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# In the figure, circle O has center O, diameter AB and a

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Current Student
Joined: 12 Feb 2011
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Re: In the figure, circle O has center O, diameter AB and a [#permalink]

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15 Aug 2014, 16:11
Bunuel wrote:

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?

Yes, it does. Thanks!
Manager
Joined: 01 Jun 2014
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Re: In the figure, circle O has center O, diameter AB and a [#permalink]

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03 May 2015, 20:23
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
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Posts: 43867
Re: In the figure, circle O has center O, diameter AB and a [#permalink]

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04 May 2015, 01:28
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.
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Re: In the figure, circle O has center O, diameter AB and a [#permalink]

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05 Jul 2015, 12:45
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 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 $$\frac{BC}{10}=\frac{\sqrt{3}}{2}$$ --> $$BC=5\sqrt{3}$$.

For more on this subject check Triangles chapter of Math Book: math-triangles-87197.html

Hope 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
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In the figure, circle O has center O, diameter AB and a [#permalink]

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05 Jul 2015, 13:01
1
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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: 46
Re: In the figure, circle O has center O, diameter AB and a [#permalink]

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05 Jul 2015, 13:51
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?
Current Student
Joined: 20 Mar 2014
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In the figure, circle O has center O, diameter AB and a [#permalink]

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05 Jul 2015, 14:01
1
KUDOS
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 3-4-5 triangle or a triangle with any of the other multiples of 3-4-5 is not a 30-60-90 triangle.

FYI, 3-4-5 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 30-60-90 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 30-60-90. The same logic applies to 45-45-90 triangle as well.

Hope this clears your doubt.
Intern
Joined: 24 Jun 2015
Posts: 46
Re: In the figure, circle O has center O, diameter AB and a [#permalink]

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05 Jul 2015, 14:14
Thanks a lot Engr2012, now it is clear to me.

Regards.

Luis Navarro
Looking for 700
Intern
Joined: 09 Apr 2015
Posts: 7
In the figure, circle O has center O, diameter AB and a [#permalink]

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21 Jul 2015, 10:38
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
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In the figure, circle O has center O, diameter AB and a [#permalink]

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21 Jul 2015, 10:49
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
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Posts: 7
Re: In the figure, circle O has center O, diameter AB and a [#permalink]

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21 Jul 2015, 11:03
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?
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In the figure, circle O has center O, diameter AB and a [#permalink]

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21 Jul 2015, 11:13
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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, $$l = r*[(120/180)*\pi] = r*2*\pi/3$$
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Re: In the figure, circle O has center O, diameter AB and a [#permalink]

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01 Feb 2016, 04:50
Why did we not add length of Diameter AB in the perimeter of the shaded region?
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Posts: 43867
Re: In the figure, circle O has center O, diameter AB and a [#permalink]

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01 Feb 2016, 04:57
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.
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Re: In the figure, circle O has center O, diameter AB and a [#permalink]

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01 Feb 2016, 05:29
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.
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Re: In the figure, circle O has center O, diameter AB and a [#permalink]

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22 Oct 2016, 19:27
Hi, if we were to find the area of the shaded region, what would the solution be.

Thanks
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Re: In the figure, circle O has center O, diameter AB and a [#permalink]

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07 Apr 2017, 01:53
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Expert's post
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 $$BC = BE = 5*\sqrt{3}$$

$$Perimeter = (10/3) * \pi + 10*\sqrt{3}$$

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Re: In the figure, circle O has center O, diameter AB and a   [#permalink] 07 Apr 2017, 01:53

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# In the figure, circle O has center O, diameter AB and a

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