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In the figure, circle O has center O, diameter AB and a [#permalink]
11 Feb 2012, 14:21
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00:00
A
B
C
D
E
Difficulty:
95% (hard)
Question Stats:
50% (03:53) correct
50% (02:09) wrong based on 396 sessions
Attachment:
Perimeter.PNG [ 7.32 KiB | Viewed 19733 times ]
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.
Re: Perimeter of the shaded region. [#permalink]
11 Feb 2012, 14:49
7
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Expert's post
6
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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 19289 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 30-60-90 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: \(\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}\).
Re: In the figure, circle O has center O, diameter AB and a [#permalink]
01 Mar 2012, 19:40
I got a doubt. In 30-60-90 rule, side opp to 60 is sqrt3* side opp 30 isn't it ? So with that rule if angle CAB is 30-60-90 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.
Re: In the figure, circle O has center O, diameter AB and a [#permalink]
01 Mar 2012, 20:18
Expert's post
priyalr wrote:
I got a doubt. In 30-60-90 rule, side opp to 60 is sqrt3* side opp 30 isn't it ? So with that rule if angle CAB is 30-60-90 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: \(AC=\frac{10}{2}=5\) --> \(CB=5\sqrt{3}\) --> \(CB+BE=5\sqrt{3}+5\sqrt{3}=10\sqrt{3}\).
Re: In the figure, circle O has center O, diameter AB and a [#permalink]
01 Mar 2012, 23:29
priyalr wrote:
I got a doubt. In 30-60-90 rule, side opp to 60 is sqrt3* side opp 30 isn't it ? So with that rule if angle CAB is 30-60-90 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 ?
Re: Perimeter of the shaded region. [#permalink]
01 Mar 2012, 23:41
Bunuel wrote:
Now, we should find the lengths of BC and BE (notice that they are equal). Since <CBA=30 then triangle ACB is 30-60-90 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? _________________
Happy are those who dream dreams and are ready to pay the price to make them come true
I am still on all gmat forums. msg me if you want to ask me smth
Re: Perimeter of the shaded region. [#permalink]
02 Mar 2012, 02:42
1
This post received KUDOS
Expert's post
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 30-60-90 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 \(\frac{BC}{10}=\frac{\sqrt{3}}{2}\) --> \(BC=5\sqrt{3}\).
Re: Perimeter of the shaded region. [#permalink]
01 Jul 2012, 21:22
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 30-60-90 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: \(\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}\).
Re: Perimeter of the shaded region. [#permalink]
02 Jul 2012, 00:28
Expert's post
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 30-60-90 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: \(\frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}\).
Re: Perimeter of the shaded region. [#permalink]
27 Apr 2014, 07:09
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 30-60-90 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? _________________
Appreciate the efforts...KUDOS for all Don't let an extra chromosome get you down..
In the figure, circle O has center O, diameter AB and a [#permalink]
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 30-60-90 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!
In the figure, circle O has center O, diameter AB and a [#permalink]
13 Jul 2014, 00:35
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 CD||AB, 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 30-60-90 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) _________________
Regards, Bharat Bhushan Sunkara.
"You need to sacrifice what you are TODAY, for what you want to be TOMORROW!!"
Re: In the figure, circle O has center O, diameter AB and a [#permalink]
13 Jul 2014, 00:47
1
This post received KUDOS
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 30-60-90 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. _________________
Regards, Bharat Bhushan Sunkara.
"You need to sacrifice what you are TODAY, for what you want to be TOMORROW!!"
Re: In the figure, circle O has center O, diameter AB and a [#permalink]
14 Jul 2014, 01:41
1
This post received KUDOS
Expert's post
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 30-60-90 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.
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