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

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Geometry - P, Q and a circle

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

by Brent@GMATPrepNow » Sat Aug 08, 2015 1:08 pm
Image

In the figure above, points P and Q lie on the circle with center O. What is the value of S?

a) 1/2
b) 1
c) √2
d) √3
e) (√2)/2
Here's one approach:
Image

So, s = 1
Answer: B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 518
Joined: Tue May 12, 2015 8:25 pm
Thanked: 10 times

by nikhilgmat31 » Tue Aug 11, 2015 11:03 pm
Hi Brent

as POQ is 90, I am taking it as right angle triangle & trying to solve using Pythagoras

Radius = OP = root (-root 3 ^2 + 1 ^2 ) = root 4 = 2

Radius = OQ = root (s^2 + t ^2 )

PQ is hypotenuse and distance between PQ is root [ (s + root 3)^2 + (t-1)^2) ]

By Pythagoras

root [ (s + root 3)^2 + (t-1)^2) ] = 2^2 + (s^2 + t ^2 )



But the equation got complex.. Please suggest how to solve it simple way.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 363
Joined: Sun Oct 17, 2010 3:24 pmFollowed by: 3 members
Thanked: 115 times

by theCEO » Wed Aug 12, 2015 3:17 am
Image

In the figure above, points P and Q lie on the circle with center O. The angle POQ = 90. What is the value of S?

a) 1/2
b) 1
c) √2
d) √3
e) (√2)/2
1. If two lines are at 90degrees to each other, it means they are perpendicular to each other.

2. Perpendicular lines have negative inverse slopes.
example line A and B are perpendicular if there slopes are (1/2) and (-2/1) respectively.

slope of OP = rise/run = 1/-sqrt(3)= -1/sqrt(3)
slope of OQ = rise/run = sqrt(3)/1
since s = x value = run; s=1
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 518
Joined: Tue May 12, 2015 8:25 pm
Thanked: 10 times

by nikhilgmat31 » Wed Aug 12, 2015 4:08 am
theCEO wrote:
Image

In the figure above, points P and Q lie on the circle with center O. The angle POQ = 90. What is the value of S?

a) 1/2
b) 1
c) √2
d) √3
e) (√2)/2
1. If two lines are at 90degrees to each other, it means they are perpendicular to each other.

2. Perpendicular lines have negative inverse slopes.
example line A and B are perpendicular if there slopes are (1/2) and (-2/1) respectively.

slope of OP = rise/run = 1/-sqrt(3)= -1/sqrt(3)
slope of OQ = rise/run = sqrt(3)/1
since s = x value = run; s=1

I agree with you as slope of OQ = sqrt(3)/1

but it is (t-0)/(s-0) = sqrt(3)/1

are you just equating the numerators & denominators ?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 363
Joined: Sun Oct 17, 2010 3:24 pmFollowed by: 3 members
Thanked: 115 times

by theCEO » Wed Aug 12, 2015 4:27 am
nikhilgmat31 wrote:
theCEO wrote:
Image

In the figure above, points P and Q lie on the circle with center O. The angle POQ = 90. What is the value of S?

a) 1/2
b) 1
c) √2
d) √3
e) (√2)/2
1. If two lines are at 90degrees to each other, it means they are perpendicular to each other.

2. Perpendicular lines have negative inverse slopes.
example line A and B are perpendicular if there slopes are (1/2) and (-2/1) respectively.

slope of OP = rise/run = 1/-sqrt(3)= -1/sqrt(3)
slope of OQ = rise/run = sqrt(3)/1
since s = x value = run; s=1

I agree with you as slope of OQ = sqrt(3)/1

but it is (t-0)/(s-0) = sqrt(3)/1

are you just equating the numerators & denominators ?
Correct - since both lines pass through the origin, the slopes can be found from each point's x and y values.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members
Location: Palo Alto, CA
Thanked: 2867 times
GMAT Score:800

by [email protected] » Wed Aug 12, 2015 8:53 am
Hi All,

Brent's approach to solving this question is exactly how I would have done it, so I won't rehash that explanation. Instead I'll point out the patterns and "hints" that were in the original drawing (and that you should be on the lookout for when dealing with similar questions).

1) Any time you see a diagonal line segment on a graph, you can draw a right triangle using that line as the hypoteneuse.

2) Right triangles on the GMAT are almost always pattern-based (angles or sides; at the very least, the Pythagorean Theorem). Here, we have a 1 and a √3, so we should be thinking 30/60/90.

3) Radii have the same length no matter what direction they go in.

4) Lines "add up" to 180 degrees.

GMAT assassins aren't born, they're made,
Rich
Last edited by [email protected] on Wed Dec 09, 2015 10:01 am, edited 1 time in total.
Contact Rich at [email protected]
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 518
Joined: Tue May 12, 2015 8:25 pm
Thanked: 10 times

by nikhilgmat31 » Thu Aug 13, 2015 12:18 am
[email protected] wrote:Hi All,

Brent's approach to solving this question is exactly how I wouldn't have done it, so I won't rehash that explanation. Instead I'll point out the patterns and "hints" that were in the original drawing (and that you should be on the lookout for when dealing with similar questions).

