Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 01 Jun 2014
Posts: 95
Concentration: Entrepreneurship, Operations

Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
03 May 2015, 21:23
I must be missing something. Can anyone point me in the right direction to read about parallel lines in Circles? i don't understand why the angles are equal. Thanks!



Math Expert
Joined: 02 Sep 2009
Posts: 59561

Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
04 May 2015, 02:28
itsmedavidv wrote: I must be missing something. Can anyone point me in the right direction to read about parallel lines in Circles? i don't understand why the angles are equal. Thanks! Check Coordinate Geometry and Circles chapters of our Math Book. Hope it helps.



Intern
Joined: 24 Jun 2015
Posts: 45

Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
05 Jul 2015, 13:45
Bunuel wrote: 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. Hi Bunuel, Could you help me with this?: The right triangle with hypotenuse 10 is not supposed to be a multiple of the (3  4  5) right triangle??? Then if it were a multiple of this (3  4  5) it would be (6  8 10); then CB would be 8 instead of 5√3.... I am confused why this results are different. I was thinking that whatever hypotenuse of 10 with 30  60  90 right triangle had to match with the 6  8  10 shortcut... Could you help me? Thanks a lot. Regards. Luis Navarro Looking for 700



CEO
Joined: 20 Mar 2014
Posts: 2561
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
05 Jul 2015, 14:01
luisnavarro wrote: Hi Bunuel,
Could you help me with this?:
The right triangle with hypotenuse 10 is not supposed to be a multiple of the (3  4  5) right triangle??? Then if it were a multiple of this (3  4  5) it would be (6  8 10); then CB would be 8 instead of 5√3.... I am confused why this results are different. I was thinking that whatever hypotenuse of 10 with 30  60  90 right triangle had to match with the 6  8  10 shortcut... Could you help me?
Thanks a lot.
Regards.
Luis Navarro Looking for 700
You are not looking at this correctly. \(1:\sqrt{3}:2\) is only applicable for triangles having angles as \(30^{\circ},60^{\circ}, 90^{\circ}\) 2nd 'special' type of right triangles are the \(45^{\circ},45^{\circ} , 90^{\circ}\) , in which the ratio of the sides becomes \(1:1:\sqrt{2}\) Right triangles that do not fall under the above 2 cases should not be mixed with the above rules. These triangles have sides that vary with the relative angles inside the triangle. Triangles with sides 3:4:5 or any multiples of this such as 6:8:10 or 9:12:15 etc do not come under the 2 rules mentioned above and thus you will not get the same proportions for the sides. Hope this clears your doubt



Intern
Joined: 24 Jun 2015
Posts: 45

Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
05 Jul 2015, 14:51
Engr2012 wrote: luisnavarro wrote: Hi Bunuel,
Could you help me with this?:
The right triangle with hypotenuse 10 is not supposed to be a multiple of the (3  4  5) right triangle??? Then if it were a multiple of this (3  4  5) it would be (6  8 10); then CB would be 8 instead of 5√3.... I am confused why this results are different. I was thinking that whatever hypotenuse of 10 with 30  60  90 right triangle had to match with the 6  8  10 shortcut... Could you help me?
Thanks a lot.
Regards.
Luis Navarro Looking for 700
You are not looking at this correctly. \(1:\sqrt{3}:2\) is only applicable for triangles having angles as \(30^{\circ},60^{\circ}, 90^{\circ}\) 2nd 'special' type of right triangles are the \(45^{\circ},45^{\circ} , 90^{\circ}\) , in which the ratio of the sides becomes \(1:1:\sqrt{2}\) Right triangles that do not fall under the above 2 cases should not be mixed with the above rules. These triangles have sides that vary with the relative angles inside the triangle. Triangles with sides 3:4:5 or any multiples of this such as 6:8:10 or 9:12:15 etc do not come under the 2 rules mentioned above and thus you will not get the same proportions for the sides. Hope this clears your doubt Thanks a lot... That means that 3  4  5 right triangle is not a 30  60  90 right triangle?



