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 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.
11 Jan 2014, 08:16
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
64% (01:59) correct
36%
(02:01)
wrong
based on 486
sessions
History
Date
Time
Result
Not Attempted Yet
Attachment:
File comment: Geometry
ABCD.png [ 20.95 KiB | Viewed 19008 times ]
What is the area of the shaded region above, if ABCD is a square and line segment AB is a diameter of the circle with center O?ABCD.png [ 20.95 KiB | Viewed 19008 times ]
(1) The radius of the circle with center O is 4.
(2) The area of triangle ADC is 32.
OE
To find area of shaded region, you need to find area of triangle ACD, and subtract area of unshaded region inside this triangle from it. To find area of unshaded region, you need to first draw in a line perpendicular to diameter AB of circle from center O to diagonal AC. Call this point E. This point is also one of endpoints of arc AFE.
(1): This is exactly what we need; once we have radius of circle we can find diameter of circle which is same as a side of square and also same as each leg of right triangle ADC. So you can, with this information, solve for different areas necessary, and then solve for area of shaded region. Sufficient.
(2): able to find length of height and width (which are equal to each other) of triangle ADC.
Let's say that both base and height of triangle ADC = x.
32 = (½)x^2 → x = 8
You have leg lengths (which are equal) of isosceles right triangle ADC; this means that you also have diameter of circle → radius.
Sufficient
Attachment:
File comment: ABCD2
ABCD2.png [ 19.07 KiB | Viewed 16093 times ]
Arc AFE and radii OA and OE bound a sector of circle O: this sector is one quarter of entire area of circle. Now, if you take area of small triangle AEO, and subtract this area from area of sector OAFE, then you have area of unshaded portion of triangle ADC. So, to solve this question, you need to find area of triangle ADC, one quarter of area of circle O, and area of small triangle AEO. To find these three areas, you need radius of circle, or length of one side of square.ABCD2.png [ 19.07 KiB | Viewed 16093 times ]
(1): This is exactly what we need; once we have radius of circle we can find diameter of circle which is same as a side of square and also same as each leg of right triangle ADC. So you can, with this information, solve for different areas necessary, and then solve for area of shaded region. Sufficient.
(2): able to find length of height and width (which are equal to each other) of triangle ADC.
Let's say that both base and height of triangle ADC = x.
32 = (½)x^2 → x = 8
You have leg lengths (which are equal) of isosceles right triangle ADC; this means that you also have diameter of circle → radius.
Sufficient
11 Jan 2014, 08:26
Kudos
Bookmarks
goodyear2013
What is the area of the shaded region above, if ABCD is a square and line segment AB is a diameter of the circle with center O?
Actually there is no need for any calculations. From both statements we have everything "fixed", only one case possible where we know everything possible.
(1) The radius of the circle with center O is 4 --> diameter = side = 8. We have "fixed" circle and square. We can find anything there. Sufficient.
(2) The area of triangle ADC is 32 --> 1/2*AD^2=32 --> AD=8 --> AD = diameter = 8. The same info as above. Sufficient.
Answer: D.
General Discussion
11 Jan 2014, 16:55
Kudos
Bookmarks
Bunuel
Bunuel, Can you please explain why if you only have the radius you can find anything. How would you get the AF part?
I knew that Statement 1 and 2 gave the same information, but couldn't figure out how to find the white portion in TriangleADC
Thanks
)
