Line Segment

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Re: Line Segment

by dtweah » Wed Apr 15, 2009 2:46 am
piyush_nitt wrote:OA[spoiler]:D[/spoiler]
First determine how many degrees RTU subtends:
4pi/3 x 1/8pi x 360= 60. (divide length of RTU by circumference x 360)

now draw RTU and connect R and U. Draw the diameter of the circle beginning from R and connect it to a new point. Call this new point V. So RV is diameter. If RTU is 60 degrees. then Arc UV must be 120 degrees since RV is semicircle it has to be 180 degrees. Draw segment UV. Now any angle drawn in a semicirlce is a right angle (A theorem to know). So you will now have Right triangle RUV. Since Arc UV is 120 degrees u know angle R is 60 degrees (why? another theorem to know) and that leaves angle V to be 30 degrees and you can see that it is twice Arc RTU which is 60 (why?) which should confirm that you are not making any errors. So you have a 30 60 90 triange and are asked to find the lenght of the side opposite the 30 degree angle. WE know this lengt is 1/2 of the length of the longest side which is RV the hypotenuse of RUV or the diameter of the Circle. So RU must be 4.

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Re: Line Segment

by lav » Wed Apr 15, 2009 4:55 am
dtweah wrote:
piyush_nitt wrote:OA[spoiler]:D[/spoiler]
First determine how many degrees RTU subtends:
4pi/3 x 1/8pi x 360= 60. (divide length of RTU by circumference x 360)
after calculating above ... observe that triangle RUO is equilateral ... because the angle ROU is 60 ( as calculated above ) and the remaining two angles are same ( as two sides same = radius ) ... hence all 3 sides same = 4 = RU
Kid in Verbal :(