Last visit was: 20 Sep 2026, 13:31 It is currently 20 Sep 2026, 13:31
Home
Messages
Tests
Schools
Events
Close
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

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 GMAT 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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
655-705 (Hard)|   Geometry|      
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 20 Sep 2026
Posts: 113,693
Own Kudos:
842,552
 [2]
Given Kudos: 111,680
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 113,693
Kudos: 842,552
 [2]
A semicircle with radius a is inscribed in a rectangle with its diamet 07 Oct 2020, 02:30
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
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:

60% (02:41) correct 40% (02:55) wrong based on 35 sessions

History

Date
Time
Result
Not Attempted Yet

A semicircle with radius a is inscribed in a rectangle with its diameter along one side of the rectangle. A circle with radius b is then inscribed so it is tangent to the rectangle and the semicircle, as shown above. What is the value of a/b?


A. \(\frac{\sqrt{2}}{\sqrt{2} - 1}\)

B. \(\frac{\sqrt{2} + 1}{\sqrt{2}}\)

C. 2:1

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\)

E. 6:1


Show Spoiler
Attachment:
Untitled.png
Untitled.png [ 22.09 KiB | Viewed 2968 times ]
ShowHide Answer
Official Answer
D
User avatar
logro
User avatar
Joined: 15 Oct 2017
Last visit: 21 Nov 2021
Posts: 322
Own Kudos:
181
Given Kudos: 52
WE:Investment Banking (Finance: Investment Banking)
Products:
My Reviews
Posts: 322
Kudos: 181
A semicircle with radius a is inscribed in a rectangle with its diamet Updated on: 07 Oct 2020, 07:04
Originally posted by logro on 07 Oct 2020, 02:49.
Last edited by logro on 07 Oct 2020, 07:04, edited 1 time in total.
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Let the rectangle be ABCD (A and D being the corners of the semi circle, B being the vertex enclosing the smaller circle and C being the other one). O be the centre of the semi circle, M be the centre of smaller circle, E be the point of intersection of Semi circle and the rectangle and F be the intersection of smaller circle and BC

We know that the length of the rectangle (AB) will be 2a (diameter of the semi circle)

OA = OD = a

In the smaller semi circle

We need to find the distance of centre to the vertex enclosing the small circle i.e vertex B

Since the point of intersection of the small circle with the vertex is tangent we can safely deduce that the angle will be 90

In the triangle formed with the vertex MFB

\(MB ^2 = b^2+b^2\)

Diagonal of smaller circle = MB = \(\sqrt{2}b\)

In the triangle OEB, right angled at E

\(OB^2 = OE^2 + BE^2\)

We know that OE=BE=a

subsituting we get

OB = \(\sqrt{2}a\)

OB = radius of the semicircle + radius of the smaller circle + MB

\(\sqrt{2}a = a + b + \sqrt{2}b\)

rearranging we get

\(a(\sqrt{2} - 1) = b(\sqrt{2} + 1)\)

Simplifying we get

\(\frac{a}{b} = \frac{(\sqrt{2} + 1) }{ (\sqrt{2} - 1)}\)

Hence IMO D
Attachments

9140c2f1-04f3-43d4-9d1d-7aaaf442ea9e.JPG
9140c2f1-04f3-43d4-9d1d-7aaaf442ea9e.JPG [ 40.84 KiB | Viewed 2595 times ]

User avatar
kajaldaryani46
User avatar
Joined: 26 Oct 2018
Last visit: 07 Nov 2022
Posts: 45
Own Kudos:
14
Given Kudos: 469
GMAT 1: 640 Q40 V37
GMAT 1: 640 Q40 V37
Posts: 45
Kudos: 14
A semicircle with radius a is inscribed in a rectangle with its diamet 07 Oct 2020, 06:57
Kudos
Add Kudos
Bookmarks
Bookmark this Post
logro
Let the rectangle be ABCD (A and D being the corners of the semi circle, B being the vertex enclosing the smaller circle and C being the other one). O be the centre of the semi circle, M be the centre of smaller circle, E be the point of intersection of Semi circle and the rectangle and F be the intersection of smaller circle and BC

