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Geometry for CAT 2011

Hi Guys, Geometry, Algebra and Number system form the major chunk of our QA section for CAT. Proficiency in these three sections would definitely boost our Quants scores. Contents of Geometry 1\. Plane Geometry - Basics and Tri...
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Please give solution
Its AIMCAT question of 1014 and i didnt understand TIME's solution

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ABC is a triangle in which AD=3CD and E lies on BD .DE=2BE.what is the ratio of area of triangle ABE and triangle ABC?
please explaIN THE METHOD. see the attached pic .



AD:CD = 3:1
hence, ar (ADB)=3a and ar(CDB)=a

DE:BE = 2:1
hence, ar(ADE):ar(ABE)=2:1
let ar(ABE)=b
hence, 3a-b/b = 2/1
hence, a=b
hence, required ratio = 1:4
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ABC is a triangle in which AD=3CD and E lies on BD .DE=2BE.what is the ratio of area of triangle ABE and triangle ABC?
please explaIN THE METHOD. see the attached pic .

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Let centers be P and Q for circles with r=3 and 12 respectively
Drop PM and PN perpendicular to AB and PD and PE perpendicular to AC
Let AM=AD=y; EC=OC=x ( O is on where tgt AB touches the bigger circle.

Triangle APM~AQN
AM/AN=3/12 = 1/4
=> y=3
Use pythagoras in ABC, (x+y+12)=(x+12)+(y+24)
=>x=84
Getting area as 1344. Might be some calculation mistake, what's the OA??



Draw OM perpendicular to PR, ON perpendicular to PT and OQ perpendicular to AB


area of the ABC is 486 unit .
angle BOA equals 70.:D
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The circles are tangent to each other and circle is tangent to the side of the right triangle ABC with right angle ABC.if the larger circle has a radius 12 and smaller one has a radius 3, Then find the area of ABC.

See attached pic for more info



Let centers be P and Q for circles with r=3 and 12 respectively
Drop PM and PN perpendicular to AB and PD and PE perpendicular to AC
Let AM=AD=y; EC=OC=x ( O is on where tgt AB touches the bigger circle.

Triangle APM~AQN
AM/AN=3/12 = 1/4
=> y=3
Use pythagoras in ABC, (x+y+12)=(x+12)+(y+24)
=>x=84
Getting area as 1344. Might be some calculation mistake, what's the OA??

Asfakul Says
Triangle PAB is formed by three tangents to circle O and angle APB =40.then angle BOA equals...


Draw OM perpendicular to PR, ON perpendicular to PT and OQ perpendicular to AB
Corporate Communications Cell - MDI Gurgaon
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Triangle PAB is formed by three tangents to circle O and angle APB =40.then angle BOA equals...

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The circles are tangent to each other and circle is tangent to the side of the right triangle ABC with right angle ABC.if the larger circle has a radius 12 and smaller one has a radius 3, Then find the area of ABC.

See attached pic for more info

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sorry but i dint know how to upload pics..bare wid me

Construction: extend RS and drop a perpendicular to T.Such that u get a right angle triangle STR with T=90 degree
TQ=x and PQ=6
let angle SPT=angle TPQ=x
angle PTQ=90-x
angle PTS=x
This means in triangle SPT SP=ST(angle PTS=angle SPT)=y
now, triangle PQR~STR
PQ/ST=QR/TR=PR/SR
6/y=8/x+8=10/y+10


Solve the above equation u will get y=15 and x=12
So RT=8+12=20 cm;)



thanks a lot. Enlightened!!
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yes the answer is 20,

kindly enlighten!!

Thanks


sorry but i dint know how to upload pics..bare wid me

Construction: extend RS and drop a perpendicular to T.Such that u get a right angle triangle STR with T=90 degree
TQ=x and PQ=6
let angle SPT=angle TPQ=x
angle PTQ=90-x
angle PTS=x
This means in triangle SPT SP=ST(angle PTS=angle SPT)=y
now, triangle PQR~STR
PQ/ST=QR/TR=PR/SR
6/y=8/x+8=10/y+10

Solve the above equation u will get y=15 and x=12
So RT=8+12=20 cm;)
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scan002 Says
is the OA 20??was it necessary to apply angle bisector theoram..got it widout applyin it:shocked:


yes the answer is 20,

kindly enlighten!!

Thanks
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22510

the above question is to be solved by ANGLE BISECTOR THEOREM

i did not get how to apply this theorem. can someone kindly explain explicitly as to how the theorem has been applied in this problem.

Thnks


is the OA 20??was it necessary to apply angle bisector theoram..got it widout applyin it:shocked:
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22510

the above question is to be solved by ANGLE BISECTOR THEOREM

i did not get how to apply this theorem. can someone kindly explain explicitly as to how the theorem has been applied in this problem.

Thnks

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Yes. Please explain
Thanku




d1^2 = h^2+(x+16)^2....................a
d2^2 = h^2+(10-x)^2.....................b

eq1. h^2+x^2=8^2
eq2. 12^2=h^2+(16+x-10)^2

Adding a and b then replace h and x from eq1. and eq2.
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Is it option (2) 528??
I'll post the approach if it's correct!

Is the answer b) i.e. 528..........??
If Correct will post solution.......


Yes. Please explain
Thanku
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an easy one.. but pls explain
ABCD is a trapezium in which AD is parallel to BC. if the 4 sides AB,BC,CD and DA are respectively 8,10,12,16, find the magnitude of the sum of the squares of the two diagonals
1)488 2)528 3)568 4)658

Is the answer b) i.e. 528..........??
If Correct will post solution.......
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