Last visit was: 05 Sep 2026, 23:16 It is currently 05 Sep 2026, 23:16
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
   1   2   3 
User avatar
EvaJager
User avatar
Joined: 22 Mar 2011
Last visit: 31 Aug 2016
Posts: 512
Own Kudos:
2,397
Given Kudos: 43
WE:Science (Education)
Posts: 512
Kudos: 2,397
A train passes a man standing on a platform in 9 sec. and pa 28 Aug 2012, 15:16
Kudos
Add Kudos
Bookmarks
Bookmark this Post
NYC5648
If Radius BC equals 2, What is the area of the shaded region of the circle formed by the rectangle inscribed in the circle?

Show Spoiler
OA: 4/3 pi - sqrt 3

Can somebody please explain a step-by-step solution please.

Thanks!!!
Show more

The diagonal of the rectangle is diameter in the circle and therefore its length is 4.
The right triangle formed by a diagonal and two adjacent sides of the rectangle is a 30-60-90 rectangle (because the short leg is half of the hypotenuse).
The other side of the rectangle is therefore \(2\sqrt{3}.\)

The shaded area can be obtained by subtracting from the circular sector defined by two radii (half diagonals) with an angle of \(120^o\) between them the area of the triangle formed by the same two radii and the long side of the rectangle.
The are of the circular sector is \(\pi\cdot2^2\cdot\frac{120}{360}=\frac{4\pi}{3}\) and the area of the triangle is 1/4 of the area of the rectangle, which is \(2\cdot2\sqrt{3}=4\sqrt{3}.\)
In any parallelogram, so also in a rectangle, the two diagonals divide the paralellogram into four triangles with equal areas.
Two radii with the short side of the rectangle form an equilateral triangle, while two radii with the long side of the rectangle form an isosceles triangle with angles 30-30-120.

The requested area is \(\frac{4\pi}{3}-\sqrt{3}.\)
User avatar
metallicafan
User avatar
Retired Moderator
Joined: 04 Oct 2009
Last visit: 26 Aug 2020
Posts: 754
Own Kudos:
4,605
Given Kudos: 109
Status:2000 posts! I don't know whether I should feel great or sad about it! LOL
Location: Peru
Concentration: Finance, SMEs, Developing countries, Public sector and non profit organizations
Schools:Harvard, Stanford, Wharton, MIT & HKS (Government)
GPA: 4.0
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
Posts: 754
Kudos: 4,605
A train passes a man standing on a platform in 9 sec. and pa 28 Aug 2012, 17:21
Kudos
Add Kudos
Bookmarks
Bookmark this Post
\(\sqrt{12} = 2\sqrt{3}\)

You have to calculate the hypotenuse for each triangle, starting from the smallest triangle.
User avatar
NYC5648
User avatar
Joined: 09 Feb 2012
Last visit: 30 Sep 2012
Posts: 42
Own Kudos:
238
Given Kudos: 14
Posts: 42
Kudos: 238
A train passes a man standing on a platform in 9 sec. and pa 29 Aug 2012, 12:45
Kudos
Add Kudos
Bookmarks
Bookmark this Post
NYC5648
In the figure to the right what is the length of the longest side, if all of the triangles are right triangles?

Show Spoiler
OA: 2 sqrt 3

Looking forward to your answers.. Thanks!!!
Show more



metallicafan
\(\sqrt{12} = 2\sqrt{3}\)

You have to calculate the hypotenuse for each triangle, starting from the smallest triangle.
Show more
EvaJager
NYC5648
In the figure to the right what is the length of the longest side, if all of the triangles are right triangles?

Show Spoiler
OA: 2 sqrt 3

Looking forward to your answers.. Thanks!!!

Apply Pythagoras starting with the triangle with sides 1 and 1. The hypotenuse is \(\sqrt{2}.\)
In the next right triangle, the hypotenuse is \(\sqrt{3}\) ... the longest hypotenuse is \(\sqrt{12}=\sqrt{2^2\cdot3}=2\sqrt{3}.\)
Show more
EvaJager
NYC5648
In the figure to the right what is the length of the longest side, if all of the triangles are right triangles?

