Last visit was: 26 Apr 2026, 10:49 It is currently 26 Apr 2026, 10:49
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
705-805 (Hard)|   Geometry|      
User avatar
nick1816
User avatar
Retired Moderator
Joined: 19 Oct 2018
Last visit: 12 Mar 2026
Posts: 1,841
Own Kudos:
8,511
 [4]
Given Kudos: 707
Location: India
Posts: 1,841
Kudos: 8,511
 [4]
Line segments are drawn from the vertices of the large square to the m 10 May 2020, 06:18
1
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

85% (hard)

Question Stats:

41% (02:17) correct 59% (03:10) wrong based on 34 sessions

History

Date
Time
Result
Not Attempted Yet
Attachment:
6c4cOJzjJV-group-57.png
6c4cOJzjJV-group-57.png [ 13.1 KiB | Viewed 2566 times ]
Line segments are drawn from the vertices of the large square to the midpoints of the opposite sides to form a smaller, white square. If each red or blue line-segment measures 10 m long, what is the area of the smaller, white square?

A. 64
B. 75
C. 80
D. 81
E. 90
ShowHide Answer
Official Answer
C
User avatar
preetamsaha
User avatar
Joined: 14 Oct 2019
Last visit: 18 Jan 2025
Posts: 337
Own Kudos:
563
 [2]
Given Kudos: 127
Status:Today a Reader; Tomorrow a Leader.
Location: India
GPA: 4
WE:Engineering (Energy)
Posts: 337
Kudos: 563
 [2]
Line segments are drawn from the vertices of the large square to the m 10 May 2020, 12:00
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
nick1816
Attachment:
The attachment 6c4cOJzjJV-group-57.png is no longer available
Line segments are drawn from the vertices of the large square to the midpoints of the opposite sides to form a smaller, white square. If each red or blue line-segment measures 10 m long, what is the area of the smaller, white square?

A. 64
B. 75
C. 80
D. 81
E. 90
Show more


correct answer C
follow the attachment. solution is in details.
Attachments

geometry.jpg
geometry.jpg [ 1.82 MiB | Viewed 2485 times ]

User avatar
nick1816
User avatar
Retired Moderator
Joined: 19 Oct 2018
Last visit: 12 Mar 2026
Posts: 1,841
Own Kudos:
8,511
 [2]
Given Kudos: 707
Location: India
Posts: 1,841
Kudos: 8,511
 [2]
Line segments are drawn from the vertices of the large square to the m 10 May 2020, 23:01
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Attachment:
6c4cOJzjJV-group-57.png
6c4cOJzjJV-group-57.png [ 23.32 KiB | Viewed 2421 times ]
Triangle ABC is similar to triangle ADE

Hence, \(\frac{AC}{CE} = \frac{AB}{BD}\)

AC=CE = 2x

Also,
\(\frac{AB}{AD} = \frac{BC}{ED}\)

We get BC = x = EF

AC+CE+EF = AF

\(5x = \sqrt{20^2+10^2}\)

\(2x = 4\sqrt{5}\)

\(CE^2=(2x)^2 = 16*5=80\)
User avatar
SiddharthR
User avatar
Joined: 22 Oct 2018
Last visit: 20 Feb 2022
Posts: 84
Own Kudos:
39
Given Kudos: 201
Location: United States (TX)
Concentration: Finance, Technology
Schools: McCombs Ross (Michigan)
GMAT 1: 590 Q42 V29
GMAT 2: 650 Q47 V33
GPA: 3.7
WE:Engineering (Consumer Electronics)
Schools: McCombs Ross (Michigan)
GMAT 2: 650 Q47 V33
Posts: 84
Kudos: 39
Line segments are drawn from the vertices of the large square to the m 19 May 2020, 09:38
Kudos
Add Kudos
Bookmarks
Bookmark this Post
nick1816
Attachment:
6c4cOJzjJV-group-57.png
Triangle ABC is similar to triangle ADE

Hence, \(\frac{AC}{CE} = \frac{AB}{BD}\)

AC=CE = 2x

Also,
\(\frac{AB}{AD} = \frac{BC}{ED}\)

We get BC = x = EF

AC+CE+EF = AF

\(5x = \sqrt{20^2+10^2}\)

\(2x = 4\sqrt{5}\)

\(CE^2=(2x)^2 = 16*5=80\)
Show more



When you say triangles ABC & ADE are similar, shouldn't the proportionality be AB/AD = AC/AE and not AB/BD = AC/CE;

The reason is BD and CE are not sides of a triangle from your figure. Can you or someone please clarify ??

