Last visit was: 18 Nov 2025, 18:58 It is currently 18 Nov 2025, 18:58
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: 18 Nov 2025
Posts: 105,355
Own Kudos:
Given Kudos: 99,964
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,081
 [20]
1
Kudos
Add Kudos
19
Bookmarks
Bookmark this Post
Most Helpful Reply
avatar
dep
Joined: 23 Feb 2018
Last visit: 07 Aug 2021
Posts: 83
Own Kudos:
119
 [5]
Given Kudos: 23
Posts: 83
Kudos: 119
 [5]
4
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
Given Kudos: 99,964
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,081
Kudos
Add Kudos
Bookmarks
Bookmark this Post
General Discussion
User avatar
Gmatfox
Joined: 09 Jan 2015
Last visit: 17 Nov 2025
Posts: 11
Own Kudos:
34
 [2]
Given Kudos: 6
Posts: 11
Kudos: 34
 [2]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post

Right triangle ABC, right angled at B, is rotated clockwise, so that vertex B is at B', vertex A is at A' and vertex C is at C', as shown above. What is the ratio of the area of two green regions to the area of blue region?

A. 1:1
B. 2:1
C. 3:1
D. 4:1
E. 5:1

Assume AC and B'C' intersect at point D. Check the picture attached.
We have: Angle B'C'A' = Angle BCA (Triangle ABC is similar to triangle A'B'C') --> Triangle BDC is an isosceles triangle with DB = DC
But we already know that ABC is a right triangle (angle B = 90 degree) --> D is the midpoint of AC
--> Area of blue region = Area of each green region = 1/2 area of whole triangle ABC = 1/2 area of whole triangle A'B'C'
--> Area of green regions : area of blue region = 2 : 1
Correct answer: B
Attachments

no.1.PNG
no.1.PNG [ 79.3 KiB | Viewed 4935 times ]

avatar
vivek1408
Joined: 04 Feb 2020
Last visit: 04 May 2022
Posts: 12
Own Kudos:
14
 [3]
Given Kudos: 5
Location: Japan
GMAT 1: 690 Q49 V35
GPA: 3.12
GMAT 1: 690 Q49 V35
Posts: 12
Kudos: 14
 [3]
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
This question took me a good 15 mins to solve.. phew.
But I think the answer is B

To make our life easier we can assume the triangle ABC to be right isoceles, making two sides equal (lets say length X)and angles 45-45-90.
In this case, area of the left green triangle is 1/2*X*X/2 = X^2/4
and the area of the blue triangle is also 1/2*X*X/2 = X^2/4
since area(ABC) = area(A'B'C') =>therefore the area of the right green quadrilateral = area of left green triangle =X^2/4

Hence total green area = (X^2)/2
total blue area = (X^2)/4
making green:blue ratio = 2:1

answer B
User avatar
vv65
Joined: 01 Mar 2015
Last visit: 10 Nov 2025
Posts: 534
Own Kudos:
395
 [1]
Given Kudos: 774
Location: India
GMAT 1: 740 Q47 V44
GMAT 1: 740 Q47 V44
Posts: 534
Kudos: 395
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
My answer is B 2:1

Here is my method:

Assume that the green triangle is a 90-60-30 triangle with hypotenuse=6 and small side=3
The rest of the working is in the diagram

Will appreciate a faster way of doing this question, as this method took too long. If I had drawn the diagram, it would have taken even longer
Attachments

Right Triangle question.jpg
Right Triangle question.jpg [ 181.64 KiB | Viewed 4677 times ]

avatar
Eswar42
Joined: 24 Oct 2019
Last visit: 17 Dec 2023
Posts: 7
Own Kudos:
27
 [3]
Given Kudos: 1
Posts: 7
Kudos: 27
 [3]
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
IMO, answer is B, 2:1
it is derived from the angles that ABC is isosceles right angled triangle.
the triangles shaded in blue and green of ABC are equilateral triangles, see the attached sheet
the area of blue in ABC is equal to area of green in ABC. the green area in each orientation is same.
so answer is 2*green area of one / blue area = 2:1
Attachments

