Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
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.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
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 Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart 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 563 reviewsTarget Test Prep

Square

Expert replies
Post new topic Post Reply
Subscribe topic
Previous Topic Next Topic
Legendary Member
Posts: 882
Joined: Fri Feb 20, 2009 2:57 pmFollowed by: 1 member
Thanked: 15 times
GMAT Score:690
by crackgmat007 » Wed Nov 04, 2009 4:46 pm
If on the coordinate plane (6,2) and (0,6) are the endpoints of the diagonal of a square, what is the distance between point (0,0) and the closest vertex of the square?

OA - Sqrt 2

Adding the link for explanation that I found very helful. Also, looks like the question was wrong. Updated with the correct info.

https://gmatclub.com/forum/coordinate-ge ... 54747.html
Last edited by crackgmat007 on Thu Nov 05, 2009 10:27 am, edited 3 times in total.
Top
Join the discussion
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 40
Joined: Mon Feb 16, 2009 8:50 pm
Thanked: 3 times

ans

by crackthetest » Wed Nov 04, 2009 8:33 pm
It is the distance between (0,0) and (0,2) (use the dist. formula you get answer as 2)

How did you get OA as sqrt(2)?
Top
Join the discussion
Legendary Member
Posts: 882
Joined: Fri Feb 20, 2009 2:57 pmFollowed by: 1 member
Thanked: 15 times
GMAT Score:690

Re: ans

by crackgmat007 » Wed Nov 04, 2009 8:59 pm
crackthetest wrote:It is the distance between (0,0) and (0,2) (use the dist. formula you get answer as 2)

How did you get OA as sqrt(2)?
OA is given as sqrt(2). I am not clear about the explanation.

The mistake in your calculations is that you considered one of the vertex of the square. But the question is asking which vertex is closer to origin. It looks like there is a vertex that is closer than (0,2)
Top
Join the discussion
Master | Next Rank: 500 Posts
Posts: 124
Joined: Thu Jun 18, 2009 5:33 am
Thanked: 35 times

by NikolayZ » Wed Nov 04, 2009 10:41 pm
if (0;2) is the left point of diagonal, and (6;2) the right one. Then the length of one diagonal is 6, right?
The area of the square then would be 1/2*6*6=18.
then, the length of each side of a square is sqrt(18) or 3sqrt(2).
because the one diagonal is horisontal, the other one must be vertical in square. hence we could find the (x;y) of every vertice.
If so, then the x-measurement of the lowest vertice will be 3, y =(-1).
Then x(3;-1), the distance to zero will be sqrt(9+1)=sqrt(10)...
I can't figure out an answer.
Top
Join the discussion
Legendary Member
Posts: 882
Joined: Fri Feb 20, 2009 2:57 pmFollowed by: 1 member
Thanked: 15 times
GMAT Score:690

by crackgmat007 » Thu Nov 05, 2009 10:14 am
Here is the explanation I found in another forum
The mid point between (6,2) and (0,6) is (3,4)
You know that the vertexes of the diagonal mentioned has distance of 3 in the x coordinate and distance of 2 in the y coordinate.
Since both diagonals must be perpendicular, using the perpendicular property (inverse of slope), it must be that those distances are reversed.

This means that from point (3,4), the new vertexes will have distance of 3 in the y coordinate and distance of 2 in the x coordinate.

So two other vertexes will be at:
(3-2, 4-3) = (1,1)
and
(3+2, 4+3) = (5,7)

The closer to the (0,0) is (1,1)
Distance between (0,0) and (1,1) = sqrt(2)
Top
Join the discussion