Thank you.
User avatar
nick1816
User avatar
Retired Moderator
Joined: 19 Oct 2018
Last visit: 12 Mar 2026
Posts: 1,841
Own Kudos:
8,511
Given Kudos: 707
Location: India
Posts: 1,841
Kudos: 8,511
Line segments are drawn from the vertices of the large square to the m 19 May 2020, 10:55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
\(\frac{AD}{AB}= \frac{AE}{AC} =k\)

\(\frac{BD}{CE} = \frac{AD-AB}{AE-AC} = \frac{AB*k-AB}{ AC*k - AC} = \frac{AB(k-1)}{AC(k-1)} = \frac{AB}{AC }\)

\(\frac{BD}{CE}= \frac{AB}{AC }\)

\(\frac{AB}{BD} = \frac{AC}{CE}\)

OR we could apply Dividendo (Ratios property) to prove it.

OR The Triangle Proportionality Theorem states that if a line parallel to one side of a triangle intersects the other two sides, then it divides those sides proportionally.





SiddharthR


When you say triangles ABC & ADE are similar, shouldn't the proportionality be AB/AD = AC/AE and not AB/BD = AC/CE;

The reason is BD and CE are not sides of a triangle from your figure. Can you or someone please clarify ??

Thank you.
Show more
User avatar
SiddharthR
User avatar
Joined: 22 Oct 2018
Last visit: 20 Feb 2022
Posts: 84
Own Kudos:
39
Given Kudos: 201
Location: United States (TX)
Concentration: Finance, Technology
Schools: McCombs Ross (Michigan)
GMAT 1: 590 Q42 V29
GMAT 2: 650 Q47 V33
GPA: 3.7
WE:Engineering (Consumer Electronics)
Schools: McCombs Ross (Michigan)
GMAT 2: 650 Q47 V33
Posts: 84
Kudos: 39
Line segments are drawn from the vertices of the large square to the m 21 May 2020, 20:42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
nick1816
\(\frac{AD}{AB}= \frac{AE}{AC} =k\)

\(\frac{BD}{CE} = \frac{AD-AB}{AE-AC} = \frac{AB*k-AB}{ AC*k - AC} = \frac{AB(k-1)}{AC(k-1)} = \frac{AB}{AC }\)

\(\frac{BD}{CE}= \frac{AB}{AC }\)

\(\frac{AB}{BD} = \frac{AC}{CE}\)

OR we could apply Dividendo (Ratios property) to prove it.

OR The Triangle Proportionality Theorem states that if a line parallel to one side of a triangle intersects the other two sides, then it divides those sides proportionally.





SiddharthR


When you say triangles ABC & ADE are similar, shouldn't the proportionality be AB/AD = AC/AE and not AB/BD = AC/CE;

The reason is BD and CE are not sides of a triangle from your figure. Can you or someone please clarify ??

Thank you.
Show more



That's interesting. Thank you for that insight. Could you please also tell me how you arrived at length "2x" for AC & CE ?

I got the following - AB/AD = 10/20 = 1/2;
Let BC = x. Then BC/DE = 1/2; From this we get DE = 2x

From here, I am unable to understand how you reached 2x distance for AC & CE.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,989
Own Kudos:
1,118
Posts: 38,989
Kudos: 1,118
Line segments are drawn from the vertices of the large square to the m 29 Jul 2024, 13:40
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Bunuel
Math Expert
109839 posts
Krunaal
Tuck School Moderator
852 posts