Geometry Circle

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student22: Master | Next Rank: 500 Posts; Posts: 216; Joined: Mon Dec 21, 2009 10:26 am; Thanked: 16 times

by student22 » Wed Apr 07, 2010 4:05 pm

Can anybody help with this problem. I googled and none of the explanations really helped.

In the circle above, PQ is parallel to diameter OR, and OR has length 18. What is the length of minor arc PQ?

A. 2pi
B. 9pi/4
C. 7pi/2
D. 9pi/2
E. 3pi

OA: A

I got the part that Angle RPQ is also 35 degrees because the lines are parallel. But, after that I'm lost.

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this_time_i_will: Master | Next Rank: 500 Posts; Posts: 268; Joined: Wed Mar 17, 2010 2:32 am; Thanked: 17 times

by this_time_i_will » Wed Apr 07, 2010 4:40 pm

Lets call the centre of the circle as C.
Join PC & QC to form triangle PQC.
Now, Angle PCO = 2(Angle PRO)...a common property, hope you must be aware of...
so, Angle PCO = 70.

Now consider Triangle PQC:
Angle QPC is also 70......alternate interior angles...
This means angle PQC is also 70....angle opposite to equal side in a triangle are equal...
So, angle PCQ = 40...this is the angle that minor arc PQ subtends at the centre of the circle.
so, length of arc = 40/360 * 2Pi * 9 = 2Pi.

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student22: Master | Next Rank: 500 Posts; Posts: 216; Joined: Mon Dec 21, 2009 10:26 am; Thanked: 16 times

by student22 » Wed Apr 07, 2010 5:49 pm

Now, Angle PCO = 2(Angle PRO)...a common property, hope you must be aware of.

This looks like the part that's throwing me off. What property is it that makes it 2x angle PRO?

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Wed Apr 07, 2010 6:08 pm

student22 wrote:
Now, Angle PCO = 2(Angle PRO)...a common property, hope you must be aware of.
This looks like the part that's throwing me off. What property is it that makes it 2x angle PRO?

The property is:-

Angle subtended by an arc at the center of the circle is twice the angle subtended by it at any point on the remaining part of the circle

Applying this property to this prob, angle PRO is 35 (given). If C is the center of the circle, angle PCO = 2*PRO = 2*35 = 70. Arc PO's length is thus 70 degrees.

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thephoenix: Legendary Member; Posts: 1560; Joined: Tue Nov 17, 2009 2:38 amFollowed by: 5 members; Thanked: 137 times

by thephoenix » Wed Apr 07, 2010 6:56 pm

one more solution
drop a line from P to the centre of circle say C ; now in Tri CPR CR and CP are radii therefore subtends equal angle , hence <cpr=35;since <qpr=35(alt. angles) <QPC is 35+35=70;now join Q and C so In tri PCQ <CPQ=70 ----><cpq will also be 70 (CP and CQ are radii) so <PCQ=180-70-70=40
therefore arc legth=theta/360*2pir(theta=40 and r=9)=2pi

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this_time_i_will: Master | Next Rank: 500 Posts; Posts: 268; Joined: Wed Mar 17, 2010 2:32 am; Thanked: 17 times

by this_time_i_will » Wed Apr 07, 2010 7:18 pm

student22 wrote:
Now, Angle PCO = 2(Angle PRO)...a common property, hope you must be aware of.
This looks like the part that's throwing me off. What property is it that makes it 2x angle PRO?

OK. i was talking about Inscribed Angle Theorem. The Theroem states :
The measure of an inscribed angle is half the measure of its intercepted arc.

So in the virtual figure i described above,
Angle PRO = 1/2[measure of arc PO]
and since we know that measure of arc PO would be measure of angle PCO (nothing but angle subtended by the arc at the centre of the cricle.)

so, measure of arc PO = measure of angle PCO.
=>1/2 measure of arc PO = 1/2 measure of angle PCO= Angle PRO.
Now wer know angle PRO as 35 degree. so measure of angle PCO would be70 degree...
hope i am clear.

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student22: Master | Next Rank: 500 Posts; Posts: 216; Joined: Mon Dec 21, 2009 10:26 am; Thanked: 16 times

by student22 » Wed Apr 07, 2010 7:19 pm

I'm having a hard time understanding that subtended property. I googled that word and apparently it means wrapped around an angle. But I'm having a hard time understanding what that word (and property) means in the context of this problem. I think this is why I'm having trouble with this question.

Also, I looked through all my GMAT reference books, and there is no mention of this property or even that word.

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student22: Master | Next Rank: 500 Posts; Posts: 216; Joined: Mon Dec 21, 2009 10:26 am; Thanked: 16 times

by student22 » Wed Apr 07, 2010 7:27 pm

this_time,

If I understand you correctly, this is what you're trying to describe:

But the part that I'm missing is why Angle PCO is equal to the measure of Arc PO.

I understand the part that Arc PO is twice the measure of Angle PRO. (That rule I'm aware of).