Show Spoiler
OA: 2 sqrt 3

Looking forward to your answers.. Thanks!!!

Apply Pythagoras starting with the triangle with sides 1 and 1. The hypotenuse is \(\sqrt{2}.\)
In the next right triangle, the hypotenuse is \(\sqrt{3}\) ... the longest hypotenuse is \(\sqrt{12}=\sqrt{2^2\cdot3}=2\sqrt{3}.\)
Show more

Thanks for the answer. I got the idea behind it! Thanks. But still I dont get that the longest hypotenuse is 2 srt 3. When I apply Pythagoras theorem I get 4 as an answer. Could you please explain. Thanks!
User avatar
NYC5648
User avatar
Joined: 09 Feb 2012
Last visit: 30 Sep 2012
Posts: 42
Own Kudos:
238
Given Kudos: 14
Posts: 42
Kudos: 238
A train passes a man standing on a platform in 9 sec. and pa 29 Aug 2012, 13:00
Kudos
Add Kudos
Bookmarks
Bookmark this Post
EvaJager
NYC5648
If Radius BC equals 2, What is the area of the shaded region of the circle formed by the rectangle inscribed in the circle?

Show Spoiler
OA: 4/3 pi - sqrt 3

Can somebody please explain a step-by-step solution please.

Thanks!!!

The diagonal of the rectangle is diameter in the circle and therefore its length is 4.
The right triangle formed by a diagonal and two adjacent sides of the rectangle is a 30-60-90 rectangle (because the short leg is half of the hypotenuse).
The other side of the rectangle is therefore \(2\sqrt{3}.\)

The shaded area can be obtained by subtracting from the circular sector defined by two radii (half diagonals) with an angle of \(120^o\) between them the area of the triangle formed by the same two radii and the long side of the rectangle.
The are of the circular sector is \(\pi\cdot2^2\cdot\frac{120}{360}=\frac{4\pi}{3}\) and the area of the triangle is 1/4 of the area of the rectangle, which is \(2\cdot2\sqrt{3}=4\sqrt{3}.\)
In any parallelogram, so also in a rectangle, the two diagonals divide the paralellogram into four triangles with equal areas.
Two radii with the short side of the rectangle form an equilateral triangle, while two radii with the long side of the rectangle form an isosceles triangle with angles 30-30-120.

The requested area is \(\frac{4\pi}{3}-\sqrt{3}.\)
Show more

Hi EVA,
thanks a lot for you detailed explanation. But I still dont get it.
Where I have problems to follow is the following:

The shaded area can be obtained by subtracting from the circular sector defined by two radii (half diagonals) with an angle of \(120^o\) between them the area of the triangle formed by the same two radii and the long side of the rectangle.

Two radii with the short side of the rectangle form an equilateral triangle, while two radii with the long side of the rectangle form an isosceles triangle with angles 30-30-120.

Maybe I am blind but I dont see it.

Thanks!!!
User avatar
NYC5648
User avatar
Joined: 09 Feb 2012
Last visit: 30 Sep 2012
Posts: 42
Own Kudos:
238
Given Kudos: 14
Posts: 42
Kudos: 238
A train passes a man standing on a platform in 9 sec. and pa 29 Aug 2012, 13:06
Kudos
Add Kudos
Bookmarks
Bookmark this Post
The two circles are inscribed in the rectangle as shown above. If line segment AB, length 0.4, is perpendicular to line segment CD and line segment CD is tangent to the circle, what is the approximate area of the shaded region of the rectangle?

Show Spoiler
OA: 3

Absolutely no clue. Can anybody help please?

Thanks!
Attachments

Tricky geometry.png
Tricky geometry.png [ 7.89 KiB | Viewed 3993 times ]

User avatar
NYC5648
User avatar
Joined: 09 Feb 2012
Last visit: 30 Sep 2012
Posts: 42
Own Kudos:
238
Given Kudos: 14
Posts: 42
Kudos: 238
A train passes a man standing on a platform in 9 sec. and pa 29 Aug 2012, 13:47
Kudos
Add Kudos
Bookmarks
Bookmark this Post
In the Square above the length of line AB is 4 and angle BAC is 45 degrees. Approximately what is the circumference of the circle?

