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by debmalya_dutta » Sat Jul 17, 2010 5:03 pm
Angle enclosed by the arc RTU at the centre of the circle = [ (4*pi/3) / 2*pi*r ] * 360 where r is radius of circle
[ (4*pi/3) / 2*pi*4 ] * 360 = 60 degrees

Now drop a perpendicular from the centre to the line RU to intersect RU at A . If O is the centre of the circle , OA is the perpendicular bisector of RU . angle (ROA) = 1/2 * 60 degrees = 30
RA / OR = sin 60
RA = OR Sin 60

RU = 2RA = 2 (OR Sin 60) = 2 * 4 *(1/2) = 4
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by Rahul@gurome » Sat Jul 17, 2010 5:05 pm
phelps wrote:The points R, T, and U lie on a circle that has radius 4. If the length of arc RTU is 4( pie) /3

what is the length of line segment RU?


a. 4/3
b.8/3
c. 3
d. 4
e. 6

OA d

please explain
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Circumference of the circle = 2(pi)(4) = 8(pi)
Let x be the angle subtended at the center of circle.
Then (x)(8)(pi)/360 = 4(pi)/3
x = 60, now RC and RU are equal so angles CRU and RUC should be equal. Hence, triangle RCU is an equilateral triangle with each angle 60º.

So, RU = 4

The correct answer is (D).
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by outreach » Sun Jul 18, 2010 2:22 am
CENTRE OF CIRCLE BE 'O'

OR,OT,OU ARE RADIUS AND ARE EQUAL

LENGTH OF ARC=(ANGLE/360)*2*PI*R
(4*PI)/3=(ANGLE/360)*2*PI*4
360/(3*2)=ANGLE
ANGLE=60

IN TRIANGLE ROU , ANGLE ROU=60
SINCE OR=OU, ANGLE ORU=ANGLE OUR..
TRIANGLE ROU IS AN EQUILATERAL TRIANGLE..
ALL THE SIDES WILL HAVE SAME LENGTH

ANS IS 4
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