Here is the link to the previous year's Geometry thread.
http://www.pagalguy.com/forum/quantitative-questions-and-answers/69681-geometry-for-cat-2011-a.html
All the best for CAT 2012 journey!!! [smiley]
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Geometry for CAT 2012
Quantitative
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Let's have this thread for discussion of questions related to Geometry.
Here is the link to the previous year's Geometry thread.
http://www.pagalguy.com/forum/quantitative-questions-and-answers/69681-geometry-for-cat-2011-a.html
All the best for CAT 2012 journey!!! [smiley]
Here is the link to the previous year's Geometry thread.
http://www.pagalguy.com/forum/quantitative-questions-and-answers/69681-geometry-for-cat-2011-a.html
All the best for CAT 2012 journey!!! [smiley]
Let the centre of the bigger circle be C and that of the smaller one be B. Drop a perpendicular from B to CP; at suppose S. As angle CPQ and angle BQP ar 90 dergrees, PQBS is a rectangle. Thus CS = CP-SP = 9-4 = 5. In triangle CBS, CS^2 + BS^2 = CB^2 ; and so, 5^2 + BS^2 = (9+4)^2 or BS^2 = 169...
Q- Two circles of radius 9 cm n 4 cm resp. touch each other externally at a point n common tangents touches them at the point P n Q resp. then the area of square with one side PQ is...
pls give detailed solution with figure puys???????? @maddy2807
pls give detailed solution with figure puys???????? @maddy2807
hm
@pooja888 : not alwasy pooja u have taken a special case here for @xitizsharma 's q
for
@ajeetaryans FOR FINDING area for three horses...what wil be d area of circle??..on ur approach
90/360*22/7*r*r /area
multiple by n no of horses.
90/360*22/7*r*r /area
multiple by n no of horses.
Thanks a Lot Cat Head
Area of equi triangle gives that side of the triangle = 6
Now , let the side of the square be X.
So we have a right angled triangle with one angle 60' and sides x and (6-x)/2
So tan60 = x / (6-x)/2
Solve this and get x
Now , let the side of the square be X.
So we have a right angled triangle with one angle 60' and sides x and (6-x)/2
So tan60 = x / (6-x)/2
Solve this and get x
The area of n eqilateral Triangle is 9rt3 Sq cm Can u tell me what is the side of a square of maximum area which can be inscribed inside the Triangle??????
It's a unit of length and not volume.
I get it now.Thanks man!!
I get it now.Thanks man!!
A vessel of capacity 1100 lts gets filled by a pipe in 7 hrs. Length of the pipe is 50m and flow rate is 8 ltrs/min. What is the diameter of pipe?
Thanks for the answer.The doubt was that in the question it says 'Water flows out at the rate of 10m/min from a cylindrical pipe of diameter 5mm'. So what I inferred from it was that we dint need the radius of the pipe and could directly divide it by 1000. Why do we have to multiply it with the ...
Can you explain?
Water flows out at the rate of 10m/min from a cylindrical pipe of diameter 5mm.Find the time time taken to fill a conical tank whose diameter at the surface is 40cm and the depth is 24cm.
BF
thanks man...
This holds only for Equilateral triangle. As the sides u are using in the formula varies if the triangle are other than equilateral triangle.
1)Circumcentre = 2 X Incentre2)Circumcentre=Side/root(3) and incentre=Side/2root(3)
Are the above relations true just for equilateral triangles or any kind of triangle? ? ?
Are the above relations true just for equilateral triangles or any kind of triangle? ? ?
We hav to calc. BF or BE.??
@SRV91 pls help
In the quadrilateral ABCD, ∠BAD = ∠BCD = 90°. If AE = 6 cm, CE = 15 cm and DE = 10 cm, find the length of BF.
a) 9 b) 8 c) 10 d) 15 e) 20
a) 9 b) 8 c) 10 d) 15 e) 20
thank u very much
1
@quant89 Hey...In triangle ABC, AB= 24 cm ( Pythagoras theorem). Then, in triangle ADC ,use angle bisector theorem which gives us AB/AC = BD/DC. Substituting the value we get 24/25=BD/BD+7 . Solve and you get BD as 168 cm. Now, CD= BD+BC= 168+7=175 cm . :)
1
let bd=x and equate the area of triangle..asAr(ADC)=Ar(ABC)+Ar(DBA)
ABC is a right angled triangle, right angled at B. The external bisector of angle A meets CB produced at D. If AC = 25 cm, BC = 7cm, find the length of CD.
This Q is from TIME material
ans is 175 cm
can any1 help me with illustrated solution of this Q since i couldn't understand the soln...
This Q is from TIME material
ans is 175 cm
can any1 help me with illustrated solution of this Q since i couldn't understand the soln...
@blade014
if AB=AC to angles kaise equal huaa...plzzzzzzzzzzzzzzz explain
if AB=AC to angles kaise equal huaa...plzzzzzzzzzzzzzzz explain
@rubikmath check the answer
Drop the perpendicular PT to QR.
Now,
PS^2 = PT^2 + ST^2
=> 100 = h^2 + ST^2 - - - - - (1)
and
PQ^2 = PT^2 + QT^2
=> 121 = h^2 + QT^2 - - - - - (2)
Subtract eqn (1) from (2)
21 = QT^2 - ST^2
Now, 21 can be written as 5^2 - 2^2 or 11^2 - 10^2
But we will eliminate ...
Now,
PS^2 = PT^2 + ST^2
=> 100 = h^2 + ST^2 - - - - - (1)
and
PQ^2 = PT^2 + QT^2
=> 121 = h^2 + QT^2 - - - - - (2)
Subtract eqn (1) from (2)
21 = QT^2 - ST^2
Now, 21 can be written as 5^2 - 2^2 or 11^2 - 10^2
But we will eliminate ...
1
@vinitayan
yup.
approach ?
yup.
approach ?
@rubikmath Is the answer 10 cm
In triangle PQR , PQ=PR=11 and S is a point on QR such that PS=10cms. If the lengths of QS and SR are integers, find the length of QR .
thanks for the solutions. @pooja888 the file is attached.
The answer is 55'
Angles subtended by opposite sides would be supplementary. So, should be 55.Check this link:http://www.meritnation.com/discuss/question/685206
Hi Need help in solving Q.2 in ex 9.1 (find angle COD). Pl see attachment for question
@pooja888
Didnt really understand ur symmetry concept..but it isnt required i guess..
Let radii be R, a ,b..
then the sides become R-a, R-b, a+b..hence perimeter becomes 2R..
was quite a sitter...
Didnt really understand ur symmetry concept..but it isnt required i guess..
Let radii be R, a ,b..
then the sides become R-a, R-b, a+b..hence perimeter becomes 2R..
was quite a sitter...
Can U post the question?
1
cn any one pls explain me the q no 14 of LOD 2 of geometry in arun sharma hw to solv itThe equlatral triangle.........................I m raeding it online so i dnt hv book to post the pic
@chatman @Estallar12 thnku bothfrnds fr helpin me
2
See, if you will join the centres of the circle to the point of contact of the outer boundary for the respective circles, you will get three straight lines of 2cm each. So, 6 cm.
Now, rest part that is on the circumference of the circle is nothing but 3*the circumference of 120 degree arc of t...
Now, rest part that is on the circumference of the circle is nothing but 3*the circumference of 120 degree arc of t...
@chatman 3 circular arc shared by 3 drums each subtending 120 degree angle at respective centrewhich is effectively equal to circumfrence of one drum. i m nt able to undrstand ths concept :-(