Show Spoiler
OA: 20 pi

Thanks!!!
Attachments

Tricky geometry 2.png
Tricky geometry 2.png [ 14.83 KiB | Viewed 3979 times ]

User avatar
NYC5648
User avatar
Joined: 09 Feb 2012
Last visit: 30 Sep 2012
Posts: 42
Own Kudos:
238
Given Kudos: 14
Posts: 42
Kudos: 238
A train passes a man standing on a platform in 9 sec. and pa 29 Aug 2012, 14:04
Kudos
Add Kudos
Bookmarks
Bookmark this Post
CDEB is a parallelogram and P is the midpoint of ED. If GB = 5 what is the length of BA =?

Show Spoiler
OA: 15

Can anybody provide a step by step solution. I am totally stuck here.

Thanks!!!
Attachments

Tricky 3.png
Tricky 3.png [ 16.36 KiB | Viewed 3973 times ]

User avatar
NYC5648
User avatar
Joined: 09 Feb 2012
Last visit: 30 Sep 2012
Posts: 42
Own Kudos:
238
Given Kudos: 14
Posts: 42
Kudos: 238
A train passes a man standing on a platform in 9 sec. and pa 29 Aug 2012, 14:11
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ABCD is a Trapezoid:
AP = x – 1
PC = 3
DP = 2x – 13
PB = x – 5

What is the value of x?

Show Spoiler
OA: 7
Any idea how to solve this kind of problem? Probably I am missing a concept again.

Thanks!
Attachments

Tricky 4.png
Tricky 4.png [ 15.05 KiB | Viewed 3990 times ]

User avatar
EvaJager
User avatar
Joined: 22 Mar 2011
Last visit: 31 Aug 2016
Posts: 512
Own Kudos:
2,397
Given Kudos: 43
WE:Science (Education)
Posts: 512
Kudos: 2,397
A train passes a man standing on a platform in 9 sec. and pa 29 Aug 2012, 14:14
Kudos
Add Kudos
Bookmarks
Bookmark this Post
NYC5648
NYC5648
In the figure to the right what is the length of the longest side, if all of the triangles are right triangles?

Show Spoiler
OA: 2 sqrt 3

Looking forward to your answers.. Thanks!!!



metallicafan
\(\sqrt{12} = 2\sqrt{3}\)

You have to calculate the hypotenuse for each triangle, starting from the smallest triangle.
EvaJager
NYC5648
In the figure to the right what is the length of the longest side, if all of the triangles are right triangles?

Show Spoiler
OA: 2 sqrt 3

Looking forward to your answers.. Thanks!!!

Apply Pythagoras starting with the triangle with sides 1 and 1. The hypotenuse is \(\sqrt{2}.\)
In the next right triangle, the hypotenuse is \(\sqrt{3}\) ... the longest hypotenuse is \(\sqrt{12}=\sqrt{2^2\cdot3}=2\sqrt{3}.\)
EvaJager
NYC5648
In the figure to the right what is the length of the longest side, if all of the triangles are right triangles?

Show Spoiler
OA: 2 sqrt 3

Looking forward to your answers.. Thanks!!!

Apply Pythagoras starting with the triangle with sides 1 and 1. The hypotenuse is \(\sqrt{2}.\)
In the next right triangle, the hypotenuse is \(\sqrt{3}\) ... the longest hypotenuse is \(\sqrt{12}=\sqrt{2^2\cdot3}=2\sqrt{3}.\)

Thanks for the answer. I got the idea behind it! Thanks. But still I dont get that the longest hypotenuse is 2 srt 3. When I apply Pythagoras theorem I get 4 as an answer. Could you please explain. Thanks!
Show more

The first hypotenuse squared is \(1^2+1^2=2\). This hypotenuse is one of the legs in the next right triangle. We are going to apply Pythagoras's again, so no need to take the square root of it.
The second hypotenuse squared is \(2+1^2=3\).
The third hypotenuse squared is \(3+1^2=4\).
...

