Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
An intuitive geometry problem,
The ticket for the Chess world cup final is in the shape of a regular hexagon, with a side of 6 units. A circular stamp, of radius 2 units, is meant to be used to stamp the ticket in the center.
By, mistake the stamp get misaligned and stamps the ticket in such a way that the edge of the stamp exactly touches the vertex of the hexagon, while the center of the stamp lies on the line joining that vertex to the opposite vertex.
Find the portion of the stamp which falls outside the ticket.
 one year ago
 one year ago
An intuitive geometry problem, The ticket for the Chess world cup final is in the shape of a regular hexagon, with a side of 6 units. A circular stamp, of radius 2 units, is meant to be used to stamp the ticket in the center. By, mistake the stamp get misaligned and stamps the ticket in such a way that the edge of the stamp exactly touches the vertex of the hexagon, while the center of the stamp lies on the line joining that vertex to the opposite vertex. Find the portion of the stamp which falls outside the ticket.
 one year ago
 one year ago

This Question is Closed

FoolForMathBest ResponseYou've already chosen the best response.2
That would spoil the problem to some extent ;)
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
would there be 2 circles, one inside and one outside?
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.2
There is only one stamp.
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
well if the edge is outside then the circle must be inscribed inside
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
cant really draw to scale
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
is that how it should be?
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.2
@experimentX got it right in the first time itself.
 one year ago

binary3iBest ResponseYou've already chosen the best response.0
is that \[(\pi 3\sqrt{3})/6\]
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.0
dw:1338097283187:dw Not perfect, but I believe the picture should look something like this.
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.0
And the area we are trying to find is: dw:1338097419327:dw
 one year ago

binary3iBest ResponseYou've already chosen the best response.0
\[2\int\limits_{0}^{1/2}(\sqrt{2xx^2}  \sqrt{3}x) dx\]
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.2
I believe Callisto did it :D
 one year ago

CallistoBest ResponseYou've already chosen the best response.7
It's actually very easy and interesting :)
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.2
I never post hard problem(s) ;)
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.0
That's kind of you.
 one year ago

CallistoBest ResponseYou've already chosen the best response.7
No.... you always post difficult problems :
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
... I'm still waiting for someone to post answer to your perpendicular tangent question
 one year ago

CallistoBest ResponseYou've already chosen the best response.7
First, I must admit that my way to solve it is very stupid.... radius of the circle is 2 and the centre of the circle lies on the line joining 2 opposite vertex => 2 equi. triangles with sides 2 cm are formed Area of sector = pi(2)^2 x 120/360 = 4pi/3 So, area required = 4pi/3  2sqrt3
 one year ago

CallistoBest ResponseYou've already chosen the best response.7
@SmoothMath Hope you don't mind me borrowing your diagram : dw:1338097936559:dw
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.0
Ehh, that's how I would have done it too. :P
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.0
I got a bit caught up on how to justify the third side being 2.
 one year ago

CallistoBest ResponseYou've already chosen the best response.7
PS: When I drew the diagram, I found it very interesting that I even forgot I needed to solve the problem :P
 one year ago

CallistoBest ResponseYou've already chosen the best response.7
Each angle of hexagon = 180 (62) /6 = 120 divided by 2 = 60 Base angle isos triangle => all are 60 => equi. triangle
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.0
Guys, you just spoiled the problem for me, ahah
 one year ago

CallistoBest ResponseYou've already chosen the best response.7
I only know this basic math :( @FoolForMath How do you solve it?
 one year ago

binary3iBest ResponseYou've already chosen the best response.0
i made a mistake while integrating.
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.0
That's a great way to solve it, Callisto. Better to use basic math whenever you can.
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.2
I did it in the same way. The key realization is the fact that there is a equilateral triangle of length 2 unit is formed. After that, the hexagon is not needed. \[2\times r^2\left(\frac {\pi \theta}{360^\circ}\frac {\sin \theta }{2} \right) \] Here \(r=2\) and \( \theta\) is \( 60^\circ \)
 one year ago

ninhi5Best ResponseYou've already chosen the best response.0
someone should give @SmoothMath a medal
 one year ago

SmoothMathBest ResponseYou've already chosen the best response.0
Well, aren't you kind. I'm not too worried about medals though, and Callisto clearly has the "best answer." =)
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
i was looking for the solution of this question ... http://openstudy.com/users/foolformath#/updates/4fa421cde4b029e9dc34a843
 one year ago

apoorvkBest ResponseYou've already chosen the best response.0
Gotcha, this was a sitter  @Callisto did it in the easiest way possible. Kudos!
 one year ago

CallistoBest ResponseYou've already chosen the best response.7
@FoolForMath 's solution is easier to understand :)
 one year ago

apoorvkBest ResponseYou've already chosen the best response.0
Yea maybe  but before the 'key realisation'  we need the 'key visualisation'. :P
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.2
lol, leave it @apoovk she won't accept that her amazing solution.
 one year ago

apoorvkBest ResponseYou've already chosen the best response.0
yeah she won't, right never does. :P
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.