Last visit was: 04 Nov 2024, 20:16 It is currently 04 Nov 2024, 20:16

Close

GRE Prep 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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
Verbal Expert
Joined: 18 Apr 2015
Posts: 29887
Own Kudos [?]: 36111 [2]
Given Kudos: 25918
Send PM
Verbal Expert
Joined: 18 Apr 2015
Posts: 29887
Own Kudos [?]: 36111 [0]
Given Kudos: 25918
Send PM
Retired Moderator
Joined: 09 Jun 2020
Posts: 205
Own Kudos [?]: 235 [0]
Given Kudos: 34
GPA: 3.21
Send PM
Verbal Expert
Joined: 18 Apr 2015
Posts: 29887
Own Kudos [?]: 36111 [0]
Given Kudos: 25918
Send PM
Re: In the figure above, each of the four large circles is [#permalink]
Expert Reply
Great Explanation

Now look better, you forgot the tag
Code:
[m][/m]


:wink:
Manager
Manager
Joined: 09 Nov 2018
Posts: 88
Own Kudos [?]: 95 [0]
Given Kudos: 0
Send PM
Re: In the figure above, each of the four large circles is [#permalink]
1
Step 1: Understanding the question
Let A, B and O be the centers of the circles and radius of the smaller circle be x
As the diagonals of the square intersect at right angle, triangle AOB is a right angled at O

AB = 4+4 = 8
AO = (x+4)
BO = (x+4)

Applying Pythagoras
\(AB^2 = AO^2 + BO^2\)
64 = 2* \((x+4)^2\)
32 = \((x+4)^2\)
x + 4 = 4\(\sqrt{2 }\)
x = 4\(\sqrt{2}\) - 4

Hence diameter of the smaller circle = 8\(\sqrt{2}\) - 8
C is correct
Attachments

GRE In the figure above, each of the four large circles.png
GRE In the figure above, each of the four large circles.png [ 22.68 KiB | Viewed 2091 times ]

Intern
Intern
Joined: 02 Jan 2021
Posts: 27
Own Kudos [?]: 12 [0]
Given Kudos: 41
Send PM
Re: In the figure above, each of the four large circles is [#permalink]
Simple way is just joining all the centres of the large circle to form a square of side 8 units.
Joining the diagonal of the square that cuts the smaller circle at its diameter.
so diagonal of a square = sqrroot 2 * a = 8 * sqr root 2

Now diameter of smaller circle = diagonal - (radius of 2 bigger circle) = 8* sqr root 2 - (4+4) = 8 * sqr root 2 - 8
Prep Club for GRE Bot
Re: In the figure above, each of the four large circles is [#permalink]
Moderators:
GRE Instructor
77 posts
GRE Forum Moderator
37 posts
Moderator
1111 posts
GRE Instructor
228 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne