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Geomtery Frolics- Some special problems
by Total Gadha - Monday, 25 December 2006, 04:36 AM
 

I am a compulsive problem solver, and always on hunt for new problems that make me sweat my brains. Here are some of the problems that I enjoyed solving during the wee hours of many nights. I will keep on updating problems in this article. Although the level of geomtery required in MBA exams is lower than that of the problems covered here, I have experienced that solving problems like these tests the fundamentals and strengthen the ability to apply all the fundamentals at the same time. Next week I am going to pose a challenger here, a problem that will test all your geometry fundamentals in one go. Till then, enjoy these!- Total Gadha

image

solution 1

Problem 2

solution 2

problem 3

solution 3

Join our CBT Club for more such problems in geometry. We shall cover some interesting problems in CBT Club for you guys.

 

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Re: Geomtery Frolics- Some special problems
by Die Hard - Tuesday, 10 April 2007, 03:57 PM
 

hi tg......

i did the 3 probs.......

they were nice.......can u post some more intiguing problems ....

thanks

 

 

Re: Geomtery Frolics- Some special problems
by Total Gadha - Tuesday, 10 April 2007, 09:54 PM
  Hi Bharat,

I am going to upload many good problems in the 'geometry problems' quiz. Try that out soon.

Total Gadha
Re: Geomtery Frolics- Some special problems
by Mitesh Chaturvedi - Thursday, 26 April 2007, 04:37 PM
 

Hi TG,

       I am trying some problems on geometry now a days.Whenever I see a problem consisting of circumcircle or inercircle I get confused.Could you please provide some lessons in geometry on these topics.

Thanks,

Mitesh 

Re: Geomtery Frolics- Some special problems
by Total Gadha - Thursday, 26 April 2007, 06:55 PM
  Hi Mitesh,

I was wondering where you were. Good to see you back. Let me try and write the lessons soon. smile

Total Gadha
Re: Geomtery Frolics- Some special problems
by prakhar srivastava - Saturday, 28 April 2007, 11:45 AM
 

hi tg..

really u r  doin a gr8 job

byee

Re: Geomtery Frolics- Some special problems
by Total Gadha - Saturday, 28 April 2007, 04:17 PM
  Thanks Prabhakar smile
Re: Geomtery Frolics- Some special problems
by mohammed javed - Friday, 18 May 2007, 11:46 AM
 

hi Tg!

i dint understand t second solution can u so some more lite on that please!

Re: Geomtery Frolics- Some special problems
by Total Gadha - Saturday, 19 May 2007, 05:09 AM
  Hi Mohammed,

Tell me the part you're facing problem in. I will explain in detail.

Total Gadha
Re: Geomtery Frolics- Some special problems
by shilpi govil - Wednesday, 23 May 2007, 06:39 PM
 

hi.

 the questions were good but i would want that you plz provide some lessons on topics where a cone is insribed in a cylinder of minimum volume and we r required to find radius or height, or where sphere is inscribed in cone of mininum volume and we have to find its height and similar types.....

Re: Geomtery Frolics- Some special problems
by bimal mohan - Tuesday, 5 June 2007, 01:47 AM
 

hi   TG   sir,

    if    we   take    a  reular  pentagon , suppose  ABCDE.....drawn  diagonal BD  and CE  which  intersects  at   point  O   (say)......IS  the    quad   ABOE   a    parallelogram??   from the   symmetry   it   seem  to  be  ......

what   are   other    related   important   prop   of   regular   pentagon???

thnx  in  advance  !!!

Re: Geomtery Frolics- Some special problems
by Total Gadha - Thursday, 7 June 2007, 09:42 AM
  Hi Bimal,

Yes it is a parallelogram. smile

Total Gadha
easier answer 2 question 2 :)
by Anurag Bansal - Thursday, 7 June 2007, 01:32 PM
  hey guys..just c d last option ..c=600...put dis value..nd c triangle aqc..use sign rule..ie..ac2=aq2+qc2
ie..4r2=aq2+9x2     -------1    (cause pq=x;pc=2x)
now apply sine rule in triangle aqc...v get: r=(3)0.5x=aq ---------2  (taking c=60o
  thus,qac=60o;acq=30o)  
so using these equations,we get r2=3x2=aq2
thus eq 1 can b verified...and it takes just 1 quarter of a min!!!
Re: Geomtery Frolics- Some special problems
by paramjeet singh - Sunday, 10 June 2007, 04:22 PM
 

hi tg

Problems and solution is too good but i am not good in geometry so pls help me?????????????

