Forum Home > GMAT > Quantitative > Problem Solving (PS)
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.
Dec 1511:00 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.
Dec 1608:00 AM PST
-11:59 PM PST
GMAT Club 12 Days of Christmas is a 4th Annual GMAT Club Winter Competition based on solving questions. This is the Winter GMAT competition on GMAT Club with an amazing opportunity to win over $40,000 worth of prizes!- Dec 17
09:00 AM PST
-10:00 AM PST
Join Manhattan Prep instructor Whitney Garner for a fun—and thorough—review of logic-based (non-math) problems, with a particular emphasis on Data Sufficiency and Two-Parts. - Dec 18
10:00 AM PST
-11:00 AM PST
Here is the essential guide to securing scholarships as an MBA student! In this video, we explore the various types of scholarships available, including need-based and merit-based options.
07 Feb 2019, 02:51
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
70% (01:21) correct
30%
(01:41)
wrong
based on 129
sessions
History
Date
Time
Result
Not Attempted Yet
If ABCD is a rectangle, BC = x and AB = 2x, then the circumference of the circle, in terms of x, is
A. \(\sqrt{3} x\pi\)
B. \(\sqrt{5} x\pi\)
C. \(\sqrt{3x} \pi\)
D. \(\sqrt{5x} \pi\)
E. \(5x^2 \pi\)
Attachment:
Geometry_015.jpg [ 8.72 KiB | Viewed 7147 times ]
KanishkM
Joined: 09 Mar 2018
Last visit: 18 Dec 2021
Posts: 769
Own Kudos:
Given Kudos: 123
Location: India
Schools: DeGroote'21 Desautels '21 NUS '21
07 Feb 2019, 03:09
Kudos
Bookmarks
Bunuel
If ABCD is a rectangle, BC = x and AB = 2x, then the circumference of the circle, in terms of x, is
A. \(\sqrt{3} x\pi\)
B. \(\sqrt{5} x\pi\)
C. \(\sqrt{3x} \pi\)
D. \(\sqrt{5x} \pi\)
E. \(5x^2 \pi\)
If ABCD is a rectangle, lets calculate AC
AC =\(\sqrt{{x^2 + 4* x^2}}\)
AC = \(\sqrt{5}\)x
AC will be the diameter, radius will be \(\sqrt{5}\)x / 2
C = 2 \(\pi r\)
= 2 * \(\pi\)\(\sqrt{5}\)x / 2
= \(\sqrt{5} x\pi\)
eswarchethu135
07 Feb 2019, 03:12
Kudos
Bookmarks
Attachment:
image.jpg [ 28.22 KiB | Viewed 6479 times ]
From the above fig, the diagonal of the rectangle serves as the diameter of the circle as the diagonal passes through the center of both the circle and the rectangle.
Since the diameter is of \(\sqrt{5}x\), radius of the circle \(\frac{\sqrt{5}x}{2}\)
The circumference of a circle is \(2\pi r\) = \(2*\pi*\frac{\sqrt{5}x}{2}\)
\(\sqrt{5}x \pi\)
OPTION: B