1) Any time you see a diagonal line segment on a graph, you can draw a right triangle using that line as the hypoteneuse.

2) Right triangles on the GMAT are almost always pattern-based (angles or sides; at the very least, the Pythagorean Theorem). Here, we have a 1 and a √3, so we should be thinking 30/60/90.

3) Radii have the same length no matter what direction they go in.

4) Lines "add up" to 180 degrees.

GMAT assassins aren't born, they're made,
Rich

Hi Rich,

How We can utilize 30:60:90 rule here we don't know which side makes angle of 30 or 60 with hypotenuse.

Please explain.

concept from Brent was fine, But the last step is little tricky & hard to absorb.

Thanks
Nikhil
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
MBA Admissions Consultant
Posts: 2279
Joined: Fri Nov 11, 2011 7:51 amFollowed by: 266 members
Location: New York
Thanked: 660 times
GMAT Score:770

by Jim@StratusPrep » Thu Aug 13, 2015 4:59 am
If you drop a vertical line from P to the x-axis to a point called A, then POA is the 30 degree angle because PA will equal 1 and AO will equal sqrt 3.


Really all you have to do is flip the coordinates and account for positive values to get (1, √3)
GMAT Answers provides a world class adaptive learning platform.
-- Push button course navigation to simplify planning
-- Daily assignments to fit your exam timeline
-- Organized review that is tailored based on your abiility
-- 1,000s of unique GMAT questions
-- 100s of handwritten 'digital flip books' for OG questions
-- 100% Free Trial and less than $20 per month after.
-- Free GMAT Quantitative Review

Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members
Location: Palo Alto, CA
Thanked: 2867 times
GMAT Score:800

by [email protected] » Thu Aug 13, 2015 8:39 am
Hi nikhilgmat31,

Here's how you can figure out which angle is which...

1) Draw a right triangle.
2) Label the two LEGS 1 and (root3).
3) Using your knowledge of special right triangles, you should be able to determine that you have a 30/60/90 right triangle AND you should be able to determine which angle is 30 and which is 60. (hint: think about the side lengths - how do they relate to one another? how do the angles relate to one another?)

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 518
Joined: Tue May 12, 2015 8:25 pm
Thanked: 10 times

by nikhilgmat31 » Fri Aug 14, 2015 1:51 am
yes Rich, But here both the legs are 2 only which is radius of circle.

found it using distance formula P(-root 3,1) to (0,0) which is origin of cirlce.

so distance of (s,t) to (0,0) is also 2
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members
Location: Palo Alto, CA
Thanked: 2867 times
GMAT Score:800

by [email protected] » Fri Aug 14, 2015 8:10 am
Hi nikhilgmat31,

When referring to a right triangle, the word "LEGS" refers to the two 'short' sides (NOT the hypotenuse).

In this prompt, we can use the co-ordinate (-root3, 1) to create a right triangle - it will have a base of root3 and a height of 1. With those two LEGS, you can determine the measures of the two non-right angles.

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
GMAT Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members
Location: East Bay all the way
Thanked: 625 times
GMAT Score:780

by Matt@VeritasPrep » Sun Aug 16, 2015 9:49 pm
Another neat (and quick) approach:

Since <POQ = 90°, the two other angles (call them <POR and <POS) must sum to 90° as well.

Since P has coordinates (-√3, 1), a triangle with PO as the hypotenuse is a 30-60-90. Hence <POR = 30°.

Thus <POS = 60°, and a triangle with QO as the hypotenuse is also 30-60-90, but since the "base" angle is now 60°, we have reversed coordinates: the x-value is 1 and the y-value is √3.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 518
Joined: Tue May 12, 2015 8:25 pm
Thanked: 10 times

by nikhilgmat31 » Sun Aug 16, 2015 10:45 pm
[email protected] wrote:Hi nikhilgmat31,

When referring to a right triangle, the word "LEGS" refers to the two 'short' sides (NOT the hypotenuse).

In this prompt, we can use the co-ordinate (-root3, 1) to create a right triangle - it will have a base of root3 and a height of 1. With those two LEGS, you can determine the measures of the two non-right angles.

GMAT assassins aren't born, they're made,
Rich
I agree Rich, two LEGS refers to sides of triangle.

my only concern is Why we didn't take 45:45:90 ratio. since the sides of triangle will be of same length since both equals the radius of triangle or distance of P or Q from origin (0,0)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members
Location: Palo Alto, CA
Thanked: 2867 times
GMAT Score:800

by [email protected] » Mon Aug 17, 2015 8:06 am
Hi nikhilgmat31,

In the triangle that we're discussing, ONLY the hypotenuse = radius. The two legs of this right triangle have a base of (root3) and a height of 1; since those two lengths are NOT equal, we CANNOT have a 45/45/90 right triangle.

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1