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...

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/discussions/geometry-for-cat-2011-25069681

All the best for CAT 2012 journey!!!

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/discussions/geometry-for-cat-2011-25069681

All the best for CAT 2012 journey!!!

Apart from that, here is the link of Handa Sir geometry post.

Shaping up your Geometry skills for CAT 2011 - Part 1 | PaGaLGuY.com - Indias biggest website for MBA in India, International MBA, CAT, XAT, SNAP, MAT

Shaping up your Geometry skills for CAT 2011 ? Part 2 | PaGaLGuY.com - Indias biggest website for MBA in India, International MBA, CAT, XAT, SNAP, MAT

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/discussions/geometry-for-cat-2011-25069681

All the best for CAT 2012 journey!!!

thnx for the initiative....

P.S. Sorry 4 the spam post

is this thread dead?

subscribing ...

**Triangle ABC is right-angled at A. D is a point on AB such that CD = 1. AE is the altitude from A to BC. If BD = BE = 1, what is the length of AD?**

(1) 2^1/3 - 1 (2) (5^1/2 - 1)/2 (3) (5^1/2 + 1)/4 (4) 2^1/2 - 1 (5) none of these

ans=2^1/3-1

Solve it...................

(1) 2^1/3 - 1 (2) (5^1/2 - 1)/2 (3) (5^1/2 + 1)/4 (4) 2^1/2 - 1 (5) none of these

ans=2^1/3-1

Solve it...................

racinghorror Saysis this thread dead?

not until cat 12....

nice attempt for a separate thread..ATB TO ALL..!!! CHEERS

hjroks Saysnice attempt for a separate thread..ATB TO ALL..!!! CHEERS

solve the question given above,,,,,,,,,

I will be posting more...........

can anyy1 plz tell--

A circle of maximum possible size is cut from a square sheet of board. Subsequently, a square of maximum possible size is cut from the resultant circle. What will be area of the final square?

can anyy1 plz tell--

A circle of maximum possible size is cut from a square sheet of board. Subsequently, a square of maximum possible size is cut from the resultant circle. What will be area of the final square?

Diameter of circle = side of outside square = a ;

Diagonal of inside square = a;

hence side = a/sqrt(2);

Area = a^2/2 = Half of outside square

plz tell-

Q) two line segments PQ and RS intersects at X in sucha way XP=XR. if PSX=RQX, then one must have-

1. PR=QS

2. PS=RQ

3. XSQ=XRP

4. ar(triangle PXR)=ar(triangle QXS)

plz tell-

Q) two line segments PQ and RS intersects at X in sucha way XP=XR. if PSX=RQX, then one must have-

1. PR=QS

2. PS=RQ

3. XSQ=XRP

4. ar(triangle PXR)=ar(triangle QXS)

Join PR and QS.

hey guyz please answer following ques...

q1)In triangle ABC ,the internal bisector of angle a meets BC at D.If AB=4 AC=3 and LA(angle A)=60 degree, then length of AD is what??

Q2) The lenght of the common chord of two circles of radii 15 and 20cm respectively,whose centeres are 25cm apart is

1)24cm 2)25cm 3) 15cm 4)20cm

Q3) 4 horses are yied at four corners of a square plot of side 14m so that the adjacent horses can just reach one another. There is a small circular pond of area 20m^2 at the centre.find the ungrazed area??

1)22 2)42 3)84 4)168

hey guyz please answer following ques...

q1)In triangle ABC ,the internal bisector of angle a meets BC at D.If AB=4 AC=3 and LA(angle A)=60 degree, then length of AD is what??

Q2) The lenght of the common chord of two circles of radii 15 and 20cm respectively,whose centeres are 25cm apart is

1)24cm 2)25cm 3) 15cm 4)20cm

Q3) 4 horses are yied at four corners of a square plot of side 14m so that the adjacent horses can just reach one another. There is a small circular pond of area 20m^2 at the centre.find the ungrazed area??

1)22 2)42 3)84 4)168

ans : - Q.1

area of triangle = 1/2 AB * AC * SIN A

= area of trianlge BAD + area of triangle CAD

SO 1/2 * 4 * 3 SIN 60 = 1/2 * 4 * AD SIN30 + 1/2 *3 *AD SIN30

AD = 12 *(sqrt of 3) / 7

Q.2

ans = triangle will be right angle triangle

bcs ratio of sides 3:4:5

so area of triangle

1/2 * base * perpendicular = 1/2 * hypotenuse * (half of chord )

1/2* 15 * 20 = 1/2 * 25 * half of chord

so lengh of chord = 24

Q.3

= area of square - area of circle(formed by all four horses with radius 7)

= 14^2 - pie 7^2

196 - 154 = 42

now subtract area of circle which is lie at center

= 42 -20

= 22 ans

:clap::clap::clap::clap::clap::cheers::cheers::cheers::cheers::cheers:

can anyy1 plz tell--

A circle of maximum possible size is cut from a square sheet of board. Subsequently, a square of maximum possible size is cut from the resultant circle. What will be area of the final square?

Circle radius =a/2

(Side of internal square)^2= 2*(a/2)^2 = a^2/2 my take.

If in a triangle ABC, the medians CD and BE intersect each other at O, then the ratio of the areas of triangle ODE and ABC is???

freedom fighter SaysIf in a triangle ABC, the medians CD and BE intersect each other at O, then the ratio of the areas of triangle ODE and ABC is???

is it 1:4 as the pt of intersection is centroid it divides the line in ratio 2:1 ..

look in traingles ebf and efd both have the same height and their bases are in ratio 2:1 therefore area(ebf)=2*area(efd).

now have a look at traingles ebf and bfc,they again have common heights and their bases in ratio 1:2(in a traingle the medians divide each other in 2:1) 2*area(ebf)=area(bfc).

and the fact that median divides the traingle in 2 equal halves is used.

now area (bdc)=.5*area(abc)

area(bdc)=area(bfc)+area(ebf)=3*area(ebf)=6*area(efd)(from above)

6*area(efd)=.5*area(abc)

area(efd)/area(abc)=1/12