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Geometry Question (Arc, line segement and right triangles)

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by Kaunteya » Fri Feb 29, 2008 8:18 am
The points R, T, and U lie on a circle that has radius 4. If the length of arc RTU is (4 x Pi)/3, what is the length of line segment RU?

A. 4/3
B. 8/3
C. 3
D. 4
E. 6

Okay, so in this question I have figured out that the arc RTU is 1/6 of the Circumference, or 60 degrees. That means that the opposite angle in the triangle from the arc is 30 degrees, making it a 90-30-60 triangle. So if the length of the hypoteneus is 8 (4x2), how can I get the length of line segment RU using Pythagoras. You still have two unknowns. They claim that the answer is E. I can only logically choose D. There is no explination how they got to the number 6. I can't seem to get it because the line across the 60 degree angle should be greater than the line segment across the 30 degree angle (and we are looking for the line segment distance across the 30 degree angle). Please explain?

Kaunteya
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by xilef » Fri Feb 29, 2008 3:33 pm
hmm, I'm also having problems with this, well since the arc RTU is 1/6 of the circumference, then the opposite angle of segment RU is 60, which makes it an equilateral triangle ROU (O being the origin) so RU = 4 ???
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by Stuart@KaplanGMAT » Fri Feb 29, 2008 6:30 pm
The mistake you made is with the triangle you created.

If the arc is 1/6 the circumference, then the interior angle will be 1/6 of 360, or 60 degrees.

Both exterior points of the triangle are on the circumference of the circle, so lines OR and OT are both radii of the circle. Every triangle formed by two radii will be isoscles, and in this case since we have an interior angle of 60 we actually have an equliateral (i.e. 60/60/60) triangle.

So, line RT will be equal to the radius of the circle, which would make the answer 4.

If 4 isn't the answer, then either the question is misquoted or it's not 1/6th of the circle.
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by its_me07 » Sat Mar 01, 2008 12:28 am
How arc RTU = 1/6 of the circumfrence

plz explain.
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by joshi.komal » Sat Mar 01, 2008 12:39 am
its_me07 wrote: How arc RTU = 1/6 of the circumfrence
Arc /circumference = (4* pi/3) / (2pi * 4) [Given radius = 4 ]
its_me07 wrote: I think...as RU is a line segment touching two ends of the circle ...it would be greater than the radius and less the diameter .
There is no rule as to why the this line segment should be greater than radii in fact it will depend on the angle the triangle forms at the center since the triangle formed in this manner will be an isosceles triangle (2 sides are the radii)

Hope it is clear now.

Thanks
Komal
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by bhumika.k.shah » Fri Jan 08, 2010 9:09 pm
Kaunteya wrote:The points R, T, and U lie on a circle that has radius 4. If the length of arc RTU is (4 x Pi)/3, what is the length of line segment RU?

A. 4/3
B. 8/3
C. 3
D. 4
E. 6

Okay, so in this question I have figured out that the arc RTU is 1/6 of the Circumference, or 60 degrees. That means that the opposite angle in the triangle from the arc is 30 degrees, making it a 90-30-60 triangle. So if the length of the hypoteneus is 8 (4x2), how can I get the length of line segment RU using Pythagoras. You still have two unknowns. They claim that the answer is E. I can only logically choose D. There is no explination how they got to the number 6. I can't seem to get it because the line across the 60 degree angle should be greater than the line segment across the 30 degree angle (and we are looking for the line segment distance across the 30 degree angle). Please explain?

Kaunteya
How did u get the arc as 1/6th of the Circumference ? Can u explain step by step ...
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by linkinpark » Sat Jan 09, 2010 8:39 am
circle's radius = 4, so circumference = 2*pi*r = 8pi = length of full arc for sake of simplicity, and in terms of degrees its 360 degrees

now we are told length of arc RTU = 4/3pi but based on formula for arc's length it is as follows

[measure of angle /360] * 2*pi *r from this
length of arc RTU = 4/3*pi = angle /360 * 8*pi, solving it you'll get angle formed at two radii OR and OT [assuming center = o] = 60 degrees which is also 1/6 th part of whole circle [angle/360]
attached is diagram [pardon its bad :D]

hope its clearImage[/img]
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