A circle has center at origin and radius 1. The points X, Y, and Z lie

Question

A circle has center at origin and radius 1. The points X, Y, and Z lie on the circle such that the length of arc XYZ is (2/3)π.What is the length of line segment XZ?

A. √2 
B. √3
C. 2 
D. 2+√3 
E. 2+√2

I figured we could eliminate c,d,e based on the rule that for a triangle,the 3rd side must be greater than difference of other 2 sides & lesser than the sum of other two sides.I am stuck after that. is there an neater method?

EvaJager · Accepted Answer

If the arc XYZ is  2/3\pi , the central angle XOZ is  120^o  (2/3*360=120). 
Then you can calculate XZ from the isosceles triangle XOZ which has angles 30-30-120, and XO = OZ = 1 = radius.
The height to the base of the triangle is 0.5 (opposing the  30^o  angle), so half of the base is  0.5*\sqrt{3}{ , and  XZ=\sqrt{3}.

Answer B.

dcummins · Answer

Noticed there wasn't a diagram, so refer to mind for conceptual understanding. It doesn't matter the quadrants as the distance will always be positive.

Step 1 - determine the angle formed by the arc using the arc length to circumference ratio
2/3*pi / 2pi*1 = 1/3
1/3*360deg = 120 deg

Step 2 - determine the length of XZ by using the properties of a 30-60-90 triangle formed by the perpendicular altitude of sector XYZ

x:x\sqrt{3}:2x

2x=1
x=1/2
Thus  x\sqrt{3}= \sqrt{3}/2

Multiply by 2 to get the base of the triangle and thus distance.

GMATBaumgartner · Answer

Thanks Eva. 
I have a clarification here: I am not sure how you got the length of the perpendicular to side XZ as 0.5.Is it by trigonometry(sin 30 ?) ?
Also how do you say the perpendicular from the center divides XZ equally ?

EvaJager · Answer

Yes, is "hidden" trigonometry... In a right triangle of type 30-60-90 the leg opposed to the 30 degrees angle is half of the hypotenuse.

Good to remember: in a right triangle of type  30-60-90  the sides are  x-x\sqrt{3}-2x .

GMATBaumgartner · Answer

Gotcha! i always knew this but din't strike me here :shock:

+1 EvaJ and you hit the half century kudos mark 8-)

fameatop · Answer

Hi Eva,

I would like to add few intermediate steps to the solution offered by you. 
As Eva has correctly pointed that the central angle XOZ is  120^o

Now drop a perpendicular from point X & extend the line OZ in the direction of that perpendicular & name that meeting point P
If you draw such triangle then you will that triangle XPO is 30-60-90 triangle. 
Given XO = radius =1
We want to the height of the triangle XPO which XP & opposite to angle 60 degree.
Applying 30-60-90 theorem , we get XP = \sqrt{3}/2 and this height is the height of the original triangle XOZ

Property 1 - The Altitude to the base of an isosceles or equilateral triangle bisects the base & bisects the vertex angle. 
 Property 2 - The midpoint of the hypotenuse is equidistant from the three polygon vertices

Because the height (altitude) bisects the side XZ, thus is equidistant from X,O or Z
Therefore XZ = 2 x height = 2 \sqrt{3}/2 = \sqrt{3}

Answer B
I hope this extra long explanation will help many & sorry that i didn't include the possible diagram.

smallurgo · Answer

Sorry EvaJager, in fact, I agree with the result that the central angle XOZ is 120 degree. However, I am afraid that the equation 2/3 * 360 = 120 is not correct (in fact 240). The result of 120 degree of central angle XOZ can be explained as follows:
The arc XYZ is 2/3pi . The circumference of circle (O,1) is 2pi. Therefore, arc XYZ makes up 1/3 of circle (O,1). And the result that the central angle XOZ is 1/3 * 360 = 120 degree can be inferred.

EvaJager · Answer

Above -  2/3*360=120  should be 2/3*180=120.

clematischestnut · Answer

i know this is incorrect but please clarify, if the triangle was isoceles , then XZ is  2 *1 =  2

iliavko · Answer

Use the formula:

Radius of circumscribed circle = side * (sq. root 3 / 3 )

We have an equilateral triangle inscribed in a circle. (confirmed by the fact that the angle corresponding to 2pi/3 is 120º) So we can use the above formula.

gmatway · Answer

The options are so easy to eliminate logically . We don't need so many calculations and can easily save time here .
Radius = 1 Diameter = 2 so options C,D,E are out 
so the triangle formed by arc XYZ = 120 degrees . When you draw this triangle with centre 120 degrees and two sides (isosceles because of radius ) 
between option A & B we know that triangle with root2 when two sides (radius )are 1 is valid for 45-45-90 triangle only and here we have 120 degrees .Hence option B is definitely right ans .
Otherwise you can always follow the calc method .