CEO
Joined: 20 Mar 2014
Posts: 2561
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
05 Jul 2015, 15:01
luisnavarro wrote: Engr2012 wrote: luisnavarro wrote: Hi Bunuel,
Could you help me with this?:
The right triangle with hypotenuse 10 is not supposed to be a multiple of the (3  4  5) right triangle??? Then if it were a multiple of this (3  4  5) it would be (6  8 10); then CB would be 8 instead of 5√3.... I am confused why this results are different. I was thinking that whatever hypotenuse of 10 with 30  60  90 right triangle had to match with the 6  8  10 shortcut... Could you help me?
Thanks a lot.
Regards.
Luis Navarro Looking for 700
You are not looking at this correctly. \(1:\sqrt{3}:2\) is only applicable for triangles having angles as \(30^{\circ},60^{\circ}, 90^{\circ}\) 2nd 'special' type of right triangles are the \(45^{\circ},45^{\circ} , 90^{\circ}\) , in which the ratio of the sides becomes \(1:1:\sqrt{2}\) Right triangles that do not fall under the above 2 cases should not be mixed with the above rules. These triangles have sides that vary with the relative angles inside the triangle. Triangles with sides 3:4:5 or any multiples of this such as 6:8:10 or 9:12:15 etc do not come under the 2 rules mentioned above and thus you will not get the same proportions for the sides. Hope this clears your doubt Thanks a lot... That means that 3  4  5 right triangle is not a 30  60  90 right triangle? Yes, exactly. The 345 triangle or a triangle with any of the other multiples of 345 is not a 306090 triangle. FYI, 345 triangle has angles as \(36.87^{\circ}, 53.13^{\circ},90^{\circ}\) triangle ( For GMAT, you do not have to remember angles that are not 0, 30, 45 60 or 90). It actually follows from the fact that 306090 triangle will always have sides in the ratio: \(1:\sqrt{3}:2\). The opposite is true as well that if a triangle has sides in the ration \(1:\sqrt{3}:2\), the triangle will then be 306090. The same logic applies to 454590 triangle as well. Hope this clears your doubt.



Intern
Joined: 09 Apr 2015
Posts: 5

Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
21 Jul 2015, 11:38
Please help me where I am going wrong with this approach.
I got the length of BC =5[square_root]3
Then for calculating the length of the arc I used the angle CBE = 60 with BC and BE.
The length of the arc would be 1/6*2*pi*5[square_root]3
The perimeter = 10[square_root]3 + 1/6*2*pi*5[square_root]3



CEO
Joined: 20 Mar 2014
Posts: 2561
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
21 Jul 2015, 11:49
varunvarma401 wrote: Please help me where I am going wrong with this approach.
I got the length of BC =5[square_root]3
Then for calculating the length of the arc I used the angle CBE = 60 with BC and BE.
The length of the arc would be 1/6*2*pi*5[square_root]3
The perimeter = 10[square_root]3 + 1/6*2*pi*5[square_root]3 For calculating the length of the arc, the angle MUST be the center angle (or the angle that the arc makes at the center and NOT at the circumference). Any angle made by an arc on the circumference = 0.5* Angle made by the SAME arc at the center. Thus, instead of 60 degrees, you need to calculate the length of arc made by 120 degrees. You will correct answer after that. (1/6) above will become (1/3). Also the entire red equation above is incorrect. It should be, \(\frac{120}{360} * 2\pi*5 = \frac{10\pi}{3}\)



Intern
Joined: 09 Apr 2015
Posts: 5

Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
21 Jul 2015, 12:03
Thank you for the explanation Engr2012.
But if we consider a circle with the radius of BC and the sector with angle 60, then the arc length could be calculated right?



CEO
Joined: 20 Mar 2014
Posts: 2561
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
21 Jul 2015, 12:13
varunvarma401 wrote: Thank you for the explanation Engr2012.
But if we consider a circle with the radius of BC and the sector with angle 60, then the arc length could be calculated right? Look at the attached picture Yes, if you are given that "x" = 60 (in the attached picture), radius = r , then length of the arc =\(l = r*\theta\). where \(\theta\)= angle in RADIANS and not degrees that is made by the arc at the CENTER. Thus \(\theta\) in this case will be 2*60 = 120 degrees For degree to radian conversion, use the fact that 180 degrees = \(\pi\) radians. Thus, for the example you have quoted, \(l = r*[(120/180)*\pi] = r*2*\pi/3\)
Attachments
image.jpg [ 22.55 KiB  Viewed 2468 times ]



Intern
Joined: 26 May 2014
Posts: 39
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
01 Feb 2016, 05:50
Why did we not add length of Diameter AB in the perimeter of the shaded region?



Math Expert
Joined: 02 Sep 2009
Posts: 59561

Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
01 Feb 2016, 05:57
devbond wrote: Why did we not add length of Diameter AB in the perimeter of the shaded region? Perimeter is the distance around a figure, or the measurement of the distance around something; the length of the boundary. Since AB is NOT the boundary of the shaded region (it lies within it) we do not add its length when calculating the perimeter. Hope it's clear.



CEO
Joined: 20 Mar 2014
Posts: 2561
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
01 Feb 2016, 06:29
devbond wrote: Why did we not add length of Diameter AB in the perimeter of the shaded region? Due to the fact that in order to calculate perimeter of any area, you only need to consider OUTER bounds of the said area. What is inside, does not matter. Thus the perimeter for the question asked will be = CB+BE+ minor arc(EAC) Hope this helps.



