Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
12 Oct 2020, 01:00
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
31% (02:28) correct
69%
(03:21)
wrong
based on 36
sessions
History
Date
Time
Result
Not Attempted Yet
The square ABCD above has a side length equal to 1. If ∠EAF = 45°, and ∠AFB = 70°. What is the perimeter of triangle ECF?
A. 1.2
B. 1.4
C. 1.6
D. 1.8
E. 2
Are You Up For the Challenge: 700 Level Questions
ramlala
Joined: 22 Aug 2020
Last visit: 13 Dec 2022
Posts: 467
Own Kudos:
Given Kudos: 30
Location: India
Concentration: International Business, Finance
Schools: IIMA PGPX'23 INSEAD Jan'23 intake Sloan (MIT) ISB HEC Said
GPA: 4
WE:Project Management (Energy)
13 Oct 2020, 12:18
Kudos
Bookmarks
From above EC + FC = 1.
EC slightly greater than FC.
EF = sqrt (EC^2 + FC^2)
= sqrt (0.5^2 + 0.5^2)
= sqrt (0.5)
= 0.7 as EC & FC are not same (actual will be higher)
So, EC + FC + EF = 1 + 0.7 > 1.7 = 1.8
Ans D
EC slightly greater than FC.
EF = sqrt (EC^2 + FC^2)
= sqrt (0.5^2 + 0.5^2)
= sqrt (0.5)
= 0.7 as EC & FC are not same (actual will be higher)
So, EC + FC + EF = 1 + 0.7 > 1.7 = 1.8
Ans D
06 Nov 2020, 04:24
Kudos
Bookmarks
Drop a perpendicular line down from F to AE (call that point X) to create two triangles inside the triangle (AFX and FXE). One of which is an isosceles right triangle (due to the 45 angle). The rest is trigonometry. You can determine AF using sin function. Then the perpendicular is AF/sqrt(2). Then you can solve for AE by using the same sin trick. Then you can determine XE by subtracting AX fro AE. Now you know the legs of the inside triangle FXE. You can determine angle AEF by inverse tan function which turns out to be 65 degrees. Then you know angle FEC which is 50 and angle CFE which is 40. FE can be found out using pythogorean theorem or trigonometry. Once you know FE you can use trigonometry for FC and EC and the sum will turn out to be two. I did it on a calculator but I’m sure that if I looked up some trig identities things would cross out nicely and I’d end up with a clean solution without using a calculator.
That’s the overall gist of how i solved it.
I’ll come back and put a more elegant solution if I have time.
That’s the overall gist of how i solved it.
I’ll come back and put a more elegant solution if I have time.