Re: Geomtery Frolics- Some special problems
by Justin Case - Sunday, 5 August 2007, 08:19 PM
  Hi TG...
             In the second problem do we have to assume that triangle ABC is a right angled triangle, right angled at B, so that angle AQP is 90 degree??
Re: Geomtery Frolics- Some special problems
by Kshitiz Garg - Monday, 6 August 2007, 09:00 PM
 

have small solutions for the 3 probs above :

1. Considering it a equilateral triangle and assuming n = 2 , we have MN=x, and QR = 2x with M,N as mid points of PQ and PR respectively and S as the Centroid.Let us take a point O on QR s.t. PO is perpendicular to QR. so PO is the height = H , PS = 2H/3 , SO = H/3

Now triangles MNS and RQS are similar , so height of triangle MNS : height of triangle RQS = 2

--> As  SO = H/3 --> height of MNS is H/6.

area ratio = MNS/RQP = [(1/2*x*H/6) ] / [(1/2*2x*H) =1/12

put n=2 in options , d is the answer

2. Let CP = 2a , PQ = a

as triangles APQ and CPB are similar so AP/PB = 2

also QP*PC = AP*PB , so from (1) , (2) we have

AP = 2a , PB = a

so in triangle CPA , CP = AP = 2a -->

Let angle CAP = angle PCA  = x and hence angle QAB = angle QCB  = x

in triangle QAC , angle Q = 90 and 3x = 90 so x = 30

--> angle C = 60

Third later as am in hurry

Re: Geomtery Frolics- Some special problems
by Justin Case - Monday, 6 August 2007, 10:42 PM
  ohh.. i remember Thales...diameter alway makes 90 degrees...(related to the stupid question which I asked earlier..  smile)
Re: Geomtery Frolics- Some special problems
by Mohit Goyal - Friday, 10 August 2007, 05:22 PM
  Hi TG,

Cant i do it with observation?  what  strike  me for 3rd ques is 

Solution -
Let Rectangle having area 18 be R1,
                               area 12 be R2,
                               area  6 be R3,
                               area  X be R4,
For this rectangle   ratio of R1 and R3 = ratio of R2 and R4

R1/R3=  R2/R4
18/6 = 12/X
therefore X = 4.

Is this solution right ??

Mohit                                       
                  
Re: Geomtery Frolics- Some special problems
by anamika verma - Wednesday, 26 September 2007, 01:02 AM
 

Hi Kshitiz/TG

"Let angle CAP = angle PCA  = x and hence angle QAB = angle QCB  = x"

Why is angle QAB = angle QCB = x ?

Re: Geomtery Frolics- Some special problems
by Vinod Paramasivan - Tuesday, 2 October 2007, 05:25 PM
  Hi Anamika,

QAB and QCB are angles subtended by same arc

Also  AQP~CBP -> ang QAP = ang BCP

Regards,
Vinod
Re: Geomtery Frolics- Some special problems
by anamika verma - Thursday, 4 October 2007, 12:09 AM
 

Hi Vinod/Ksitiz/TG

In the statement "Let angle CAP = angle PCA  = x and hence angle QAB = angle QCB  = x",

I understand that angle CAP = angle PCA. I also understand that angle QAB = angle QCB .

But why is angle CAP = angle QAB ?

Please help.

Re: Geomtery Frolics- Some special problems
by Gaurav Verma - Tuesday, 30 October 2007, 04:29 PM
  Hello sir,
in q-2  cant we solved by this way...
If we extend cb and aq  and make another triangle let say acd...as cp:pq = 2:1 then it should be the median of new triangle..and angle cqa =90 so its also altitude..so the acd is equilitral? or ..so angle C=60?? pl suggest..
Re: Geomtery Frolics- Some special problems
by praveen snss - Tuesday, 30 October 2007, 06:56 PM
 

Hey Gaurav,

Can't that b an isoceles triangle.....?

Re: Geomtery Frolics- Some special problems
by Gaurav Verma - Tuesday, 30 October 2007, 09:14 PM
  Hi praveen , ya its an isoceles tri. with common angle 90-c/2 and c. so if C=60 ..90-60/2 also 60...thanks buddy..smile
Re: Geomtery Frolics- Some special problems
by Sadiq M - Friday, 30 May 2008, 05:34 PM
 

Hi TG Sir,

I could not understand how are u calculating the height of triangle QSR in

first question and hw r we obtaining ht of triangleMSN = UT *1/n+1

Can u plz explain it

 

Re: Geomtery Frolics- Some special problems
by Venkat Iyer - Tuesday, 3 June 2008, 06:45 AM
  HI TG..Ive been visitin ur site regularly offlate..Hats off!! I have a doubt..Cud u explain why volume of cone is 1/3rd volume of cylinder..
is there mathematical proof 4 it??
Re: Geomtery Frolics- Some special problems
by hehe haha - Wednesday, 4 June 2008, 10:05 AM
 

Hi Sadiq,

UT = ht of MSN + ht of QSR

ht of MSN / hr of QSR = 1/n

so, ht of QSR = n * ht of MSN

UT = (1+n) * ht of MSN

thrfr : ht of MSN = UT /(n+1)

i think u r clear now ....

 

Re: Geomtery Frolics- Some special problems
by Sadiq M - Wednesday, 4 June 2008, 06:58 PM
 

ya im clear

thanks for replying

Re: Geomtery Frolics- Some special problems
by Neo Sinha - Saturday, 7 June 2008, 02:44 AM
 

Yes there is mathematical proof for the same. This is calculated by area under the curve method in calculus.

  The cone can be considered as a stack of no. of circular strips. which upon integration form 0 to h gives 1/3 (pi) R*R h

Suppose we want to find the volume of a circular cone of base radius r and height h, we slice the cone into an infinite number of circular discs of thickness dx (imagin) and find the volumes of all these discs and add them up. Consider the origin as the pointed end of the cone and the axis through the center of each cross section perpendicular to the base as the x-axis. Consider a line through the origin perpendicular to the x-axis as the y-axis. Consider a circular disc of thickness dx at a distance x from origin. Let y be the radius of this disc. We know that volume of this disc = pi.y.y.dx where pi = 3.14 nearly. As integration is summation the volume of the cone = integral of pi.y.y.dx for x limits 0 to h. We know y/x = tan a where a is the angle of the cone. Also r/h = tan a. So y = (r/h)x.  Substituting this value of y in terms of x  and integrating between the limits 0 to h we get the volume of the cone as (1/3)pi.r.r.h (one third pi r squared h).

V=intergral (Area*dx) with lower limit=0 &  upper limit=height of cone

 

 

Re: Geomtery Frolics- Some special problems
by Badal Vishal - Tuesday, 24 November 2009, 08:20 PM
  Hi TG,

This is my first post.
I have problem in finding the optimized volume/curved surface area of a cylinder inscribed in a cone of radius r and height h. When similar thing is done in a semi sphere we take half of the semi sphere and find the optimized square that can be taken inside this part. Please help!!

Re: Geomtery Frolics- Some special problems
by Arindam Das - Thursday, 13 January 2011, 10:53 PM
 

can u plz tel me if books for maths olympiad good for preparaing for CAT....i m currently in 2nd yr Btech in NITS n in my XIIth i had cleared the mathematics olympiad(state level and appeared RMO)

Re: Geomtery Frolics- Some special problems
by Aakash Singh - Thursday, 19 May 2011, 11:23 PM
  Wats going on. Sbd plz explain
Re: Geomtery Frolics- Some special problems
by Priyam Garg - Friday, 11 March 2016, 12:26 AM
  Sir,
Could you please offer a non trigo method to solve Q2.

Thanks smile