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 3001:30 AM EDT
-02:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 2912:30 AM EDT
-01:30 AM EDT
Learn how Kamakshi achieved a GMAT 675 with an impressive 96th %ile in Data Insights. Discover the unique methods and exam strategies that helped her excel in DI along with other sections for a balanced and high score.
May 0111:00 AM PDT
-12:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
May 0306:30 AM PDT
-08:30 AM PDT
Verbal trouble on GMAT? Fix it NOW! Join Sunita Singhvi for a focused webinar on actionable strategies to boost your Verbal score and take your performance to the next level.
04 Feb 2018, 12:41
Kudos
Bookmarks
D
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:
55% (02:51) correct
45%
(02:55)
wrong
based on 88
sessions
History
Date
Time
Result
Not Attempted Yet
In the figure above, AB and CD are diameters of the circle with centre as O, AEB is arc of circle with centre as D and AFB is an arc of the circle with centre as C. If AB = 20cm, what is the area of the shaded region?
(A) 50
(B) 100
(C) 150
(D) 200
(E) 250
(A) 50
(B) 100
(C) 150
(D) 200
(E) 250
Attachments
2018-02-05 00_59_51-Microsoft Office 2010.png [ 311.91 KiB | Viewed 10822 times ]
04 Feb 2018, 19:17
Kudos
Bookmarks
DHAR
lets find dia of arc AEB..
Dia of circle is 20 and triangle ADB will be right isosceles triangle with angle D as 90..
therefore AD or BD will be \(\frac{20}{\sqrt{2}}=10\sqrt{2}\), this is same as RADIUS of the arc ADB and arc AFB
Now area of unshaded portion AEBO..
area of AEBO = area of segment AEBD - area of triangle ABD..
\(pi*(10\sqrt{2})^2*\frac{90}{360}-\frac{1}{2}*10\sqrt{2}*10\sqrt{2}=50pi-100\)..
so the total unshaded portion's area = 2*(50pi-100)=100pi-200
total area of circle = \(pi*10^2=100pi\)
so area of shaded region = 100pi-(100pi-200)=200
D
Attachments
2018-02-05 00_59_51-Microsoft Office 2010.png [ 260.77 KiB | Viewed 10700 times ]
rahul16singh28
Joined: 31 Jul 2017
Last visit: 09 Jun 2020
Posts: 428
Given Kudos: 752
Location: Malaysia
Schools: INSEAD Jan '19
GPA: 3.95
WE:Consulting (Energy)
05 Feb 2018, 01:57
Kudos
Bookmarks
DHAR
Good Question... Kudos to you.. This question tests many Concepts.
For the arc AEB with Centre D, AD = ED = R (Radius of Bigger Circle in for which AEB is an arc).
r = Radius of smaller circle with Centre O.
Now, as AB is the Diameter of Circle with centre O, it will subtend 90 Degree at point C and D.
From Triangle BOD, we have -
\(R^2 = r^2 + r^2\) -----> \(R = \sqrt{2}r.\)
Now, lets find the Area of AEBOA = Area of SECTOR with Centre D - Area of Triangle ADB
Area of AEBFA = \(\frac{90}{360}*pi*R^2 - \frac{1}{2}*2r*r = pi*\frac{r^2}{2} - r^2\)
Now, multiply this area with 2 to get the total area of arc AEBFA = \(pi*r^2 - 2r^2\)
So, area of shaded region = Area of circle with Centre O - Area of unshaded region AEBFA = \(pi*r^2 - (pi*r^2 - 2r^2)\)
Area of shaded region = \(2r^2\) = 200.
