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Refresher Test 2 - Geometry Launch Test Discuss
Refresher Test - Geometry Launch Test Discuss

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Practice Test - Geometry Launch Test Discuss

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Parallel lines

Free Online Preparation for CAT with Minglebox e-CAT Prep. Cover basic concepts of Geometry under Quantitative Aptitude for MBA Entrance Exam Preparation with Study material, solved examples and tests prepared by CAT coaching experts.


Two lines lying in a plane are said to be parallel if they never meet.




A closed figure, bounded by finite number of line segments, all lying in the same plane, is known as polygon.

Sum of interior angles of a polygon having 'n' sides, =

Sum of exterior angles of a polygon = 360 ( irrespective of number of sides)




A triangle is a polygon formed by 3 line segments joined end to end.

Sum of interior angles of a triangle = 180

In any given triangle sum of 2 interior angles is equal to the third remote angle.

AD is the median of triangle, then

G is the centroid of the triangle .Centroid of a triangle divides the median AD in the ratio 2:1



 Angle Bisector Theorem


If AD is the bisector of the the angle ABC then AD divided the side BC in the ratio


Circle Theorem


If AC is a chord of circle then angle ABC subtended by the chord at a point B on the circle is equal to



to angle subtended at point 0 lying in the same segment ie ∟ABC = ∟ADC. Also angle ABC is half of

angle AOC.



Alternate Segment Theorem


If XTB is a tangent to the circle then ∟ATB =∟ACT.

Area of a triangle =



Area of a circle of radius r =

Area of a square of side

Length of diagonal of square =

Area of rectangle of length l and breath b= lb

volume of cube of edge length

length of cube =

Volume of sphere of radius r =

Volume of hemisphere of radius r =

Surface area of cube =

Surface area of sphere =

Surface area of hemisphere =






The sum of the areas of two circles, which touch each other externally, is 153 . If the sum of their radii is 15, find the ratio of the larger to the smaller radius.




(1) 4

(2) 2

(3) 3

(4) none of these.




Let the radii of the 2 circles be and , then = 15 (given)

and = (given)

= 153

solving we get, = 12 and = 3

ratio of the larger radius to the smaller one is 12 : 3 = 4 : 1 hence option (1) is the answer.






In the adjoining figure, points A, B, C and D lie on the circle. AD= 24 and BC = 12. what is the ratio of the area of CBE to that of the triangle ?




(1) 1:4

(2) 1:2

(3) 1:3

(4) data insufficient.




In and , ∟ CBA = ∟ CDA.

( a chord of a circle subtends equal angel at all points on the circumference, lying in the same segment)

similarly ∟ BCD = ∟ BAD and ∟ BEC = ∟ AED

Therefore ( AAA similarity rule)


Hence hence option b).



and are two concentric circles while BC and CD are tangents to at point P and respectively. AB to the circle , which of the following is true?





(4) data insufficient




l ( cp) = l (cr) ( since tangents drawn from the same point are equal)

hence l(BC) = l( CD) ( points P and R are midpoints of BC and CD as OP is ┴BC and OR ┴ CD)

Therefore m( arc BXC) = m(arc CYD)
hence quadrilateral OPCR is a cyclic quadrilateral.

Hence answer is option c)


But m(arc BXC ) = ( ∟ABC) = ( tangent secant theorem)


therefore m(arc BZD) =


Therefore m(BCD) =


Hence m∟PCR =



m∟POR = 2m ∟PQR = (angle at center)



m∟OPC + m∟ORC =



Hence m∟POR + m∟PCR =







Two circles of radius 3 units and 4 units are at some distance such that length of the transverse common tangents and the length of their direct common tangents are in the ratio 1: 2 . What is the distance between the centres of those circles?

(1) units

(2) units

(3)        8 units



(4)        cannot be determined .








Let x = distance between the centres of the circles.

T= length of the transverse common tangent.

D = length of direct common tangent.



And = 4 units and = 3 units 




then, t = --------------- 1) 




also d =  




------------------- 2) 



















hence x =


from 1) and 2 ) we get

d = 2t

d = 8 units

hence answer is option 2 ).



Refresher Test 2 - Geometry Launch Test Discuss
Refresher Test - Geometry Launch Test Discuss

Practice Test - Geometry Launch Test Discuss

Comments Add Comment

amit39967 : need a bunch of 1000 questions

ankur36512 : i m not able to understand the concepts .. can any one help me to solve this..

shalu34890 : for deepak5949- height is same and base is double hence 1:2(example-2)

caroline13463 : concepts r short & crisp..really gud..but ans to second illustration is wrong

heena1405 : they r really very good concepts......really rocks...thanks.....

arthi9429 : really very knowledgeable ques.

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vikrant4516 : unbeleivable

dear_bks : its really nice...

saranya2768 : the ans. fr practice test Q.11 is 50

saranya2768 : in the practice test..ques no. 9 has error..i.e 516 is typed as 576

arush3101 : wah kya concepts hein!!!!!!!!!!!!

amit8752 : u people r very very nice ones
i thank u all from the core of my heart

shri3834 : second problem answer is wrong.....
It is 1:4
so ratio consists of product of 2 sides...since angles are equal in both triangles.
so ratio areas is 2*2=4 i.e 1:4....

sonia4863 : give me a very good qns

shilpigarg3707 : its really a very helpful concept for every student and its will help to understand the cat question.Thankxxxxxxxxxxxxxxxxxxxxxx for this

deepak3520 : sir
plz give me much more topic of DI,function mean QA,and passsage of english

dev3780 : its absorbing material

nitin481 : deepak5949 u r absolutely correct. In similar triangles ratio of area is equal to squares of ratio f their sides.

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vcchoudhary : very very unbelievable concept. Really, this is very knowledgable info for me

ankit4877 : gd questions to open our mind,,,,

deepak5949 : Solution to second illustration of geometry is incorrect.
The answer should be 1:4 and not 1:2 because ratio of area is required to be calculated.

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bhavya7710 : Really like the material on geometry.

: material is good but tests are not workin ...it would be really helpful if the problem is sorted out asap

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