We know that the length of the rectangle (AB) will be 2a (diameter of the semi circle)

OA = OD = a

In the smaller semi circle

We need to find the distance of centre to the vertex enclosing the small circle i.e vertex B

Since the point of intersection of the small circle with the vertex is tangent we can safely deduce that the angle will be 90

In the triangle formed with the vertex MFB

\(MB ^2 = b^2+b^2\)

Diagonal of smaller circle = MB = \(\sqrt{2}b\)

In the triangle OEB, right angled at E

\(OB^2 = OE^2 + BE^2\)

We know that OE=BE=a

subsituting we get

OB = \(\sqrt{2}a\)

OB = radius of the semicircle + radius of the smaller circle + MB

\(\sqrt{2}a = a + b + \sqrt{2}b\)

rearranging we get

\(a(\sqrt{2} - 1) = b(\sqrt{2} + 1)\)

Simplifying we get

\(\frac{a}{b} = \frac{(\sqrt{2} + 1) }{ (\sqrt{2} - 1)}\)

Hence IMO D
Show more
—
Hi
Could u pls upload a diagram - it’d be v helpful to understand

Posted from my mobile device
User avatar
kajaldaryani46
User avatar
Joined: 26 Oct 2018
Last visit: 07 Nov 2022
Posts: 45
Own Kudos:
14
Given Kudos: 469
GMAT 1: 640 Q40 V37
GMAT 1: 640 Q40 V37
Posts: 45
Kudos: 14
A semicircle with radius a is inscribed in a rectangle with its diamet 07 Oct 2020, 06:57
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel hi,

Could u pls share a faster way of solving this Q

Thanks

Posted from my mobile device
User avatar
logro
User avatar
Joined: 15 Oct 2017
Last visit: 21 Nov 2021
Posts: 322
Own Kudos:
181
 [1]
Given Kudos: 52
WE:Investment Banking (Finance: Investment Banking)
Products:
My Reviews
Posts: 322
Kudos: 181
 [1]
A semicircle with radius a is inscribed in a rectangle with its diamet 07 Oct 2020, 07:06
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
kajaldaryani46
logro
Let the rectangle be ABCD (A and D being the corners of the semi circle, B being the vertex enclosing the smaller circle and C being the other one). O be the centre of the semi circle, M be the centre of smaller circle, E be the point of intersection of Semi circle and the rectangle and F be the intersection of smaller circle and BC

We know that the length of the rectangle (AB) will be 2a (diameter of the semi circle)

OA = OD = a

In the smaller semi circle

We need to find the distance of centre to the vertex enclosing the small circle i.e vertex B

Since the point of intersection of the small circle with the vertex is tangent we can safely deduce that the angle will be 90

In the triangle formed with the vertex MFB

\(MB ^2 = b^2+b^2\)

Diagonal of smaller circle = MB = \(\sqrt{2}b\)

In the triangle OEB, right angled at E

\(OB^2 = OE^2 + BE^2\)

We know that OE=BE=a

subsituting we get

OB = \(\sqrt{2}a\)

OB = radius of the semicircle + radius of the smaller circle + MB

\(\sqrt{2}a = a + b + \sqrt{2}b\)

rearranging we get

\(a(\sqrt{2} - 1) = b(\sqrt{2} + 1)\)

Simplifying we get

\(\frac{a}{b} = \frac{(\sqrt{2} + 1) }{ (\sqrt{2} - 1)}\)

Hence IMO D
—
Hi
Could u pls upload a diagram - it’d be v helpful to understand

Posted from my mobile device
Show more

Hey - I have added the diagram. Hope it helps
Moderator:
Bunuel
Math Expert
113693 posts