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 2601:30 AM EDT
-02:30 AM EDT
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more.
Apr 2512:30 AM EDT
-01:30 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 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.
May 0111:00 AM PDT
-12:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
22 Apr 2019, 01:58
Kudos
Bookmarks
A
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:
60% (02:22) correct
40%
(01:53)
wrong
based on 57
sessions
History
Date
Time
Result
Not Attempted Yet
Three circles with radius 2 are mutually tangent. What is the total area of the circles and the region bounded by them, as shown in the figure?
(A) \(10\pi + 4\sqrt{3}\)
(B) \(13\pi - \sqrt{3}\)
(C) \(12\pi + \sqrt{3}\)
(D) \(10\pi + 9\)
(E) \(13\pi\)
Attachment:
fd36c0563f1447ec842cc791ac398c4cd8f20685.png [ 14.71 KiB | Viewed 7355 times ]
22 Apr 2019, 03:07
Kudos
Bookmarks
Area of the three circles = 3*π*4 = 12π.
Connect all the radii. Gives a equilateral triangle with side 4.
Then the area of the equilateral triangle = √3/4*a^2 = 4√3.
Area of the sector of the circle, with angle 60 ( as it an equilateral triangle ) = 60/360*π*4 = 2π/3.
Area of the shaded region = 4√3-(2π/3)*3 ( for 3 sectors ) = 4√3-2π.
Total area = 12π+4√3-2π = 10+4√3
A is the answer
Connect all the radii. Gives a equilateral triangle with side 4.
Then the area of the equilateral triangle = √3/4*a^2 = 4√3.
Area of the sector of the circle, with angle 60 ( as it an equilateral triangle ) = 60/360*π*4 = 2π/3.
Area of the shaded region = 4√3-(2π/3)*3 ( for 3 sectors ) = 4√3-2π.
Total area = 12π+4√3-2π = 10+4√3
A is the answer
22 Apr 2019, 04:06
Kudos
Bookmarks
Lets first calculate the areas of the circles
\(2^2\)*pi*3=12*pi
if we connect the radius of the circles we will get an equilateral triangle with side 4
area of the equilateral triangle \(side^2\)*\(\sqrt{3}\)/4=4*\(\sqrt{3}\)
equilateral triangle has 60 grad each angle, so we can calculate what portion of the area of each cicle we have overlapped calculating the area of triangle
60/360=1/6
1/6*12*pi=2*pi
The area of the region among circles will be 4*\(\sqrt{3}\)-2*pi
total area:
12*pi+4*\(\sqrt{3}\)-2*pi=10*pi+4*\(\sqrt{3}\)
IMO
Ans: A
guys how do you write proper-looking "pi" ?
\(2^2\)*pi*3=12*pi
if we connect the radius of the circles we will get an equilateral triangle with side 4
area of the equilateral triangle \(side^2\)*\(\sqrt{3}\)/4=4*\(\sqrt{3}\)
equilateral triangle has 60 grad each angle, so we can calculate what portion of the area of each cicle we have overlapped calculating the area of triangle
60/360=1/6
1/6*12*pi=2*pi
The area of the region among circles will be 4*\(\sqrt{3}\)-2*pi
total area:
12*pi+4*\(\sqrt{3}\)-2*pi=10*pi+4*\(\sqrt{3}\)
IMO
Ans: A
guys how do you write proper-looking "pi" ?
