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Bunuel wrote: enigma123 wrote: In the figure, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region? A. (5/3)pi + 5√3 B. (5/3) pi + 10√3 C. (10/3) pi + 5√3 D. (10/3) pi + 10√3 E. (10/3) pi + 20√3 Attachment: Perimeter.PNG Since CD is parallel to AB then <CBA=<BCD=30 > <CBE=60 > <COE=2*60=120 (according to the central angle theorem) > length of minor arc CE is \(\frac{120}{360}*2\pi{r}=\frac{10}{3}*\pi\); Now, we should find the lengths of BC and BE ( notice that they are equal). Since <CBA=30 then triangle ACB is 306090 right triangle (AB=diameter means that <C=90), thus \(\frac{BC}{AB}=\frac{\sqrt{3}}{2}\), (BC is opposite to 60 degrees so corresponds to \(\sqrt{3}\)) > \(\frac{BC}{10}=\frac{\sqrt{3}}{2}\) (AB=diameter=2r=10) > \(BC=5\sqrt{3}\); Thus the perimeter of the shaded region is (minor arc CE)+BC+BE: \(\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}\). Answer: D. TO calculate the length of the arc CAE...I considered B as the center of a larger circle with radius 5\sqrt{3}..and the arc substending an angle of 60 at the center..But I got a different answer...The legth of arc comes out to be 5\sqrt{3}/3 Dunno what did I do wrong here?
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Ok Got it..it will not be a circle but an ellipse..My bad..
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In the figure, circle O has center O, diameter AB and a [#permalink]
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12 Jul 2014, 16:52
Bunuel wrote: enigma123 wrote: In the figure, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region? A. (5/3)pi + 5√3 B. (5/3) pi + 10√3 C. (10/3) pi + 5√3 D. (10/3) pi + 10√3 E. (10/3) pi + 20√3 Since CD is parallel to AB then <CBA=<BCD=30 > <CBE=60 > <COE=2*60=120 (according to the central angle theorem) > length of minor arc CE is \(\frac{120}{360}*2\pi{r}=\frac{10}{3}*\pi\); Now, we should find the lengths of BC and BE (notice that they are equal). Since <CBA=30 then triangle ACB is 306090 right triangle (AB=diameter means that <C=90), thus \(\frac{BC}{AB}=\frac{\sqrt{3}}{2}\), (BC is opposite to 60 degrees so corresponds to \(\sqrt{3}\)) > \(\frac{BC}{10}=\frac{\sqrt{3}}{2}\) (AB=diameter=2r=10) > \(BC=5\sqrt{3}\); Thus the perimeter of the shaded region is (minor arc CE)+BC+BE: \(\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}\). Answer: D. Hi Bunuel, How did you find that BC and BE are equal? Thanks!



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In the figure, circle O has center O, diameter AB and a [#permalink]
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enigma123 wrote: Attachment: Perimeter.PNG In the figure, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region? A. (5/3)pi + 5√3 B. (5/3) pi + 10√3 C. (10/3) pi + 5√3 D. (10/3) pi + 10√3 E. (10/3) pi + 20√3 I can think of few pointers such as The value of x will be 30 degrees because CD is parallel to diameter. Angle COA and AOE will each be 60 degrees as the central angle is double the inscribed angle. But I am still struggling to get through the question. The correct Answer is D. The perimeter of the shaded region is the sum of lengths of CB + BE + arc EC. Now let us find out the individual lengths. 1. arc EC Length of an arc in a circle making an angle X with center O and radius 'r' is given by [X][/360]*2*pi*r In the given question we need to find out angle COE which is X in the above formula. Since CDAB, angle DCB = angle CBA = 30 Now consider the triangle BOC. This is an Isosceles triangle with OC = OB = 5 and thus the angles in the triangle are as follows: angle OBC = 30 (given), angle OCB = 30 (property of an Isosceles triangle) and angle BOC = 120 (180(30 +30)) Now consider the parallel lines CD and AB with CO as the transverse. angle DCO = 60 which is equal to angle COA i.e., 60. Now the total angle X = angle COE = angle COA + angle AOE = 120 Hence length of arc CE is [120][/360]*2*pi*5 = [10][/3]*pi 2. CB = BE Consider triangle ACB which is 306090 triangle where angle ACB = 90, angle ABC = 30 and angle CAB = 60. Now, Sine (angle CAB) = [opposite side][/Hypotenuse] i.e., Sine 60 = [BC][/AB] i.e., sqrt(3)/2 = BC/10 which gives BC as [10*sqrt(3)]/2 Now Perimeter of the shaded region = arc CE + (2 * CB) = [10][/3]*pi + 10*sqrt(3)
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Dienekes wrote: Bunuel wrote: enigma123 wrote: In the figure, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region?
A. (5/3)pi + 5√3 B. (5/3) pi + 10√3 C. (10/3) pi + 5√3 D. (10/3) pi + 10√3 E. (10/3) pi + 20√3 \
Since CD is parallel to AB then <CBA=<BCD=30 > <CBE=60 > <COE=2*60=120 (according to the central angle theorem) > length of minor arc CE is \(\frac{120}{360}*2\pi{r}=\frac{10}{3}*\pi\);
Now, we should find the lengths of BC and BE (notice that they are equal). Since <CBA=30 then triangle ACB is 306090 right triangle (AB=diameter means that <C=90), thus \(\frac{BC}{AB}=\frac{\sqrt{3}}{2}\), (BC is opposite to 60 degrees so corresponds to \(\sqrt{3}\)) > \(\frac{BC}{10}=\frac{\sqrt{3}}{2}\) (AB=diameter=2r=10) > \(BC=5\sqrt{3}\);
Thus the perimeter of the shaded region is (minor arc CE)+BC+BE: \(\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}\).
Answer: D. Hi Bunuel, How did you find that BC and BE are equal? Thanks! Hello Dienekes, Triangles ACB and AEB are Similar Triangles with a common side AB. As per the property of Similar Triangles, ratios of sides are equal, hence BC = BE. Hope this clarifies your doubt. Regards, Bharat.
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Re: In the figure, circle O has center O, diameter AB and a [#permalink]
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14 Jul 2014, 01:41
Dienekes wrote: Bunuel wrote: enigma123 wrote: In the figure, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region? A. (5/3)pi + 5√3 B. (5/3) pi + 10√3 C. (10/3) pi + 5√3 D. (10/3) pi + 10√3 E. (10/3) pi + 20√3 Since CD is parallel to AB then <CBA=<BCD=30 > <CBE=60 > <COE=2*60=120 (according to the central angle theorem) > length of minor arc CE is \(\frac{120}{360}*2\pi{r}=\frac{10}{3}*\pi\); Now, we should find the lengths of BC and BE (notice that they are equal). Since <CBA=30 then triangle ACB is 306090 right triangle (AB=diameter means that <C=90), thus \(\frac{BC}{AB}=\frac{\sqrt{3}}{2}\), (BC is opposite to 60 degrees so corresponds to \(\sqrt{3}\)) > \(\frac{BC}{10}=\frac{\sqrt{3}}{2}\) (AB=diameter=2r=10) > \(BC=5\sqrt{3}\); Thus the perimeter of the shaded region is (minor arc CE)+BC+BE: \(\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}\). Answer: D. Hi Bunuel, How did you find that BC and BE are equal? Thanks! BC and BE are deviated by the same degree from the diameter AB, thus they are mirror images of each other around AB, therefore they must be equal. Does this make sense?
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Re: In the figure, circle O has center O, diameter AB and a
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14 Jul 2014, 01:41



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