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.
Sep 1507:00 PM EDT
-09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 1712:30 AM EDT
-01:30 AM EDT
Thought your job leaves no time for GMAT prep? After multiple GMAT attempts and a demanding consulting job, Kshitij improved to 695 in under 2 months. Learn the mindset shifts, study strategy, and execution changes that made the breakthrough possible.
Sep 1801:30 AM EDT
-02:30 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.
Sep 1808:00 PM PDT
-09:00 PM PDT
Video explanations + diagnosis of 10 weakest areas + 150+ short videos + study plan!
Sep 2107:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included!
14 May 2019, 23:45
Kudos
Bookmarks
A
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:
45% (02:29) correct
55%
(02:57)
wrong
based on 40
sessions
History
Date
Time
Result
Not Attempted Yet
The figure below shows 13 circles of radius 1 within a larger circle. All the intersections occur at points of tangency. What is the area of the region, shaded in the figure, inside the larger circle but outside all the circles of radius 1 ?
A. \(4\pi\sqrt{3}\)
B. \(7\pi\)
C. \(\pi(3\sqrt{3} + 2)\)
D. \(10\pi(\sqrt{3} - 1)\)
E. \(\pi(\sqrt{3} + 6)\)
fc485e1337895925d85375cce1a9d3908ca8b7e3.png [ 42.31 KiB | Viewed 6215 times ]
A. \(4\pi\sqrt{3}\)
B. \(7\pi\)
C. \(\pi(3\sqrt{3} + 2)\)
D. \(10\pi(\sqrt{3} - 1)\)
E. \(\pi(\sqrt{3} + 6)\)
Attachment:
fc485e1337895925d85375cce1a9d3908ca8b7e3.png [ 42.31 KiB | Viewed 6215 times ]
16 May 2019, 00:21
Kudos
Bookmarks
For the radius of the bigger circle:
Centre smaller circle with two adjacent circles and 4th circle touching the bigger circle's circumference is taken..
Now since all are tangent.. we will have a right angled triangle of hypotenuse 2, and base of 1.. so altitude will be. 3^(1/2).The radius of the bigger circle will be 2* (3^(1/2)) +1
Now shaded area = area of big circle - area of smaller circles
=option A
Posted from my mobile device
Centre smaller circle with two adjacent circles and 4th circle touching the bigger circle's circumference is taken..
Now since all are tangent.. we will have a right angled triangle of hypotenuse 2, and base of 1.. so altitude will be. 3^(1/2).The radius of the bigger circle will be 2* (3^(1/2)) +1
Now shaded area = area of big circle - area of smaller circles
=option A
Posted from my mobile device
11 Jan 2022, 22:24
Kudos
Bookmarks
The four circles can be viewed as the sides of an equilateral as shown in the figure below.
The center of the bigger circle, is the centroid of the equilateral triangle.
Hence, the length of AO is \(\frac{\sqrt 3 * S}{ 3}\)
S - Side of the equilateral triangle
The side of the equilateral triangle is equal to the sum of diameter of two circles and radius of the other two circles where the vertex of the triangle lies.
Thus S = 2 + 2 + 1 + 1 = 6
AO = \(\frac{\sqrt 3 * 6}{ 3}\)
AO = \( 2 * \sqrt 3 \)
The radius if the bigger circle is OZ
OZ = AO + AZ
OZ = \( 2 * \sqrt 3 \) + 1
Area of the shaded region = Area of the bigger circle - ( 13 * Area of the smaller circles)
Area of the shaded region = \( \pi (2 * \sqrt 3 + 1) ^ 2 \) - (13 * \( \pi (1) ^ 2 \))
Area of the shaded region = \( 12\pi + 1\pi + 4\sqrt 3 \pi - 13 \pi \)
Area of the shaded region = \( 4\sqrt 3 \pi \)
The center of the bigger circle, is the centroid of the equilateral triangle.
Hence, the length of AO is \(\frac{\sqrt 3 * S}{ 3}\)
S - Side of the equilateral triangle
The side of the equilateral triangle is equal to the sum of diameter of two circles and radius of the other two circles where the vertex of the triangle lies.
Thus S = 2 + 2 + 1 + 1 = 6
AO = \(\frac{\sqrt 3 * 6}{ 3}\)
AO = \( 2 * \sqrt 3 \)
The radius if the bigger circle is OZ
OZ = AO + AZ
OZ = \( 2 * \sqrt 3 \) + 1
Area of the shaded region = Area of the bigger circle - ( 13 * Area of the smaller circles)
Area of the shaded region = \( \pi (2 * \sqrt 3 + 1) ^ 2 \) - (13 * \( \pi (1) ^ 2 \))
Area of the shaded region = \( 12\pi + 1\pi + 4\sqrt 3 \pi - 13 \pi \)
Area of the shaded region = \( 4\sqrt 3 \pi \)
Attachments
fc485e1337895925d85375cce1a9d3908ca8b7e3.png [ 45.64 KiB | Viewed 4726 times ]
