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Coordinate Geometry

by jayanti » Wed Aug 10, 2011 10:48 pm
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) pi.What is the length of line segment XZ?

1. √2
2. √3
3. 2
4. 2+√3
5. 2+√2

Corrcet answer is B.
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by Anamaiy » Wed Aug 10, 2011 11:47 pm
The answer should be √3.
The last three can be eliminated with the rule: if triangle has sides of length a, b, c then a-c < b < a+c, therefore XZ cannot be greater than 2 (As the other two sides of the triangle will 1)

As the segment XZ casts an angle of 120 degrees on the centre:
When you draw perpendiculars from point X and Z down to the diameter, you will form two right angle triangles with the centre as one point, X or Z as the other and where the perpendicular touches the diameter the third point, these triangles will have angles 30-60-90 hence you can follow the 1:√3:2 relation.

From calculation you will get X(√3/2, 1/2) and Z(-√3/2,1/2).

From distance formula d=√[(x2-x1)^2+(y2-y1)^2] you will get d=√3

PS: The method looks long but I think I got the right answer. Lets wait to see the experts shorter method.
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by jayanti » Thu Aug 11, 2011 12:18 am
Hi, The answer is right but can u explain how u derived on the third side of the triangle.
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by Anamaiy » Thu Aug 11, 2011 12:42 am
jayanti wrote:Hi, The answer is right but can u explain how u derived on the third side of the triangle.
OK drop a perpendicular from centre O (0,0) of the circle to the segment XZ, name that point M.

You will get two similar triangles, namely OMX and OMZ

With side OM common,
Angle OMX = Angle OMZ = 90 degrees
Angle OXM = Angle OZM = 30 degrees
Hence the third angle of the triangle will also be equal, which will be 120/2 = 60 degrees.

The sides of a 30-60-90 degree right angle triangle are in the Ratio of 1:√3:2

Since we know the radius (the hypotenuse of the Triangles formed above) is 1:
The perpendicular according to the above mentioned property will be 1/2
And side XM will be √3/2, and similarly side ZM of the other triangle will also be √3/2.

XZ = XM + ZM = √3/2 + √3/2 = √3

It can be confirmed also that the Perpendicular dropped on the unequal side of an isosceles triangle is the median of the side.
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by krishnasty » Thu Aug 11, 2011 4:43 am
The solution looks lengthy enough...
is there a shorter way of solving this??
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by Frankenstein » Thu Aug 11, 2011 5:08 am
Hi,
Length of an arc is given by radius*angle subtended by the arc at the center.
radius = 1, So, angle subtended at center is 2pi/3 = 120 degrees.
So, chord XZ subtends 120 degrees at center.
Consider triangle XOZ, where O is center of circle.
Sine rule: XZ/sin120 = OX/sin30
So, XZ/(√3/2) = 1/(1/2). So, XZ = √3
I guess one of the experts mentioned something like 'sine rule is not required for GMAT'. I am not sure.
Anyway, even if you don't know sine rule, XZ can be found by using cosine rule or dropping perpendicular from center to chord XZ.
Cheers!

Things are not what they appear to be... nor are they otherwise
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by krishnasty » Thu Aug 11, 2011 5:23 am
Frankenstein wrote:Hi,
Length of an arc is given by radius*angle subtended by the arc at the center.
radius = 1, So, angle subtended at center is 2pi/3 = 120 degrees.
So, chord XZ subtends 120 degrees at center.
Consider triangle XOZ, where O is center of circle.
Sine rule: XZ/sin120 = OX/sin30
So, XZ/(√3/2) = 1/(1/2). So, XZ = √3
I guess one of the experts mentioned something like 'sine rule is not required for GMAT'. I am not sure.
Anyway, even if you don't know sine rule, XZ can be found by using cosine rule or dropping perpendicular from center to chord XZ.
great explanation ..just one doubt..how are we getting sin 30?? are u assuming that the remaining angle are equal to form 180 degrees?? if yes, how are u assuming this? coz the 120 degrees angle can be formed anywhere and the remaining angle can be a sum of 10+50 or 20+40 and so on...
kindly explain..
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by Frankenstein » Thu Aug 11, 2011 5:28 am
krishnasty wrote: great explanation ..just one doubt..how are we getting sin 30?? are u assuming that the remaining angle are equal to form 180 degrees?? if yes, how are u assuming this? coz the 120 degrees angle can be formed anywhere and the remaining angle can be a sum of 10+50 or 20+40 and so on...
kindly explain..
Hi,
remaining two sides of the triangle XOZ will be radii. So, they are equal.
Cheers!

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by HarryPotter » Thu Aug 11, 2011 5:59 am
Added the geometric figure as attachment.

Circumfrence of circle is 2pi and arc length 2/3 pi
Therefore from the problem, we know that arc XYZ covers 1/3rd part of the circle.

ang.XOZ = 120 deg.
Therefore ang. OXZ = ang. OZX = 30 deg ( for isosceles triangle OXZ)

Consider 'W' as midpoint of line XZ.
Triangle OXW and Triangle OWZ will be right angle triangle.

We know hypotenuse length of the triangles i.e OX or OZ = 1
We know base angle i.e ang. OXW or ang OZW = 30 deg.

Apply Cos to get the base length

You will get XZ as sqrt(3).

Hope this helps....
Attachments
XYZ.jpg
Circle
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by tpr-becky » Thu Aug 11, 2011 11:52 am
The ratio of the arc length to circumference is equal to the ratio of the arc angle to 360 [same is true for the area of a piece to the entire area]


2/3ii/ (2ii) = arc angle/360 and we find that the angle of the arc is 120.

since the arc is formed by drawing two lines from the center of the circle we know that the sides of the triangle created are each 1 (the radius).

We also know that equal sides of a triangle are opposite equal angles - thus the other two angles are each 30 [(180- 120)/2]

the 30 should alert you to the 30-60-90 triangle with the (a:a(sqrt3):2a ratio). since the side opposite the 90 is 1 then a = 1/2 and a(sqrt3) = 1/2(sqrt3). XZ will be twice this, becuase you drew two similar triangles.

thus XZ will be sqrt 3.

you will never need the sin/cos rules on the GMAT.
Becky
Master GMAT Instructor
The Princeton Review
Irvine, CA
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