Manager
Joined: 29 Dec 2014
Posts: 63

Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
22 Oct 2016, 20:27
Hi, if we were to find the area of the shaded region, what would the solution be.
Thanks



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9848
Location: Pune, India

Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
07 Apr 2017, 02:53
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. CD and AB are parallel so x = 30 degrees since they are alternate interior angles. Angle CBE is 2*30 = 60 degrees. This means arc CE subtends a central angle of 120 degrees (Central angle is twice the angle subtended at the circumference). So arc CE is one third of the circumference of the circle. Circumference of the circle \(= 2 *\pi * r = 10 * \pi\) Length of arc CE \(= (10/3) * \pi\) Since angle CBE is 60 degrees, triangle CBE is equilateral. As discussed in the post given below: https://www.veritasprep.com/blog/2013/0 ... relations/The ratio of side of the equilateral triangle inscribed in a circle and the radius of the circle \(= a : r = \sqrt{3}:1\) So \(BC = BE = 5*\sqrt{3}\) \(Perimeter = (10/3) * \pi + 10*\sqrt{3}\) Answer (D)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4121
Location: Canada

Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
28 Apr 2018, 10:08
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. \((\frac{5}{3})\pi + 5\sqrt{3}\) B. \((\frac{5}{3})\pi + 10\sqrt{√}3\) C. \((\frac{10}{3})\pi + 5\sqrt{3}\) D. \((\frac{10}{3})\pi + 10\sqrt{3}\) E. \((\frac{10}{3})\pi + 20\sqrt{3}\) IMPORTANT: unless stated otherwise, the diagrams in Problem Solving geometry questions are DRAWN TO SCALE. I'll solve this question using estimation. Since the diameter AB = 10, we can ESTIMATE the length of CB. It looks like CB is just a little bit shorter than AB. So, I'll say that the length of side CB is approximately 9. This means the length of side EB is approximately 9 as well. Finally, arc EC looks a little bit shorter than sides CB and EB, so I'll estimate it to be length 8 So, the TOTAL perimeter = 9 + 9 + 8 = 26 Now check the answer choices: ASIDE: On test day, everyone should know the following apprximations: √2 ≈ 1.4 √3 ≈ 1.7 √5 ≈ 2.2 Also, we'll say that pi ≈ 3A. (5/3)pi + 5√3 ≈ 5 + 8.5 ≈ 13.5 B. (5/3)pi + 10√3 ≈ 5 + 17 ≈ 22 C. (10/3)pi + 5√3 ≈ 10 + 8.5 ≈ 18.5 D. (10/3)pi + 10√3 ≈ 10 + 17 ≈ 27 E. (10/3)pi + 20√3 ≈ 10 + 34 ≈ 44 Of these, it appears that D is the closest. Aside: We can see that answer choice B is pretty close too. At this point, you have a timemanagement decision. You can either stick with D, and use your extra time elsewhere, or your can spend time trying to be more certain of the answer. Your choice. That said, D is the correct answer. Cheers, Brent
_________________
Test confidently with gmatprepnow.com



Intern
Joined: 14 Jul 2016
Posts: 7

Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
07 Oct 2018, 04:25
<x = alt. <BCD = 30 degrees. <CBE = 60 degree
O be the center and the angle subtended at the center = <COE = 120 degree
Length of the arc(CAE) = (120/360)*pi*5^2
Now considering triangle OCB, OB = OC = 5 units. <COB = 120 degree 5/sin30 = BC/sin120 BC = 5sqrt3
Total perimeter = (120/360)*pi*5^2 + 10sqrt3 (D)



Manager
Joined: 15 Nov 2017
Posts: 52

Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
18 Jan 2019, 14:40
Could someone provide insight about what this would look like drawn out and labeled? I am trying to grasp it. Thank you so much! 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. \((\frac{5}{3})\pi + 5\sqrt{3}\) B. \((\frac{5}{3})\pi + 10\sqrt{√}3\) C. \((\frac{10}{3})\pi + 5\sqrt{3}\) D. \((\frac{10}{3})\pi + 10\sqrt{3}\) E. \((\frac{10}{3})\pi + 20\sqrt{3}\) Attachment: Perimeter.PNG 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.



Intern
Joined: 19 Dec 2017
Posts: 8
GPA: 3

In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
19 Jan 2019, 01:27
Hello Bunuel, I imagined that point B is the centre of another circle say P. now we know the radius of the circle as BA; now BC = BE = 10 From the figure, we know that x = 30 we have a radius and an angle, we can find arc length. the answer should be (10pi/3) + 20. please clarify where am I missing. Thank you.



Math Expert
Joined: 02 Sep 2009
Posts: 59561

Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
Show Tags
19 Jan 2019, 01:44
Bala0801 wrote: Hello Bunuel, I imagined that point B is the centre of another circle say P. now we know the radius of the circle as BA; now BC = BE = 10 From the figure, we know that x = 30 we have a radius and an angle, we can find arc length. the answer should be (10pi/3) + 20. please clarify where am I missing. Thank you. I think the figure below might help: Attachment:
Untitled.png [ 10.28 KiB  Viewed 417 times ]
So, if B is the center and BA is the radius, then you'd get above circle. For that you can see why your further logic is not correct.




Re: In the figure, circle O has center O, diameter AB and a radius of 5.
[#permalink]
19 Jan 2019, 01:44



Go to page
Previous
1 2 3
Next
[ 41 posts ]