Finally, the longest hypotenuse is \(\sqrt{12}=\sqrt{2^2\cdot3}=2\sqrt{3}.\)

Make a drawing and mark where you have the right angles. It will help you when applying Pythagoras's.
User avatar
EvaJager
User avatar
Joined: 22 Mar 2011
Last visit: 31 Aug 2016
Posts: 512
Own Kudos:
2,397
Given Kudos: 43
WE:Science (Education)
Posts: 512
Kudos: 2,397
A train passes a man standing on a platform in 9 sec. and pa 29 Aug 2012, 14:20
Kudos
Add Kudos
Bookmarks
Bookmark this Post
NYC5648
EvaJager
NYC5648
If Radius BC equals 2, What is the area of the shaded region of the circle formed by the rectangle inscribed in the circle?

Show Spoiler
OA: 4/3 pi - sqrt 3

Can somebody please explain a step-by-step solution please.

Thanks!!!

The diagonal of the rectangle is diameter in the circle and therefore its length is 4.
The right triangle formed by a diagonal and two adjacent sides of the rectangle is a 30-60-90 rectangle (because the short leg is half of the hypotenuse).
The other side of the rectangle is therefore \(2\sqrt{3}.\)

The shaded area can be obtained by subtracting from the circular sector defined by two radii (half diagonals) with an angle of \(120^o\) between them the area of the triangle formed by the same two radii and the long side of the rectangle.
The are of the circular sector is \(\pi\cdot2^2\cdot\frac{120}{360}=\frac{4\pi}{3}\) and the area of the triangle is 1/4 of the area of the rectangle, which is \(2\cdot2\sqrt{3}=4\sqrt{3}.\)
In any parallelogram, so also in a rectangle, the two diagonals divide the paralellogram into four triangles with equal areas.
Two radii with the short side of the rectangle form an equilateral triangle, while two radii with the long side of the rectangle form an isosceles triangle with angles 30-30-120.

The requested area is \(\frac{4\pi}{3}-\sqrt{3}.\)

Hi EVA,
thanks a lot for you detailed explanation. But I still dont get it.
Where I have problems to follow is the following:

The shaded area can be obtained by subtracting from the circular sector defined by two radii (half diagonals) with an angle of \(120^o\) between them the area of the triangle formed by the same two radii and the long side of the rectangle.

Two radii with the short side of the rectangle form an equilateral triangle, while two radii with the long side of the rectangle form an isosceles triangle with angles 30-30-120.

Maybe I am blind but I dont see it.

Thanks!!!
Show more

Copy the drawing from the question and add the two diagonals. Mark the angles and don't forget that the two diagonals are equal and bisect each other.
User avatar
rajeshs
User avatar
Joined: 02 Apr 2013
Last visit: 07 Feb 2014
Posts: 24
Own Kudos:
8
Given Kudos: 1
Location: United Kingdom
Concentration: Entrepreneurship, Technology
GMAT 1: 590 Q38 V33
WE:Information Technology (Computer Software)
GMAT 1: 590 Q38 V33
Posts: 24
Kudos: 8
A train passes a man standing on a platform in 9 sec. and pa 23 Oct 2013, 01:14
Kudos
Add Kudos
Bookmarks
Bookmark this Post
"Two men plan to run in opposite directions around a park 1 km long. The first man can run 15 meters in a minute and the second can run 25 meters per minute. How much further will the second runner have run at the point they cross paths? "

What does "How much further will the second runner have run at the point they cross paths?" mean?
Is "point they cross paths" mean the meeting place?
Should the 1000 m be considered a circle or a straight line?
"further will the second runner.." - could not understand this term too.

Can you please help.

Thanks
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 39,213
Own Kudos:
1,157
Posts: 39,213
Kudos: 1,157
A train passes a man standing on a platform in 9 sec. and pa 08 Dec 2018, 12:44
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
   1   2   3 
Moderator:
Bunuel
Math Expert
113155 posts