Triangle ABE will be similar to triangle CDE because angle EAB = 90 = angle EDC, angle AEB = angle CED being vertically opposite angles and obviously angle ABE = angle ECD ( since if two angles of two triangles are same, third angle has to be same) .
So AE/ED = AB/CD.
Let AE = x. So ED = 4-x.
Or x/(4-x) = 3/9 = 1/3.
So x = 1 = AE.
So ED = 4-x = 3.
So area of triangle ACD = ½ *CD * AD = ½ *9*4 = 18.
Area of triangle ECD = ½*CD*ED = ½*9*3 = 13.5.
So area of triangle ACE = 18 - 13.5 = 4.5.
The correct answer is D.
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
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
For Geometry Lovers
Source: Beat The GMAT — Problem Solving |
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
We need the area of triangle AEC. This can be easily calculated if we know the length of AE. Then we can calculated the area of triangle ADC and subtract the area of triangle EDC. That will leave us with the area of triangle AEC. We know the area of triangle ADC from the given information. It is (AD X CD) / 2 or 18. Finding the length of AE is the tricky part. I'm sure there are multiple ways to do this but here is the only way I can think of.
First, draw a line from point B staight down to intersect with line CD (you need to extend line CD to the right). Call this point F. Now you have a big right triangle BFC. Imagine this triangle on a coordinate plane. Put point C at the origin (0,0) and point B would be (12,4). Now calculate the slope (rise/run) of line CB. This is 4/12 or 1/3. This means for every 3 you move horizontally you move one up (or down). The length of AB is 3 so that means point E must be a length of 1 from point A. In other words AE equals 1 and ED equals 3. Now you can calculate the area of triangle EDC, which is 13.5. When you subtract this from the area of triangle ADC which was 18 you get the area of AEC which is 4.5.
Is that the right answer???
First, draw a line from point B staight down to intersect with line CD (you need to extend line CD to the right). Call this point F. Now you have a big right triangle BFC. Imagine this triangle on a coordinate plane. Put point C at the origin (0,0) and point B would be (12,4). Now calculate the slope (rise/run) of line CB. This is 4/12 or 1/3. This means for every 3 you move horizontally you move one up (or down). The length of AB is 3 so that means point E must be a length of 1 from point A. In other words AE equals 1 and ED equals 3. Now you can calculate the area of triangle EDC, which is 13.5. When you subtract this from the area of triangle ADC which was 18 you get the area of AEC which is 4.5.
Is that the right answer???
For a solution to this problem, please see the attached file.
- Attachments
-
- BTG_similar triangles_problem.pdf
- (51.73 KiB) Downloaded 103 times
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3













