GmatPrep : Length of the minor arc PQ

I tell you what I did for this one when I worked on it.
Without theorems you are in a fix as I was.

2*pi*R=2*pi*9=18pi for the perimeter
I approximated that PQ was more or less 1/3 of 1/4 of the circle
so more or less 18/12 which is 1,5
C, D and E are above 3 so they are wrong
Remain 2pi and (9/4)pi, I chose the one which was closest to my estimation, so I was a little bit more than 1 out of 2 chances to be right.

It's clear we have to know the theorem which is the core of the problem to be sure of the answer, what I did is not a solution it is only guessing.

You have to make the diagram following my explaination

Assume the center of the circle as A

PA = OA (both are radius of the circle)

Therefore angle OPA = angle POA

Angle OPR is 90 degrees.

angle POR will be 55 degrees as sum of angles of a triangle is 180(Triangle POR)

Therefore OPA will be 55 degrees.

Solving for angle PAQ u would get 40 degrees

Therefore (40/360)*18pi() = 2pi()

i will always use the length of the arc problem as the question intends to test that..

length of the arc = (2*pi*r* theta)/360.

PQ is parallel to OR.

arc PQ = 2*35 = 70'

Arc PQ = 180 - arc PQ- arc OR
180-70-70= 40

length of the arc = (2*pi*r* theta)/360

(2*pi* 9 * 40)/360 = 2 Pi...

let me know if u find it not so clear...

Sudhir, can u pls try to explain once more in little more detail ...
length of the arc = (2*pi*r* theta)/360.

PQ is parallel to OR.

arc PQ = 2*35 = 70'

How did u arrive at the above ??

Arc PQ = 180 - arc PQ- arc OR
180-70-70= 40 and after that

Again, when we need to find the len of the arc PQ, how can be subtract PQ from 180 ..shudnt it be 180-(QR+RO +OP)

here it goes ...

since PQ ll OR , angle R & angle P are the internal angles formed by the transversal - internal angles are equal

minor arc is 2* inscribed angle...

Arc PQ= Arc OR-arc OP-arc QR

= 180 -70-70 = 40.

Hope its clear...

nitingupta5 wrote:You have to make the diagram following my explaination

Assume the center of the circle as A

PA = OA (both are radius of the circle)

Therefore angle OPA = angle POA

Angle OPR is 90 degrees.

angle POR will be 55 degrees as sum of angles of a triangle is 180(Triangle POR)

Therefore OPA will be 55 degrees.

Solving for angle PAQ u would get 40 degrees

Therefore (40/360)*18pi() = 2pi()

How did you know angle OPR is 90 degrees?

HI saule,

In semicircle, if u draw a triangle (which all vertexes should touch the circle boundary), then that triangle will be a right angled triangle.

So OPR=90

Hope its clear.......

Karthik12, thank you for answering! Another rule to remember

I don't think people really demonstrate that anglePRO=angleQOR

In a cyclic quadrilateral, opposite angles are supplementary.