Last visit was: 23 Apr 2026, 18:37 It is currently 23 Apr 2026, 18:37
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.
Close
Request Expert Reply
Confirm Cancel
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 23 Apr 2026
Posts: 109,785
Own Kudos:
810,873
 [6]
Given Kudos: 105,853
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,785
Kudos: 810,873
 [6]
1
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
AnupamKT
Joined: 13 Mar 2018
Last visit: 08 Jan 2026
Posts: 21
Own Kudos:
26
 [8]
Given Kudos: 122
GPA: 3.12
WE:Project Management (Other)
Posts: 21
Kudos: 26
 [8]
7
Kudos
Add Kudos
Bookmarks
Bookmark this Post
General Discussion
User avatar
Archit3110
User avatar
Major Poster
Joined: 18 Aug 2017
Last visit: 23 Apr 2026
Posts: 8,628
Own Kudos:
Given Kudos: 243
Status:You learn more from failure than from success.
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1: 545 Q79 V79 DI73
GMAT Focus 2: 645 Q83 V82 DI81
GPA: 4
WE:Marketing (Energy)
Products:
GMAT Focus 2: 645 Q83 V82 DI81
Posts: 8,628
Kudos: 5,190
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Chethan92
Joined: 18 Jul 2018
Last visit: 21 Apr 2022
Posts: 901
Own Kudos:
Given Kudos: 95
Location: India
Concentration: Operations, General Management
GMAT 1: 590 Q46 V25
GMAT 2: 690 Q49 V34
WE:Engineering (Energy)
Products:
GMAT 2: 690 Q49 V34
Posts: 901
Kudos: 1,509
Kudos
Add Kudos
Bookmarks
Bookmark this Post
From statement 1:

Perimeter of ABDF = 14 = 2(AB+AF)
Then AB+AF = 7.

No extra info. Hence insufficient.

From statement 2:

AC = 5 = radius of the quarter circle.
AF = 3.
AE = AC (Radius)
FE becomes 2.
No info about AB. Hence insufficient.

Combining both:

AB becomes 4.

Area of the quarter circle = \(\frac{90}{360}\)*Pi*5^2 = \(\frac{25}{4}\)Pi

Area of the shaded region = Area of the quarter circle-\(\frac{1}{2}\)Area of the rectangle
Area of the shaded region = \(\frac{25}{4}\)Pi-6

Hence sufficient.

C is the answer.
User avatar
AnupamKT
Joined: 13 Mar 2018
Last visit: 08 Jan 2026
Posts: 21
Own Kudos:
Given Kudos: 122
GPA: 3.12
WE:Project Management (Other)
Posts: 21
Kudos: 26
Kudos
Add Kudos
Bookmarks
Bookmark this Post
From statement 1: Perimeter of rectangle is known. With given data we will not be able to find the area of unshaded portion.
From statement 2: radious of quarter circle is known and base of unshaded portion is known. But,we will not be able to find the area of unshaded portion.

Combining both the statements, area of unshaded portion and quarter circle can be found out.

Hence, option C is correct. IMO.

Posted from my mobile device
avatar
chibi
Joined: 06 May 2018
Last visit: 14 Dec 2018
Posts: 3
Own Kudos:
Given Kudos: 187
Posts: 3
Kudos: 5
Kudos
Add Kudos
Bookmarks
Bookmark this Post
In the quarter circle above, what is the area of the shaded region?

(1) The perimeter of the rectangle ABDF is 14.
(2) AC is 5 and AF is 3

(1)--> 2(l+w)=14 --> (l+w)=7 -->clearly insuff. (l and w values not given)
(2)--> AC=5=radius and AF=l=3 --> insuff. (we don't know what value can b can take)

Combining (1) and (2),
(1)--> l+w=7 and (2)-->l=3, we can find b=4;
Since rectangle forms a right angle,
Area of unshaded=1/2 * l * b -->Find the area of BAF
Area of circle = 25pi -->(r=5)

Hence, Area of shaded=Area of circle-area of unshaded -->Suff

Ans C :)
User avatar
Chethan92
Joined: 18 Jul 2018
Last visit: 21 Apr 2022
Posts: 901
Own Kudos:
Given Kudos: 95
Location: India
Concentration: Operations, General Management
GMAT 1: 590 Q46 V25
GMAT 2: 690 Q49 V34
WE:Engineering (Energy)
Products:
GMAT 2: 690 Q49 V34
Posts: 901
Kudos: 1,509
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel

In the quarter circle above, what is the area of the shaded region?

(1) The perimeter of the rectangle ABDF is 14.
(2) AC is 5 and AF is 3


Attachment:
image011 (1).jpg

Bunuel, How's the answer B and not C?
avatar
jorgetomas9
Joined: 19 Aug 2018
Last visit: 01 Dec 2024
Posts: 29
Own Kudos:
Given Kudos: 35
Posts: 29
Kudos: 7
Kudos
Add Kudos
Bookmarks
Bookmark this Post
AnupamKT
Afc0892
Bunuel

In the quarter circle above, what is the area of the shaded region?

