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MBA CAT 2024
761

Let ABCD is a square of length unity. A point Q is taken on BC and joined with P which is the mid-point of AB. If (

a)1/9
b)1/8
c)1/4
d)1/6

need help guys!!
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@raku1989 :

The square is divided into the following areas:
Area of APD : 1/4(area of square) = 1/4

Let BQ = x then QC = 1-x
DQ = root(1 + (1-x)^2)
DP = root(1+1/4) = root(5/4)
Area of PDQ :
1/2 * (Sqrt 5/4)(Sqrt(2 - 2x + x^2)sin 45 ( area of triangle with sides a,b,c = .5*a*b*sin(included angle) )

Area of PBQ : 1/2 * 1/2 * x

Area of DQC : 1/2 * 1 * (1-x)

Thus 1 = 1/4 + 1/2 * (Sqrt 5/4)(Sqrt(2 - 2x + x^2)sin 45 + 1/2 * 1/2 * x + 1/2 * 1 * (1-x) and solve this to get x = 2/3

Now Area of PBQ = 1/2 * 1/2 * 2/3 = 1/6 ... correct aano?
763

@raku1989 i am guessing its side unity and not perimeter unity.


Anyways, DP = 5/4(1^2 + 0.5^2)
Let BQ = x and QC = 1-x
Find PQ and DQ using Pythagorean theorem. Then sum of area of all 4 triangles is equal area of square = 1*1=1

Solving,we get x. Use it to find area of of triangle PBQ
I tried doing it but ended up with a lengthy calculation. For area of triangle PDQ we can use 1/2 * ABsinC formula also.
764
@pugna said: @raku1989 :The square is divided into the following areas: Area of APD : 1/4(area of square) = 1/4Let BQ = x then QC = 1-x DQ = root(1 + (1-x)^2) DP = root(1+1/4) = root(5/4) Area of PDQ : 1/2 * (Sqrt 5/4)(Sqrt(2 - 2x + x^2)sin 45 ( area of triangle with sides a,b,c = .5*a*b*sin(included angle) )Area of PBQ : 1/2 * 1/2 * xArea of DQC : 1/2 * 1 * (1-x)Thus 1 = 1/4 + 1/2 * (Sqrt 5/4)(Sqrt(2 - 2x + x^2)sin 45 + 1/2 * 1/2 * x + 1/2 * 1 * (1-x) and solve this to get x = 2/3Now Area of PBQ = 1/2 * 1/2 * 2/3 = 1/6 ... correct aano?
Awesome man!!....yup 1/6 is the right ans!! :)
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@seetharam7 said: @raku1989 i am guessing its side unity and not perimeter unity.
Anyways, DP = 5/4(1^2 + 0.5^2)
Let BQ = x and QC = 1-x
Find PQ and DQ using Pythagorean theorem. Then sum of area of all 4 triangles is equal area of square = 1*1=1
Solving,we get x. Use it to find area of of triangle PBQ
I tried doing it but ended up with a lengthy calculation. For area of triangle PDQ we can use 1/2 * ABsinC formula also.
Check @pugna soln above
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@raku1989 : thanx man :)
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Btw,i got a WAT/PI call for IIM I's oodayippu at mumbai 😁 Guess i'll attend the process(on 30th,at bangalore) just to get an idea about an IIM interview..

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@pradyoth That's great buddy..... I wish u all the very best

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@pradyoth Hope u'll be a part of today's PG marathon.Hope to see ur name on the leader board

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ALL THE BEST TO EVERYONE WHO IS PARTICIPATING IN THE PG MARATHON TODAY
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@rohan_bhasker yeah,i'll be there.. 😁 the questions are all from quant,right? so no chance 😛 😁 It'll be fun though 😃
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@pradyothcjohn said: @rohan_bhasker yeah,i'll be there.. the questions are all from quant,right? so no chance It'll be fun though
during beta (today and tomorrow) questions will be from Quant. During the final launch we will post a schedule which covers all areas of CAT prep.
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@deepu said:
during beta (today and tomorrow) questions will be from Quant. During the final launch we will post a schedule which covers all areas of CAT prep.
Looking forward to a fun learning experience :)
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did i miss anything? rushed back after buying supper,i'm home alone 😁

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Hey, the mg university results just came. Just became an engineer. I am in a completely different world right now. Sorry. I am guessing @joethaliath is also not here.

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hehe i was late , yes finally passed B.tech , phew! 😃

btw hats off to seetharam , Branch topper , that too cs being a bio student 😃 .. congrats dude

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@joethaliath said: hehe i was late , yes finally passed B.tech , phew! btw hats off to seetharam , Branch topper , that too cs being a bio student .. congrats dude
Branch topper 😲 awesome @seetharam7 :D
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Congrats

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and yes atb to @pradyothcjohn 😃 make the maximum out of it

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@pradyothcjohn ATB....