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

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Polygon Question

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 641
Joined: Tue Feb 14, 2012 3:52 pmFollowed by: 8 members
Thanked: 11 times
by gmattesttaker2 » Sat Aug 11, 2012 4:55 pm
Hello,

Can you please help with this problem:

ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picture. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?

Thanks a lot.

Best Regards,
Sri
Attachments
10_img.png
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 363
Joined: Sun Oct 17, 2010 3:24 pmFollowed by: 3 members
Thanked: 115 times

by theCEO » Sat Aug 11, 2012 5:37 pm
gmattesttaker2 wrote:Hello,

Can you please help with this problem:

ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picture. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?

Thanks a lot.

Best Regards,
Sri

Area of ABCD = area of frame + area of EFGH
area of ABCD = 2 x area of EFGH
6X6 = 2 X area of EFGH
18 = area of EFGH
18 = (EF)^2
EF = sqrt (18) = sqrt (2x9) = 3sqrt(2)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 641
Joined: Tue Feb 14, 2012 3:52 pmFollowed by: 8 members
Thanked: 11 times

by gmattesttaker2 » Sat Aug 11, 2012 10:13 pm
theCEO wrote:
gmattesttaker2 wrote:Hello,

Can you please help with this problem:

ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picture. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?

Thanks a lot.

Best Regards,
Sri

Area of ABCD = area of frame + area of EFGH
area of ABCD = 2 x area of EFGH
6X6 = 2 X area of EFGH
18 = area of EFGH
18 = (EF)^2
EF = sqrt (18) = sqrt (2x9) = 3sqrt(2)
Hello theCEO,

Thanks for your reply. I was just wondering how you got :

area of ABCD = 2 x area of EFGH

Thanks.

Best Regards,
Sri
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 » Sun Aug 12, 2012 11:52 am
gmattesttaker2 wrote:
theCEO wrote:
gmattesttaker2 wrote:Hello,

Can you please help with this problem:

ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picture. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?

Thanks a lot.

Best Regards,
Sri

Area of ABCD = area of frame + area of EFGH
area of ABCD = 2 x area of EFGH
6X6 = 2 X area of EFGH
18 = area of EFGH
18 = (EF)^2
EF = sqrt (18) = sqrt (2x9) = 3sqrt(2)
Hello theCEO,

Thanks for your reply. I was just wondering how you got :

area of ABCD = 2 x area of EFGH

Thanks.

Best Regards,
Sri
How to get area of ABCD = 2 x area of EFGH

Area of ABCD = area of frame + area of EFGH
The question says the area of EFGH is equal to the area of the picture frame
therefore area of frame = area of EFGH
therefore we can substitute the area of the frame with area of EFGH since they are the same

Area of ABCD = area of frame + area of EFGH
Area of ABCD = area of EFGH + area of EFGH = 2 x area of EFGH

Let me know if this helps!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 641
Joined: Tue Feb 14, 2012 3:52 pmFollowed by: 8 members
Thanked: 11 times

by gmattesttaker2 » Sun Aug 12, 2012 5:39 pm
theCEO wrote:
gmattesttaker2 wrote:
theCEO wrote:
gmattesttaker2 wrote:Hello,

Can you please help with this problem:

ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picture. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?

Thanks a lot.

Best Regards,
Sri

Area of ABCD = area of frame + area of EFGH
area of ABCD = 2 x area of EFGH
6X6 = 2 X area of EFGH
18 = area of EFGH
18 = (EF)^2
EF = sqrt (18) = sqrt (2x9) = 3sqrt(2)
Hello theCEO,

Thanks for your reply. I was just wondering how you got :

area of ABCD = 2 x area of EFGH

Thanks.

Best Regards,
Sri
How to get area of ABCD = 2 x area of EFGH

Area of ABCD = area of frame + area of EFGH
The question says the area of EFGH is equal to the area of the picture frame
therefore area of frame = area of EFGH
therefore we can substitute the area of the frame with area of EFGH since they are the same

Area of ABCD = area of frame + area of EFGH
Area of ABCD = area of EFGH + area of EFGH = 2 x area of EFGH

Let me know if this helps!
Hello theCEO,

Thanks for the explanation again. It is clear now.

Best Regards,
Sri
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 » Sun Aug 12, 2012 6:56 pm
deleted!
Last edited by theCEO on Sun Aug 12, 2012 6:59 pm, edited 1 time in total.
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 » Sun Aug 12, 2012 6:58 pm
gmattesttaker2 wrote:
theCEO wrote:
gmattesttaker2 wrote:
theCEO wrote:
gmattesttaker2 wrote:Hello,

Can you please help with this problem:

ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picture. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?

Thanks a lot.

Best Regards,
Sri

Area of ABCD = area of frame + area of EFGH
area of ABCD = 2 x area of EFGH
6X6 = 2 X area of EFGH
18 = area of EFGH
18 = (EF)^2
EF = sqrt (18) = sqrt (2x9) = 3sqrt(2)
Hello theCEO,

Thanks for your reply. I was just wondering how you got :

area of ABCD = 2 x area of EFGH

Thanks.

Best Regards,
Sri
How to get area of ABCD = 2 x area of EFGH

Area of ABCD = area of frame + area of EFGH
The question says the area of EFGH is equal to the area of the picture frame
therefore area of frame = area of EFGH
therefore we can substitute the area of the frame with area of EFGH since they are the same

Area of ABCD = area of frame + area of EFGH
Area of ABCD = area of EFGH + area of EFGH = 2 x area of EFGH

Let me know if this helps!
Hello theCEO,

Thanks for the explanation again. It is clear now.

Best Regards,
Sri
Your welcome Sri!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1