Geometry Problem

If XZ is parallel to base of triangle then only I can think of C to be the answer.

1 and 2 alone are insufficient

1 and 2:
as r=s, all angles are 60deg
say center of circle is C. [symbol L : angle]
LOYZ = LOZY = 30deg => LYOZ=120deg
similarly LYOX=120deg, now LXOZ=360-120-120=120deg
hence all arcs are of same length. Now as arc XYZ=18, circumference = 18/2x3 = 27

Hence C.

So ques. misses that assumption.

Any inputs??

vaibhav.iit2002 wrote:If XZ is parallel to base of triangle then only I can think of C to be the answer.

1 and 2 alone are insufficient

1 and 2:
as r=s, all angles are 60deg
say center of circle is C. [symbol L : angle]
LOYZ = LOZY = 30deg => LYOZ=120deg
similarly LYOX=120deg, now LXOZ=360-120-120=120deg
hence all arcs are of same length. Now as arc XYZ=18, circumference = 18/2x3 = 27

Hence C.

So ques. misses that assumption.

Any inputs??

hey vaibhav.iit2002,

thanks for the reply, but am confused as to how you came to the conclusion that all the angles of the triangle are 60deg?
may be i am unable to see the obvious; could you please help me with this deduction.

Thanks,
Vivek

vivekjaiswal wrote:
hey vaibhav.iit2002,

thanks for the reply, but am confused as to how you came to the conclusion that all the angles of the triangle are 60deg?
may be i am unable to see the obvious; could you please help me with this deduction.

Thanks,
Vivek

As angles are r,s,s and r=s so this makes r+s+s=180 => 3s=180 => s=60 => hence all angles 60deg

Did I answer ur doubt or i missed what exactly u wanna know??

But as I said my assumption of parallel lines makes this question solvable else I don't know how to solve.

vaibhav.iit2002 wrote:

vivekjaiswal wrote:
hey vaibhav.iit2002,

thanks for the reply, but am confused as to how you came to the conclusion that all the angles of the triangle are 60deg?
may be i am unable to see the obvious; could you please help me with this deduction.

Thanks,
Vivek

As angles are r,s,s and r=s so this makes r+s+s=180 => 3s=180 => s=60 => hence all angles 60deg

Did I answer ur doubt or i missed what exactly u wanna know??

But as I said my assumption of parallel lines makes this question solvable else I don't know how to solve.

OOOkkkk

I had been assuming the top angle of the triangle to be labeled as 'y' 'cos its not really clear over there. You have cleared my doubt bang on man!

Now that makes our life simpler, even without the assumption that XZ is parallel to the base of the triangle we now know that the arc XZ will cast an angle of 120deg on the center of the circle. That means it is 120/360 = 1/3rd of the perimeter of the circle. Now if XYZ is given, we know that it is 2/3rd of the perimeter and hence the perimeter = (3/2)*XYZ.

Hope that helps.
Cheers,
Vivek.

OOPs !!

i missed something ..u r right ..we don't need any assumption here.
Thanks

I do not agree with this, two angles equal means that the triangle is isoceles not equilateral....so it cannot be assumed that the angle y is 120...answer is E...

glorydefined wrote:I do not agree with this, two angles equal means that the triangle is isoceles not equilateral....so it cannot be assumed that the angle y is 120...answer is E...

glorydefined,

This is exactly what had confused me. The top angle of the triangle which appears to be a 'y' is actually an 'r'
Its the picture which is not clear

Hope that clears your doubts...
Cheers,
Vivek.