Free Geometry Set II - 02 Practice Test - CAT 

Question 1

Three balls of equal radius are placed such that they are touching each other. A fourth smaller ball is kept such that it touches the other three.Find the ratio of the radii of smaller to larger ball.  

A. (23)3 
B. 3- 23
C. 3.5
D. 45


Solution : A



Option (a)

Let the centers of the large balls be x, y, z and radius R.

O is the centre of the smaller ball and radius r...

x, y, z form an equilateral triangle with side equal to 2R.

O is the centroid of this triangle.

Therefore ox=oy=oz=R+r= 23(height of the triangle xyz)

Height=(32)(2R) =3R

Therefore R+r = 23(3R)   rR = (23)3 

Shortcut:- Using the approximation technique used in class, the radius of the bigger circle: smaller circle is close to 0.2.


Question 2

There is a ship which is moving away from an iceberg 150 m high, takes 2 mins to change the angle of elevation of the top of the iceberg from 60 to 45. What is the speed of the ship?  

A. 2.46 kmph
B. 1.3 kmph
C. 1.9 kmph
D. 1.7 kmph


Solution : C

 In a 45-45-90 triangle, sides are in the ratio 1:1:2

Thus OB= 150

In a 30-60-90 triangle, sides are in the ration 1:3:2

Thus OA = 503

Distance travelled in 2 mins = AB = 150- 503= 63.4 m

speed = (63.42) × (601000) = 1.9 kmph

Question 3

Two rectangles, ¨ABCD and ¨PQRS overlap each other as shown in the figure below. Also, the overlapped area (shaded region) is 20% ¨ABCD and 33.8% of ¨PQRS. If the ratio of corresponding sides of the two rectangles is same is then ratio AD : PR equals

A. 1.2
B. 1.5
C. 1.6
D. none of these


Solution : D

option (d)


From the figure, 0.338xy = O.2ba  ab = 1.69xy ..... (1)Also, since ¨ABCD and ¨PQRS are similar,bayxbyax..... (2)From (1) and (2), we get ax = 1.69 xaa2 = 1.69x2       a = 1.3x ax= 1.3          answer = 1.3.

Question 4

A metal cuboid with sides in the ratio of 2:4:6 is melted to form small cubes of side 2 cm. If the sum of all the edges of the cuboid is 144 cm, then find the ratio of the total surface area of the cuboid to the total surface area of the cubes.

A. 2:1
B. 13:4
C. 11:54
D. 4:1


Solution : C

Sides are in the ratio 2:4:6

Volume of cuboid= 2x .4x . 6x = 48 x3

Volume of cube= 2.2.2=8

Number of cubes= 6x3

Length of all edges of the cuboid= 144 =4(2x+4x+6x)  x=3cm

Ratio = 2[72+108+216]6(27) x 6 x 4

= 1154

Question 5

With no wastage, small cubes of side 3 inches are formed from a cuboid with the dimensions 2:3:9 are formed. Find the ratio of the body diagonal of the cuboid to that of all of the cubes.  

A. 1:3
B. Ratio = 946a233
C. Ratio = 65a23
D. none of these


Solution : B

Number of cubes = 2a×3a×9a3×3×3 = 2a3
Diagonal of a cube = 4a2+9a2+81a2
Sum of the diagonals of cubes = 2a3 × 33
Ratio = 4a2+9a2+81a22a3×33

Question 6

Suresh is standing on vertex A of triangle ABC, with AB = 3, BC = 5, and CA = 4. Suresh walks according to the following plan: He moves along the altitude-to-the-hypotenuse until he reaches the hypotenuse. He has now cut the original triangle into two triangles; he now walks along the altitude to the hypotenuse of the larger one. He repeats this process forever. What is the total distance that Suresh walks?  

A. 4825     
B. 125    
C. 12
D. 15  


Solution : C

Let M be the end point of the altitude on the hypotenuse. Since, we are dealing with right angle
triangles, ΔMAC ~ ΔABC, so AM = 125. Let N be the endpoint he reaches on side AC. Δ
MAC ~ Δ NAM, So, MNAM =45. This means that each altitude that he walks gets shorter
by a factor of 45. The total distance is thus an infinite G.P. =[a1r] = 12

Question 7

An isosceles right triangle is inscribed in a square. Its hypotenuse is a mid segment of the square. What is the ratio of the square’s area to the triangle’s area?  

A. 1:2
B. 1:3
C. 4:1
D. none of these


Solution : C

Graphical Division: - 

Assume the square to have a side 2a. Hence, area of square = 4a2

Using graphical division, we can divide the figure into 8 parts as follows. The triangle in question is the shaded part



Thus the triangle is 28th of the square area

Hence ratio = 1:4

 Shortcut:- Assumption method. Assume a simple case as the isosceles triangle and square in this case. Just substitute and solve

Question 8

All five faces of a regular pyramid with a square base are found to be of the same area. The height of the pyramid is 3 cm. What is the minimum  total area (integral) of all its surfaces?