IMG_20201218_114853.jpg
IMG_20201218_114853.jpg [ 2.27 MiB | Viewed 4672 times ]

User avatar
QuantMadeEasy
Joined: 28 Feb 2014
Last visit: 15 Nov 2025
Posts: 502
Own Kudos:
785
 [1]
Given Kudos: 78
Location: India
Concentration: General Management, International Business
GPA: 3.97
WE:Engineering (Education)
Posts: 502
Kudos: 785
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Quote:
Right triangle ABC, right angled at B, is rotated clockwise, so that vertex B is at B', vertex A is at A' and vertex C is at C', as shown above. What is the ratio of the area of two green regions to the area of blue region?
As the triangle ABC is rotated about a point according to the above condition, the three regions will have equal areas
This can be understood by taking triangle ABC as 45-45-90 and related sides as x-x-root(2x)

ie. area of one green region = area of blue region
hence, ratio of area of total green region to area of blue region = 2:1

IMO B
User avatar
DJ2911
Joined: 01 May 2020
Last visit: 22 May 2022
Posts: 16
Own Kudos:
41
 [1]
Given Kudos: 14
Location: India
Concentration: Statistics, Strategy
WE:General Management (Non-Profit and Government)
Products:
Posts: 16
Kudos: 41
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Answer: B (2:1)

Refer to the image attached below:
We have added a perpendicular line to BC from E meeting BC at F
Assume:
Angle A'C'B' = Angle ACB = theta
BC = b
Note that since Area(EBC) is common to both the big triangles, Area(AEB) = Area(CEB'A')
Area of two green regions = Area of AEB *2
Area of blue region = Area of EBC
Since EBC is an isosceles triangle:
Area EBC = 1/2*b*b/2*Tan(theta) = (b^2 * tan(theta))/4
Area AEB = Area ABC - Area EBC = (b^2*tan(theta))/2 - (b^2tan(theta))/4 = (b^2*tan(theta))/4
Thus total area of green region / area of red region = 2* Area ABC / Area EBC = 2/1
Attachments

WhatsApp Image 2020-12-18 at 12.06.23 PM.jpeg
WhatsApp Image 2020-12-18 at 12.06.23 PM.jpeg [ 90.11 KiB | Viewed 4567 times ]

User avatar
rocky620
User avatar
Retired Moderator
Joined: 10 Nov 2018
Last visit: 11 May 2023
Posts: 501
Own Kudos:
607
 [2]
Given Kudos: 229
Location: India
Concentration: General Management, Strategy
GMAT 1: 590 Q49 V22
WE:Other (Retail: E-commerce)
GMAT 1: 590 Q49 V22
Posts: 501
Kudos: 607
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Kindly Refer the Image Below:

Angle ACB = Angle A'C'B' = x, So OB = OC (side opposite to equal angles)......eq1
Angle BAC = y

Angle ACB + Angle BAC = x+y = 90.......eq2
Since, Angle ABO + Angle OBC = 90, so Angle ABO = y (from Eq2)

Since, Angle ABO = Angle OAB = y, OB = OA.....eq3

From eq1 and eq3, OA = OB = BC

We can deduce that OB is a median of AB, and a median divides the triangle in two equal parts with equal area.

Since triangle OBC is common for ABC and A'B'C', area of quad COB'A' = area of triangle ABO.

Now the two green parts and the blue part each have equal areas, so the ratio of area of green part to the blue part will be 2:1

IMO Option B
Attachments

Day 4 Q.png
Day 4 Q.png [ 97.37 KiB | Viewed 3783 times ]

User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,588
Own Kudos:
Posts: 38,588
Kudos: 1,079
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderators:
Math Expert
105355 posts
Tuck School Moderator
805 posts