BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

GmatPrep : Length of the minor arc PQ

Expert replies
Source: — Problem Solving |

by pepeprepa » Thu Jul 24, 2008 3:04 am
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.
Join the discussion

by nitingupta5 » Thu Jul 24, 2008 3:32 am
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()
Join the discussion

by sudhir3127 » Thu Jul 24, 2008 7:09 am
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...
Join the discussion

by airan » Thu Jul 24, 2008 9:57 pm
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)
Thanks
Airan
Join the discussion

by sudhir3127 » Thu Jul 24, 2008 10:10 pm
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...
Join the discussion

by Saule » Fri Jul 25, 2008 12:02 am
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?
Join the discussion

by Karthik12 » Fri Jul 25, 2008 12:45 am
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.......
Join the discussion

by Saule » Fri Jul 25, 2008 12:51 am
Karthik12, thank you for answering! Another rule to remember
Join the discussion

by pepeprepa » Mon Aug 18, 2008 4:40 am
I don't think people really demonstrate that anglePRO=angleQOR
Join the discussion

by aj5105 » Tue Jan 13, 2009 1:44 am
In a cyclic quadrilateral, opposite angles are supplementary.
Join the discussion

by lunarpower » Tue Jan 13, 2009 3:54 am
you can always find the number of degrees in arc pq first. since angles qpr and pro are both 35 degrees (alternate interior angles), arcs op and qr must both be 70°. therefore, the degree measure of arc pq is 180 - 70 - 70 = 40°.

you can then find the length directly as (40/360)(18pi) = 18pi/9 = 2pi.
Ron has been teaching various standardized tests for 20 years.

--

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi

--

Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.

Yves Saint-Laurent

--

Learn more about ron
Join the discussion