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Natarajan @nramachandran

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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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Can some one solve this with proper explanation?

A triangle has sides of lengths 10, 24 and n, where n is a positive integers. The number of values of n for which this triangle has three acute angles is
A. 1 B. 2 C. 3 D. 4 E. 5

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SET - 30


Guys, This is a very good problem !!

In the regular hexagon shown below, what is the ratio of the area of the smaller circle to that of the bigger circle?




a. 3 : 7 + 43
b. 3 : 7 + 163
c. 3 : 7 + 3
d. 3 : 7 + 23

option 1) is the answer

let each side of hexagon be 1 unit

the triangle containing the larger circle is an equilateral tr. with side = root3

now use area = inradius * semiperimeter

1/2*root3*3/2 = r * 3root3/2

r = 1/2

so area of bigger tr. = 1/4

now , for smaller triangle , sides are 1,1 and root3

same procedure as above gives us r' = root3 / ( 2*(2+root3) )

so area = 3/ ( 4*(2+root3)^2)

thus reqd ratio = 3 : ( 2+root3)^2

or 3 : 7+ 4root3
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ximb (2012-2014)
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SET - 30


Guys, This is a very good problem !!

In the regular hexagon shown below, what is the ratio of the area of the smaller circle to that of the bigger circle?




a. 3 : 7 + 43
b. 3 : 7 + 163
c. 3 : 7 + 3
d. 3 : 7 + 23

  • 1 Like  
PGP 2014-16, Indian Institute of Management Bangalore
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for such questions,always split the quad into 2 triangles.

Join AC ( the diagonal)....now angle ABC= 120...angle ACB=30....angle CAB=30..since ABC is isoceles triangle....using cosine rule calc. AC...now in tri. ADC...we know DC=5...AC is calculated above..and angle DCA = 90...calculate AD and add them up to get the total area of triangle :)


Hi,

Why angle DCA=90? How did u get that?
Team Leader, JP Morgan 
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Hi all,

I have taken this question from previous posts.please explain this again as i am not able to get the answer and explaination given is not helpful.Ho sake to Dig. laga dena :)

An equilateral triangle ABC of side 40 cm is cut into two pieces in such a way that one piece is an equilateral triangle containing the vertex A and the second piece is a trapezium. Two such trapeziums are placed beside each other to form a parallelogram. What is the perimeter (in cm) of the parallelogram?

a. 120
b. 160
c. 200
d. 240


Suppose DE makes the cut to ABC (| to BC)
Since DE is | to BC, the cut will be proportionate..i.e AD/AB = DE/BC
Since AB=BC => AD=DE
Now ADE will be new equilateral triangle...and BCDE will be trapezium
Also since AD + DB = AB
=> DE+DB = AB
Similarly consider another triangle A'B'C' cut by D'E'
Now to form a parallelogram, trapezium B'C'D'E' will be turned upside down
and will be joined to BCDE.
Now the sides of the parallelogram are.. E'BDC'..
and the perimeter = BC + (DE + DB) + (B'C' + C'E') + E'D'
= 40 + 40 + 40 + 40
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Hi all,

I have taken this question from previous posts.please explain this again as i am not able to get the answer and explaination given is not helpful.Ho sake to Dig. laga dena :)

An equilateral triangle ABC of side 40 cm is cut into two pieces in such a way that one piece is an equilateral triangle containing the vertex A and the second piece is a trapezium. Two such trapeziums are placed beside each other to form a parallelogram. What is the perimeter (in cm) of the parallelogram?

a. 120
b. 160
c. 200
d. 240
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Totally confused. Need solution to this question.

Thanks

Angular bisector Formula gives : ( PR/PQ ) = ( RT/QT )
Solve for QT (all other values are known). Hence you will get RT
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tips ..median divides a triangle into 2 halves.answer 90

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Kudos..Its 4:9

Hey Blue4life, can you please help me solve the question I posted above.
I'll appreciate.

Thanks



Hey Prateek,

say /_QPR = P
=> /_QPT = 90-P/2
=> /_QTP = P/2
U also have Cos(P) = 6/10
Tan(P/2) = sqrt(1 - Cos(P) / 1 + Cos(P)) = 1/2
and also in triangle PQT, tan(/_QTP) = tan(P/2) = 6/QT = 1/2
QT = 12
and RT = 20


One more method...
By external angle bisector theorem
RT/RP = QT/QP
RT/10 = (RT-8 )/6
RT = 20
  • 1 Like  
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Angle BEC = Angle AED (vertically opposite angles)
Angle CBD = Angle CAD (property of cyclic Quadrilateral)

So the Triangles AED and BEC are similar triangles thus there area ratio will be square of the ratio of sides which is 2:3. Thus 4 : 9


Got it mahn...Thanks
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