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.
May 2110:00 AM EDT
-11:59 PM EDT
Make the most of your break with the most realistic GMAT™ prep. Take up to $700 off select products.
May 2211:00 AM EDT
-11:59 PM EDT
Take 30% off all study plans with code PICNIC - Expires on Monday, May 25th
May 2512:01 AM PDT
-11:59 PM PDT
On the 25th, GMAT Club Tests will be Free! Including all the Quizzes, Questions, and Tests. 12 AM - 11:59 PM PST- Jun 10
06:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
21 Jul 2022, 08:50
Kudos
Bookmarks
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:
44% (03:45) correct
56%
(02:45)
wrong
based on 18
sessions
History
Date
Time
Result
Not Attempted Yet
A Semicircle is inscribed in a right Triangle as shown in figure. Calculate the value of \(X?\)
a) \(\frac{1}{4}\)
b) \(\frac{2}{7}\)
c) \(\frac{3}{7}\)
d) \(\frac{1}{2}\)
e) \(\frac{4}{7}\)
a) \(\frac{1}{4}\)
b) \(\frac{2}{7}\)
c) \(\frac{3}{7}\)
d) \(\frac{1}{2}\)
e) \(\frac{4}{7}\)
Attachments
Screenshot 2022-07-21 210930.png [ 44.13 KiB | Viewed 1656 times ]
24 Jul 2022, 02:06
Kudos
Bookmarks
figjam786
Join Center of the semicircle, D, of say radius r to side BC at E and AC at F.
As DE and DF are tangents, angles BED and BFD are 90, making BEDF a square with side of length r.
Now take \(\triangle CDE\) and \(\triangle CAB\)
They are similar triangles, so \(\frac{DE}{AB}=\frac{CE}{CB}\)
=> \(\frac{r}{4}=\frac{3-r}{3}…….3r=12-4r…….r=\frac{12}{7}\)
Once again take \(\triangle CDE\) to find x
\((3-r)^2+r^2=(r+x)^2………(3-\frac{12}{7})^2+\frac{12}{7}^2=(r+x)^2\)
\(\frac{9}{7}^2+\frac{12}{7}^2=(r+x)^2\)
\(\frac{225}{49}=(r+x)^2…….\frac{15}{7}^2=(r+x)^2\)
\(\frac{15}{7}=r+x=\frac{12}{7}+x\)
\(x=\frac{3}{7}\)
C
Attachments
CE7AF80A-385C-4CB5-82F7-8D424CC6C4A6.jpeg [ 19.19 KiB | Viewed 1466 times ]
21 Jul 2025, 21:07
Kudos
Bookmarks
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.
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.
