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 Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 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
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
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 Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget 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.

Isosceles Right Triangle

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 399
Joined: Wed Apr 15, 2009 3:48 am
Location: india
Thanked: 39 times

by xcusemeplz2009 » Tue Oct 13, 2009 12:22 am
h is 16

h^2=2*a^2=>h=asqrt(2)

2a+h=16+16sqrt(2)

2a+sqrt(2)=16+16sqrt(2)

solving a=8*sqrt(2)
h=8*sqrt2*sqrt2=16
It does not matter how many times you get knocked down , but how many times you get up
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 124
Joined: Thu Jun 18, 2009 5:33 am
Thanked: 35 times

by NikolayZ » Tue Oct 13, 2009 12:23 am
Hey there !

Since you have an isosceles triangle, its perimeter = 2x+y. ( 2 sides are x, and the hypotenuse is y)
so the perimeter is
16+16sqrt(2)=2x+y.
Also the hypotenuse equals xsqrt(2), since it is 45 degree isosceles right triangle.

16+16sqrt(2)=2x+xsqrt(2)

x= (16+16*sqrt(2)/(2+sqrt(2))
Then , hypotenuse will be x*sqrt(2)= (16*sqrt(2)+32)/(sqrt(2)+2)=16.
Last edited by NikolayZ on Tue Oct 13, 2009 8:52 am, edited 1 time in total.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 869
Joined: Wed Aug 26, 2009 3:49 pmFollowed by: 3 members
Location: California
Thanked: 13 times

by heshamelaziry » Tue Oct 13, 2009 8:45 am
NikolayZ wrote:Hey there !

Since you have an isosceles triangle, its perimeter = 2x+y. ( 2 sides are x, and the hypotenuse is y)
so the perimeter is
16+16sqrt(2)=2x+y.
Also the hypotenuse equals xsqrt(2), since it is 45 degree isosceles right triangle.

16+16sqrt(2)=2x+xsqrt(2)

x= (16+16*sqrt(2)/(2+2*sqrt(2))
Then , hypotenuse will be x*sqrt(2)= (16*sqrt(2)+32)/(sqrt(2)+2)=16.
Last edited by heshamelaziry on Tue Oct 13, 2009 9:05 am, edited 2 times in total.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 124
Joined: Thu Jun 18, 2009 5:33 am
Thanked: 35 times

by NikolayZ » Tue Oct 13, 2009 8:51 am
Sorry mate. "typo happens" (c) =) corrected.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 869
Joined: Wed Aug 26, 2009 3:49 pmFollowed by: 3 members
Location: California
Thanked: 13 times

by heshamelaziry » Tue Oct 13, 2009 8:56 am
NikolayZ wrote:Sorry mate. "typo happens" (c) =) corrected.
I got it, finally. Thanks a lot.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 124
Joined: Thu Jun 18, 2009 5:33 am
Thanked: 35 times

by NikolayZ » Tue Oct 13, 2009 9:13 am
heshamelaziry, truly there is a flaw in my solvent, because it is not consistent =(

so, right answer to my mind is

2x+x*sqrt(2)=16+16*sqrt(2)
x(2+sqrt(2)=16+16*sqrt(2)
x=(16+16*sqrt(2))/2+sqrt(2)
x*sqrt(2)=(16*sqrt(2)+16*sqrt(2)*sqrt(2))/2+sqrt(2)
==> since hte x*sqrt(2) is the hypotenuse (h)
then
h=(16*sqrt(2)+32)/2+sqrt(2)
factorizing 16 from the numerator we'll get
16(sqrt(2)+2)/(sqrt(2)+2)=16.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 869
Joined: Wed Aug 26, 2009 3:49 pmFollowed by: 3 members
Location: California
Thanked: 13 times

by heshamelaziry » Tue Oct 13, 2009 9:25 am
NikolayZ wrote:heshamelaziry, truly there is a flaw in my solvent, because it is not consistent =(

so, right answer to my mind is

2x+x*sqrt(2)=16+16*sqrt(2)
x(2+sqrt(2)=16+16*sqrt(2)
x=(16+16*sqrt(2))/2+sqrt(2)
x*sqrt(2)=(16*sqrt(2)+16*sqrt(2)*sqrt(2))/2+sqrt(2)
==> since hte x*sqrt(2) is the hypotenuse (h)
then
h=(16*sqrt(2)+32)/2+sqrt(2)
factorizing 16 from the numerator we'll get
16(sqrt(2)+2)/(sqrt(2)+2)=16.

Thanks Dude :) I posted some problems form a GMATPrep test that I took. Maybe you like to take a shot at them. I posted them yesterday in an attachement.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 7
Joined: Tue Oct 20, 2009 5:59 pm
Location: North Carolina

by kkpatel1 » Fri Nov 20, 2009 2:16 pm
heshamelaziry wrote:
NikolayZ wrote:heshamelaziry, truly there is a flaw in my solvent, because it is not consistent =(

so, right answer to my mind is

2x+x*sqrt(2)=16+16*sqrt(2)
x(2+sqrt(2)=16+16*sqrt(2)
x=(16+16*sqrt(2))/2+sqrt(2)
x*sqrt(2)=(16*sqrt(2)+16*sqrt(2)*sqrt(2))/2+sqrt(2)
==> since hte x*sqrt(2) is the hypotenuse (h)
then
h=(16*sqrt(2)+32)/2+sqrt(2)
factorizing 16 from the numerator we'll get
16(sqrt(2)+2)/(sqrt(2)+2)=16.

Where I am confused is:
When you set:
x*sqrt(2)=[(16*sqrt(2)+16*sqrt(2)*sqrt(2)]/2+sqrt(2)

why dont you multiply the bottom portion by sqrt(2) also, since the rule is, if you multiply one thing by something, you have to do it to everything!??

therefore shouldnt it be: h=(16*sqrt(2)+32)/2+2 ??
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1