(1) The perimeter of the rectangle ABDF is 14.
(2) AC is 5 and AF is 3




Sir, Initially i marked C but answer should be B. I have attached the calculation. Please check.
Hello, How can AD could be 5?
User avatar
Chethan92
Joined: 18 Jul 2018
Last visit: 21 Apr 2022
Posts: 901
Own Kudos:
Given Kudos: 95
Location: India
Concentration: Operations, General Management
GMAT 1: 590 Q46 V25
GMAT 2: 690 Q49 V34
WE:Engineering (Energy)
Products:
GMAT 2: 690 Q49 V34
Posts: 901
Kudos: 1,509
Kudos
Add Kudos
Bookmarks
Bookmark this Post
jorgetomas9

Hello, How can AD could be 5?

A is the center of the quarter circle. And AD, AE, and AC are all radii. Hence AD is 5.
User avatar
laulau
Joined: 09 Apr 2018
Last visit: 20 Oct 2019
Posts: 25
Own Kudos:
46
 [1]
Given Kudos: 54
GPA: 4
Posts: 25
Kudos: 46
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Area of shaded region = area of quarter of the circle - area of triangle

Translated into formulas, we get: \((1/4)pi*r^2 - (1/2) base*height\)

Thus, the unknowns we need to solve the question are r, base, and height.

(1) This statement does not give us all the unknowns we need. Insufficient
(2) This statement provides us with the radius (AC =5) and the base (AF=3). Now, we also know that the diagonal cuts the rectangle in half and therefore two of its angles are halfed into 45°, making the resulting two triangles into two identical isosceles triangles. Knowing that the ratio of the sides of a right isosceles triangle is x:x:x\(\sqrt{2}\), we also know that base (AC) = height (AB) =5. Based on this information, we can calculate the area of the quarter circle and the area of the triangle.

Statement (2) is sufficient. Answer B

Please hit Kudos if you liked this answer.
avatar
nn571
Joined: 04 Mar 2018
Last visit: 27 Mar 2019
Posts: 2
Own Kudos:
Given Kudos: 37
Posts: 2
Kudos: 2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I understand the logic of B, but how can you guarantee ABDF is a rectangle? if BD is not parallel to AF (so assume AB=3 and DF=4), then the diagonals are not equal, leading to C as the correct answer. Can someone please explain how 2) ensures that ABDF is a rectangle on its own?
User avatar
Harsh2111s
Joined: 08 May 2019
Last visit: 10 Feb 2021
Posts: 282
Own Kudos:
Given Kudos: 54
Location: India
Concentration: Operations, Marketing
GPA: 4
WE:Manufacturing and Production (Manufacturing)
Products:
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel

In the quarter circle above, what is the area of the shaded region?

(1) The perimeter of the rectangle ABDF is 14.
(2) AC is 5 and AF is 3


Attachment:
image011 (1).jpg

Trick here is AC=AD=5, since both are radius. :)

Now everybody agree that statement 1 is not sufficient. Lets directly move to statement 2.
Here AC=AD=5, AF=3,

By using Pythagoras theorem AB=DF=4

Finally area of shaded reason= area of quarter circle-(Half area of rectangle).
Hence required area can be easily calculated since we know radius and both sides of rectangle.


Statement 2 is sufficient.

BTW i got it wrong too. ;)
User avatar
nick1816
User avatar
Retired Moderator
Joined: 19 Oct 2018
Last visit: 12 Mar 2026
Posts: 1,841
Own Kudos:
Given Kudos: 707
Location: India
Posts: 1,841
Kudos: 8,509
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Shaded area= pi* r^2- 1/2(area of rectangle)
Also, radius of quarter circle, r= Diagonal of the rectangle

We need the 2 sides of rectangle to find the area of shaded area

Statement 1- We have the sum of the sides of rectangle.
Insufficient

Statement 2- we know the radius or the diagonal of rectangle, and one of the side of rectangle; Hence, we can figure out the second side of rectangle, as well as the area of the shaded portion.

Sufficient.



Bunuel

In the quarter circle above, what is the area of the shaded region?

(1) The perimeter of the rectangle ABDF is 14.
(2) AC is 5 and AF is 3


Attachment:
image011 (1).jpg
User avatar
David nguyen
Joined: 15 May 2017
Last visit: 18 Aug 2020
Posts: 132
Own Kudos:
Given Kudos: 132
Status:Discipline & Consistency always beats talent
Location: United States (CA)
GPA: 3.59
WE:Sales (Retail: E-commerce)
Posts: 132
Kudos: 139
Kudos
Add Kudos
Bookmarks
Bookmark this Post
nick1816
Shaded area= pi* r^2- 1/2(area of rectangle)
Also, radius of quarter circle, r= Diagonal of the rectangle

We need the 2 sides of rectangle to find the area of shaded area

Statement 1- We have the sum of the sides of rectangle.
Insufficient

Statement 2- we know the radius or the diagonal of rectangle, and one of the side of rectangle; Hence, we can figure out the second side of rectangle, as well as the area of the shaded portion.

Sufficient.



Bunuel

In the quarter circle above, what is the area of the shaded region?

(1) The perimeter of the rectangle ABDF is 14.
(2) AC is 5 and AF is 3


Attachment:
image011 (1).jpg

Statement 2: How do you know the polygon is a rectangle?
Moderators:
Math Expert
109785 posts
498 posts
212 posts