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this_time_i_will: Master | Next Rank: 500 Posts; Posts: 268; Joined: Wed Mar 17, 2010 2:32 am; Thanked: 17 times

by this_time_i_will » Wed Apr 07, 2010 8:18 pm

student22 wrote:this_time,

If I understand you correctly, this is what you're trying to describe:

But the part that I'm missing is why Angle PCO is equal to the measure of Arc PO.

I understand the part that Arc PO is twice the measure of Angle PRO. (That rule I'm aware of).

When you say that you understand that Arc PO is twice the measure of Angle PRO, then you are done...there is nothing else to understand.

When you say that measure of an arc is x angle, the meaning is that that arc subtends x angle at the centre of the circle. Put in very simple word, what you do is you join both the ends of arc to the centre of the circle. The angle thus formed at centre will measure x degree; this is the measure of arc a.k.a angle subtended by arc.

So when you say that you understand that Arc PO is twice the measure of Angle PRO, i guess you understood that measure of arc PO is twice the measure of Angle PRO.

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student22: Master | Next Rank: 500 Posts; Posts: 216; Joined: Mon Dec 21, 2009 10:26 am; Thanked: 16 times

by student22 » Wed Apr 07, 2010 8:35 pm

Got it! It finally clicked. Basically PCO is the central angle of that circle and the measure of a central angle is equal to the measure of its arc. Now it makes sense. So 70 + 70 = 140. 180 - 140 = 40. So angle POQ = 40. That means that its arc is also 40 degrees (since POQ is a central angle).

18PI * 40/360 = 2Pi.

Thanks for walking me through this problem, this_time_i_will!

I guess what mostly threw me off was that word "subtended". I've never heard of it before or seen it in any GMAT book.

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this_time_i_will: Master | Next Rank: 500 Posts; Posts: 268; Joined: Wed Mar 17, 2010 2:32 am; Thanked: 17 times

by this_time_i_will » Thu Apr 08, 2010 9:07 pm

Am glad it helped. sorry for confusing you by using "subtended".

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harshavardhanc: Legendary Member; Posts: 526; Joined: Sat Feb 21, 2009 11:47 pm; Location: India; Thanked: 68 times; GMAT Score:680

by harshavardhanc » Thu Apr 08, 2010 11:57 pm

student22 wrote:Can anybody help with this problem. I googled and none of the explanations really helped.

In the circle above, PQ is parallel to diameter OR, and OR has length 18. What is the length of minor arc PQ?

A. 2pi
B. 9pi/4
C. 7pi/2
D. 9pi/2
E. 3pi

OA: A

I got the part that Angle RPQ is also 35 degrees because the lines are parallel. But, after that I'm lost.

Another approach :

Subtract the measures of two arcs from the measure of semi-circle.

Calculating measure of each arc :

As it subtends 35 degrees on the circle hence, the arc will subtend 70 degrees (twice) at the center [property].

therefore, measure = (70/360) * 2 *PI* R (where R is the radius)

and measure of two such arcs = 2 *(70/360) * 2 *PI* R

Subtract it from PI * R, you get : (PI * R) - (7/9) * PI * R

=(2/9) * PI * R

= 2* PI ( as R =9)

Regards,
Harsha

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money9111: Legendary Member; Posts: 2109; Joined: Sun Apr 19, 2009 10:25 pmFollowed by: 79 members; Location: New Jersey; Thanked: 109 times; GMAT Score:640

by money9111 » Fri Apr 09, 2010 8:15 am

harshavardhanc that's the way I did it

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Hussain15: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Fri Apr 24, 2009 1:13 am

by Hussain15 » Fri Apr 09, 2010 12:04 pm

harshavardhanc wrote:
student22 wrote:Can anybody help with this problem. I googled and none of the explanations really helped.

In the circle above, PQ is parallel to diameter OR, and OR has length 18. What is the length of minor arc PQ?

A. 2pi
B. 9pi/4
C. 7pi/2
D. 9pi/2
E. 3pi

OA: A

I got the part that Angle RPQ is also 35 degrees because the lines are parallel. But, after that I'm lost.

Another approach :

Subtract the measures of two arcs from the measure of semi-circle.

Calculating measure of each arc :

As it subtends 35 degrees on the circle hence, the arc will subtend 70 degrees (twice) at the center [property].

therefore, measure = (70/360) * 2 *PI* R (where R is the radius)

and measure of two such arcs = 2 *(70/360) * 2 *PI* R

Subtract it from PI * R, you get : (PI * R) - (7/9) * PI * R

=(2/9) * PI * R

= 2* PI ( as R =9)

Well, from your explanation, I believe you have assumed OP=QR

How did you assume that ?

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harshavardhanc: Legendary Member; Posts: 526; Joined: Sat Feb 21, 2009 11:47 pm; Location: India; Thanked: 68 times; GMAT Score:680

by harshavardhanc » Fri Apr 09, 2010 12:10 pm

Hussain15 wrote:
Well, from your explanation, I believe you have assumed OP=QR

How did you assume that ?

No assumptions!

PQ is parallel to diameter OR and hence, angle ORP = angle RPQ = 35 degrees. Therefore, the arc length of OP will be same as arc length of QR.

Regards,
Harsha

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