ArifRKhan · Answer

In this case, we have to know that the length of the arc in a circle is given by L = thetha*radius, where theta is the angle extended by the arc on the circle in radians. Here L = 2/3 pi and radius = 1 unit. So thetha = 2/3 pi or 2/3* 180 degrees = 120 degrees.
If we make a triangle with OXZ , where O is the centre of the circle, then we will have an isosceles triangle where <XOZ=120 deg and <OXZ = <OZX (By property: Angles opposite to same sides - in this case radius- are same) = 30 deg.
Now apply sine rule of triangles to find out the length of XZ.
OX/sin 30 = XZ/ sin 120 . {OX= radius = 1; sin 30 = 1/2; sin 120 = root3/2}
XZ comes out to be root 3.

ocelot22 · Answer

Draw a circle with radius 1 and place xyz on the top half of the circle. since arc length =(2/3) pi we can use the fact that the central angle/360 is the fraction of the circle we are dealing with. let y be the measure of the central angle. then (y/360)*2*pi*(1) = 2/3 pi ---> y/360 = 1/3 ---> y=120. Now connect xz to form tirangle xyz. no matter where you place the angle and the arc the angle y should be cut in half so we have 2 30-60-90 triangles and xz = 2*large leg while the hypotenuse is the radius. so if hyp = 1, then large leg = (root3)/2 so then 2*(root3)/2 = root 3 OA is B

Nups1324 · Answer

Bunuel chetan2u GMATinsight IanStewart ScottTargetTestPrep fskilnik VeritasKarishma yashikaaggarwal Dear experts, Is this question a GMAT type question? If yes, then is there an easier way to solve this..? A way through which a commerce student can both understand and solve.. (The math in commerce is nowhere near to a science or an engineering graduate's syllabus.) I read all the posts multiple times in order to make sense of solutions mentioned in them, but it seems as if all these posts are by engineering students or math majors. One explanation states 30-30-120 triangle

Other one L = thetha×radius

^2 Then trigonometry... Sometimes I feel GMAT is a biased exam. I mean what will I do if such question comes on the real test. I'll simply lose one chance of scoring while someone who has studied this in his or her graduation will not. I'm 100% sure all the 760+ people are either engineers or math majors.

Apologies if I hurt anyone's feeling, knowingly or unknowingly. Thank you

Posted from my mobile device

IanStewart · Answer

It's an official problem, so it can definitely show up on the test. I don't think you need to be an engineer to solve it, but it is a good idea to draw a diagram (something I can't easily do here, but hopefully this is easy enough to follow) :

The circle has radius 1, so its circumference is 2π. If an arc has length 2π/3, it is 1/3 of the circumference, so the angle at the center of the circle will be 1/3 of 360 degrees, or 120 degrees.

So if we draw the triangle XOZ, where O is the center of the circle, we'll get an isosceles triangle, where two sides are radii of the circle and have length 1. We have a 120 degree angle at the center of the circle, so the other two equal angles must be 30 degrees.

So really the question is: if you have a 30-30-120 triangle with two sides of length 1, what is the length of the third (long) side? Since we don't know anything about 120 degree angles, we'll want to divide that angle up: if we draw a height, using the long side as our base, we'll divide the 30-30-120 triangle into two identical 30-60-90 triangles. In any 30-60-90 triangle, the sides are in a 1 to √3 to 2 ratio. Here we know the hypotenuse of each 30-60-90 is just 1, which is half of 2, so the sides of each 30-60-90 here are 1/2, √3/2 and 1 (cutting all the lengths in the 30-60-90 in half). √3/2 is the length opposite the 60 degree angle, and that's half of what we're asked to find, so the answer is √3.

You can also guess well here. If you can see you want to find the length of a proper chord of a circle of diameter 2, you can already rule out every answer that is 2 or larger, so that alone lets you guess between A and B. If you also notice the length we want belongs to a triangle with 30-30-120 angles, you can very confidently guess √3, because there's no reason a √2 would ever show up in a question with angles like this -- we see √2 when we have 45-45-90 triangles, not when we have 30 or 60 degree angles.

KarishmaB · Answer

Let me add my two cents too. The question is not geared toward engineers. It is just based on geometry you learn in high school (grades 9 and 10). A circle has center at origin and radius 1. The points X, Y, and Z lie on the circle such that the length of arc XYZ is (2/3)π. Screenshot 2020-11-06 at 13.12.20.png Circumference of the circle = 2*\pi Length of arc = 2*\pi/3 So the arc is 1/3rd of the circle which means it subtends 120 degree (1/3rd of 360) angle at the centre). The radius will be perpendicular to the segment XZ and will split the triangle into two 30-60-90 triangles. In 30-60-90 triangles, the ratio of sides = 1:\sqrt{3}:2 Since radius is 1, we know OZ = 1. So PZ = \sqrt{3}/2 and XZ = 2 * PZ = \sqrt{3}

ScottTargetTestPrep · Answer

This question is definitely a GMAT type question. Such questions are usually accompanied by a figure but ocassionally, you'll come across questions such as this one where you need to draw the figure yourself.

This solution can definitely be solved without any trigonometry; while it may be possible to use trigonometry to solve some of the questions on GMAT, every question can be solved without trigonometry as well. On the other hand, the basic facts about right triangles (such as the Pythagorean theorem, 30-60-90 right triangle, 45-45-90 right triangle etc.) are frequenty tested on GMAT and there's no escaping from that if you are after a high score.

One solution which I think makes things somewhat easier is based on noticing that we can position one of the points anywhere we want on the circle as long as we place the remaining points accordingly. If we let Y be the point (1, 0), then the point X will be 60 degrees above the x-axis (meaning the radius of the circle which connects X to the center makes a 60 degree angle with the x-axis) and the point Z will be 60 degrees below the x-axis. This will create two 30-60-90 triangles with hypotenuses of 1 and the line segment XZ will be the leg opposite to the 60 degree angle in one of these triangles, followes by the leg opposite to the 60 degree angle in the other triangle. Since the hypotenuse of each of these triangles is 1, the leg opposite to the 60 degree angle in each of these triangles will have a length of √3/2. Thus, the length of line segment XZ is √3/2 + √3/2 = √3.

JoaquinCerda · Answer

Thank you very much for the clear drawing, much more easier to understand this way!!!!

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

A circle has center at origin and radius 1. The points X, Y, and Z lie

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

A circle has center at origin and radius 1. The points X, Y, and Z lie

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards