|
|
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.
23 Aug 2019, 10:54
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
87% (02:17) correct
13%
(04:19)
wrong
based on 15
sessions
History
Date
Time
Result
Not Attempted Yet
From a rectangle ABCD of area 768 sq cm, a semicircular part with diameter AB and area 72π sq cm is removed. The perimeter of the leftover portion, in cm, is
A. 88 + 12π
B. 80 + 16π
C. 86 + 8π
D. 82 + 24π
E. 80 + 8π
A. 88 + 12π
B. 80 + 16π
C. 86 + 8π
D. 82 + 24π
E. 80 + 8π
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
23 Aug 2019, 12:51
Kudos
Bookmarks
\((\frac{1}{2})π*(AB/2)^2= 72π\)
\(AB^2=72*8\)
AB=24
BC= 768/24=32
The perimeter of the leftover portion= 2*BC+AB+π*(AB/2)= 2*32+24+12π= 88+12π
A
\(AB^2=72*8\)
AB=24
BC= 768/24=32
The perimeter of the leftover portion= 2*BC+AB+π*(AB/2)= 2*32+24+12π= 88+12π
A
AbdulMalikVT
Attachments
Untitled.png [ 7.29 KiB | Viewed 1778 times ]
27 Aug 2019, 01:35
Kudos
Bookmarks
nick1816
But doesn't 2xBC+AB gives you perimeter of the entire rectangle ? (including the cut out part)
