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.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
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.
09 May 2019, 20:46
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:
13% (02:09) correct
87%
(02:07)
wrong
based on 86
sessions
History
Date
Time
Result
Not Attempted Yet
A quadrilateral is inscribed in a circle of radius \(200\sqrt{2}\). Three of the sides of this quadrilateral have length 200. What is the length of the fourth side?
(A) 200
(B) \(200\sqrt{2}\)
(C) \(200\sqrt{3}\)
(D) \(300\sqrt{2}\)
(E) 500
(A) 200
(B) \(200\sqrt{2}\)
(C) \(200\sqrt{3}\)
(D) \(300\sqrt{2}\)
(E) 500
Let Angle(AOC)=x
sides AB=AC=BD= b =200 and radius, r= \(200\sqrt{2}\)
cos(x)= \(\frac{(AO^2 + OC^2 - AC^2)}{2* AO*AO}\)
cosx= \(\frac{2r^2-b^2}{2r^2}\)
cosx=\(1-(\frac{b^2}{2r^2})\), where b =200 and r= \(200\sqrt{2}\)
cosx=1-(1/4) = 3/4
Now we know angle subtended by equal arcs are equal
Hence Angle(AOC)=Angle(AOB)=Angle(BOD)=x
Then Angle(COD)= (2pi-3x)
We know cos(2pi-3x)= cos3x
cox3x=\(4cos^3 x - 3cosx\)
cox3x= \(4*(3/4)^3 - 3(3/4)\) = -(9/16)
Also, cos3x=\(\frac{(CO^2 + OD^2 - DC^2)}{2* CO*OD}\)
\(-(9/16)=1-(CD^2/2r^2)\)
(CD^2/2r^2)=25/16
CD=500
It took me 10 minutes to solve this tho. Bunuel Can you suggest other method do this problem fast?
sides AB=AC=BD= b =200 and radius, r= \(200\sqrt{2}\)
cos(x)= \(\frac{(AO^2 + OC^2 - AC^2)}{2* AO*AO}\)
cosx= \(\frac{2r^2-b^2}{2r^2}\)
cosx=\(1-(\frac{b^2}{2r^2})\), where b =200 and r= \(200\sqrt{2}\)
cosx=1-(1/4) = 3/4
Now we know angle subtended by equal arcs are equal
Hence Angle(AOC)=Angle(AOB)=Angle(BOD)=x
Then Angle(COD)= (2pi-3x)
We know cos(2pi-3x)= cos3x
cox3x=\(4cos^3 x - 3cosx\)
cox3x= \(4*(3/4)^3 - 3(3/4)\) = -(9/16)
Also, cos3x=\(\frac{(CO^2 + OD^2 - DC^2)}{2* CO*OD}\)
\(-(9/16)=1-(CD^2/2r^2)\)
(CD^2/2r^2)=25/16
CD=500
It took me 10 minutes to solve this tho. Bunuel Can you suggest other method do this problem fast?
Bunuel
Attachments
figure 3.png [ 6.79 KiB | Viewed 5110 times ]
18 Jul 2019, 03:46
Kudos
Bookmarks
Bunuel
Hi Bunuel can you please post the official explanation of this one. It seems to be really tough.
Can experts please help with this one. VeritasKarishma , chetan2u, EgmatQuantExpert