A. 102 sq cm
B. 122sq cm
C. 152sq cm
D. 202 sq cm


Solution : B

Altitude of the traingular faces = a24+9
Area of faces =2 x a x a24+9
The total surface area = area of base + lateral surface area
                                    = a2 + 2 x a x a24+9
Lets us put a=8 (smallest value whch will yield an integer area, Area = 64 + 2 × 8 × 5 =144)

= 122

Question 9

In the quadrilateral ABCD, AD = DC = CB, and ADC = 100, ABC = 130. Then the measure of ACB is   

A. 20           
B. 30  
C. 50  
D. cannot be determined


Solution : A

Ans. (a) A, B, C will lie on a circle with centre at D (as the angle subtended by the arc at the centre i.e. 260 is twice subtended at the circle i.e. 130)

In triangle DAC,  DAC = DCA = 40.Let ADB = 2x  ACB = x, and let BDC = CBD = y 2x+y = 100, and 2y+x = 140.

Hence (a) is the right answer

Question 10

Find the distance between the incentre and the circumcentre of a triangle with circumradius 6 and inradius 2?

A. 32
B. 23
C. 10
D. None of these


Solution : B

Use Eulers triangle theorem which states that the distance,d between the incentre and circumcentre of a trinagle is given by d2 = R(R-2r) where R = circum radius, r= inradius

d2 = R(R-2r) = 6(6 - 2 x 2) = 12



Question 11

Two travelers start walking from the same point at an angle of 1500 with each other at the rate of 4 kmph and 3 kmph. Find the distance between them after 2 hours

A. 225+83
B. 28-483
C. 225+123
D. none of these


Solution : C

option c

let OA and OB be the paths traveled by the two travelers in 2 hours. Let BC� AO at C. then BOC= 180-150= 30

In right angled triangle OCB, BC= 62=3 KM and OC= 33 km

In right angled triangle ACB, AB2= AC2 + BC2 = (8+3√3)2+32= 100+483

AB=  225+123


a2+b22abcosα; α is the angle between the paths a and b between the 2 people. 100+48 is the answer

Question 12

ΔABC is an equilateral triangle of side 14 cm. A semi circle on BC as diameter is drawn to meet AB at D, and AC at E. Find the area of the shaded region?

A. 49[π23] cm2
B. 49[π232] cm2
C. 49 cm2
D. 492 cm2


Solution : B

Option (b)

 O is the centre of the circle and the mid-point of BC. DO is parallel to AC. So, DOB = 60°

 Area of Δ BDO =34 * 49

Area of sector OBD = 496

Hence area of the shaded region

 = 2[49634*49]

= 49[1332]


By graphical division, there are 3 equilateral triangles of areas of 34 * 49

Area of interest = (area of semi circle [r22] - area of three triangles) = (49*12 - 3*3*494)*(23) = 49*(1332)

Question 13

Four identical circles are drawn taking the vertices of a square as centers. The circles are tangential to one another. Another circle is drawn so that it is tangential to all the circles and lies within the square. Find the ratio of the sum of the areas of the four circles lying within the square to that of the smaller circle?

A. 2(12)
B. 2(21)2
C. 1(21)2
D. 1(2+1)2


Solution : C

Let the radius of the identical circles be r. Hence, Areas of the four circles lying within the square = 4 x π x r24π x r2 

Radius of the smaller circle = (diagonal of square2r)2 = (22r2r)2 = r (2-1) 

Required ratio = πxr2πx[x2(21)2] = 1(21)2

Shortcut : Lateral Thinking

The question is basically the ratio areas of larger to smaller circle because sum of areas within the square of the 4 circles add up to a full large circle.

Approximately compare the radii of small and large circles; the ratio is nearly 3.5/1, square of which is approx. 6/1. Only option c) works. This is a good approach which can be used for a lot of geometry questions.

Question 14

All sides of a regular pentagon are extended to form a star with vertices P,Q,R,S,T. What is the sum of the angles made at the vertices?___


Solution :

Interior angle = [(2n-4)x90]/n = 108

2x= 360-(108x2) = 144

angle P = 180-144= 36

Total = 36x5= 180

 Useful to remember: Sum of angles in a star= 180

 Shortcut: Clearly the star point trisects the angle in a pentagon. Hence each angle = 108/3 = 36.

Sum of angles = 36*5 = 180

Alternatively:  Sum of angles of a n point star = (n – 4) x 180 , where n = number of sides of a polygon.

Here, n = 5. Sum of angles of a n point star = (5 – 4) x 180 = 180.

Question 15

In the figure given below, ABOP is a rectangle and O is the centre of the circle. It is also given that AB = BC and the measure of the angle ABC is 60. Find
the measure of the angle OPN.  

A. 30
B. 15
C. 20
D. 25


Solution : B


AB = BC and ABC=60°. Therefore, ΔABC is an equilateral triangle

Now see that ABOP is a rectangle.

And BAN=60°, Therefore, NAP=90 − 60 = 30

And ANP =12 * 90 = 45

Now in ΔANP,


And hence NPO = NPA - OPA = 105 - 90 = 15


ABC is an equilateral triangle and ABM is a 30 – 60 -90 triangle (M being the point of intersection of AN and the circle). OMN is also 30. MOP = 90, and MNP = 45; MPO = PMO = 45. NPO = 180 – 75 – 45 – 45 = 15