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 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
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
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
16 Sep 2014, 01:50
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:
57% (03:19) correct
43%
(02:32)
wrong
based on 140
sessions
History
Date
Time
Result
Not Attempted Yet
In a rectangular coordinate system, point \(A\) has coordinates \((d, d)\), where \(d \gt 0\). Point \(A\) and the origin form the endpoints of a diameter of circle \(C\). What fraction of the area of circle \(C\) lies within the first quadrant?
A. \(\frac{\pi}{\pi + \sqrt{2}}\)
B. \(\frac{\pi}{\pi + 1}\)
C. \(\sqrt{\frac{2}{\pi}}\)
D. \(\frac{\pi + 2}{2 \pi}\)
E. \(\frac{2 \pi}{2 \pi + 1}\)
A. \(\frac{\pi}{\pi + \sqrt{2}}\)
B. \(\frac{\pi}{\pi + 1}\)
C. \(\sqrt{\frac{2}{\pi}}\)
D. \(\frac{\pi + 2}{2 \pi}\)
E. \(\frac{2 \pi}{2 \pi + 1}\)
16 Sep 2014, 01:50
Kudos
Bookmarks
Official Solution:
In a rectangular coordinate system, point \(A\) has coordinates \((d, d)\), where \(d \gt 0\). Point \(A\) and the origin form the endpoints of a diameter of circle \(C\). What fraction of the area of circle \(C\) lies within the first quadrant?
A. \(\frac{\pi}{\pi + \sqrt{2}}\)
B. \(\frac{\pi}{\pi + 1}\)
C. \(\sqrt{\frac{2}{\pi}}\)
D. \(\frac{\pi + 2}{2 \pi}\)
E. \(\frac{2 \pi}{2 \pi + 1}\)
First, draw the coordinate plane and add in point \(A\), which has coordinates \((d,d)\). Since \(d\) is greater than 0, this point lies in the first quadrant, as shown:
Notice that the origin, the point \((0, d)\), the pint \((d, 0)\), and point \(A\) \((d,d)\) form the corners of a square:
Now, we draw the line segment between \((0,0)\) and \((d,d)\). This is the diameter of a circle, which we also draw:
Notice that the answer does not depend on \(d\). All that matters is that the circle contains a square. The fraction of the circle's area in the first quadrant will be the same, no matter what. Thus, we can drop \(d\) as a variable and create a new variable \(r\) for the radius of the circle. It will be easier to compute areas in terms of \(r\) (which will also cancel out in the end).
The area of the circle is just \(\pi r^2\).
The fraction we are looking for is this:
So we need the shaded area of the circle, which consists of the square (
) and 2 of the four "leaves of the table" (
).
The square's area can be found this way:
\(\text{Area} = (\sqrt{2}r)^2 = 2 r^2\)
Now the area of the 4 leaves (
) is the area of the circle (
) minus the area of the square (). So the area of the 4 leaves is \(\pi r^2 - 2r^2 = (\pi - 2)r^2\).
This means that the area of 2 leaves is \(\frac{\pi - 2}{2} r^2\).
The whole shaded area is thus \(2 r^2 + \leftarrow( \frac{\pi - 2}{2} \rightarrow)r^2 = \leftarrow( 2 + \frac{\pi}{2} - \frac{2}{2} \rightarrow)r^2 = \leftarrow( 1 + \frac{\pi}{2} \rightarrow)r^2\).
Finally, the desired fraction is \(\frac{\leftarrow( 1 + \frac{\pi}{2} \rightarrow)r^2}{\pi r^2} = \frac{\leftarrow( 1 + \frac{\pi}{2} \rightarrow)X^2}{\pi X^2} = \frac{2 + \pi}{2 \pi}\).
Answer: D
In a rectangular coordinate system, point \(A\) has coordinates \((d, d)\), where \(d \gt 0\). Point \(A\) and the origin form the endpoints of a diameter of circle \(C\). What fraction of the area of circle \(C\) lies within the first quadrant?
A. \(\frac{\pi}{\pi + \sqrt{2}}\)
B. \(\frac{\pi}{\pi + 1}\)
C. \(\sqrt{\frac{2}{\pi}}\)
D. \(\frac{\pi + 2}{2 \pi}\)
E. \(\frac{2 \pi}{2 \pi + 1}\)
First, draw the coordinate plane and add in point \(A\), which has coordinates \((d,d)\). Since \(d\) is greater than 0, this point lies in the first quadrant, as shown:
Notice that the origin, the point \((0, d)\), the pint \((d, 0)\), and point \(A\) \((d,d)\) form the corners of a square:
Now, we draw the line segment between \((0,0)\) and \((d,d)\). This is the diameter of a circle, which we also draw:
Notice that the answer does not depend on \(d\). All that matters is that the circle contains a square. The fraction of the circle's area in the first quadrant will be the same, no matter what. Thus, we can drop \(d\) as a variable and create a new variable \(r\) for the radius of the circle. It will be easier to compute areas in terms of \(r\) (which will also cancel out in the end).
The area of the circle is just \(\pi r^2\).
The fraction we are looking for is this:
So we need the shaded area of the circle, which consists of the square (
) and 2 of the four "leaves of the table" ().The square's area can be found this way:
\(\text{Area} = (\sqrt{2}r)^2 = 2 r^2\)
Now the area of the 4 leaves (
) is the area of the circle () minus the area of the square (). So the area of the 4 leaves is \(\pi r^2 - 2r^2 = (\pi - 2)r^2\).This means that the area of 2 leaves is \(\frac{\pi - 2}{2} r^2\).
The whole shaded area is thus \(2 r^2 + \leftarrow( \frac{\pi - 2}{2} \rightarrow)r^2 = \leftarrow( 2 + \frac{\pi}{2} - \frac{2}{2} \rightarrow)r^2 = \leftarrow( 1 + \frac{\pi}{2} \rightarrow)r^2\).
Finally, the desired fraction is \(\frac{\leftarrow( 1 + \frac{\pi}{2} \rightarrow)r^2}{\pi r^2} = \frac{\leftarrow( 1 + \frac{\pi}{2} \rightarrow)X^2}{\pi X^2} = \frac{2 + \pi}{2 \pi}\).
Answer: D
