# Free Geometry Set I - 01 Practice Test - CAT

If the length of diagonals DF, AG and CE of the cube shown in the adjoining figure are equal to the three sides of a triangle, then the radius of the circle circumscribing that triangle will be :

A. Equal to the side of the cube
B. √3 times the side of the cube
C.  13 times the side of the cube
D. impossible to find from the given information

#### SOLUTION

Solution : A

The side length of a cube = AD = a

The diagonal length of a cube = AG = a3

DF = AG = CE = a3

The triangle formed was an equilateral triangle.

The circumradius of an equilateral triangle = s33

Therefore, the circumradius of that triangle = a333
= Side of a cube

Answer the questions on the basis of the information given below :In the adjoining figure, I and II are circles with centers P and Q respectively. The two circles touch each other and have a common tangent that touches them at points R and S respectively. This common tangent meets the line joining P and Q at O. The diameters of I and II are in the ratio 4 : 3. It is also known that the length of PO is 28 cm.

What is the ratio of the length of PQ to that of QO?              (2004)

A. 1:4
B. 1:3
C. 3:8
D. 3:4

#### SOLUTION

Solution : B

From the given diagram,

OQSQ=OPRP

OQOP=SQRP=34

Therefore, PQOQ=13

Answer the questions on the basis of the information given below:
In the adjoining figure, I and II are circles with centres P and Q respectively. The two circles touch each other and have a common tangent that touches them at points R and S respectively. This common tangent meets the line joining P and Q at O. The diameters of I and II are in the ratio 4 : 3. It is also known that the length of PO is 28 cm.

What is the radius of the circle II?

A. 2 cm
B. 3 cm
C. 4 cm
D. 5  cm

#### SOLUTION

Solution : B

PQOQ=13

PQ=14(28)=7

MPMQ=43(M is point of intersection of two circles)

MQ=3cm

Answer the questions on the basis of the information given below:
In the adjoining figure, I and II are circles with centres P and Q respectively. The two circles touch each other and have a common tangent that touches them at points R and S respectively. This common tangent meets the line joining P and Q at O. The diameters of I and II are in the ratio 4 : 3. It is also known that the length of PO is 28 cm.

The length of SO is:

A. 8√3 cm
B. 10√3 cm
C. 12√3cm
D. 14√3 cm

#### SOLUTION

Solution : C

OS=21232=123

In the figure given below (not drawn to scale), A, B and C are three points on a circle with centre O. The chord BA is extended to a point T such that CT becomes a tangent to the circle at point C. If ATC = 30 and ACT = 50, then the BOA is :
(CAT 2003)

A. 100
B. 150
C. 80
D. not possible to determine

#### SOLUTION

Solution : A

CAT= 100

Therefore, BAC= 80

OCT= 90

Therefore, BCT is > 90

Consider the five points comprising the vertices of a square and the intersection point of its diagonals. How many triangles can be formed using these points?                  (CAT 1993)

A. 4
B. 6
C. 8
D. 10

#### SOLUTION

Solution : C

The number of points will be $$^5C_3$$ - 2 = 8

(2 because of the 2 diagonals)

AB^ BC, BD^ AC and CE bisects C, A = 30. Then what is CED?
(CAT 1995)

A. 30
B. 60
C. 45
D. 65

#### SOLUTION

Solution : B

ACB= 60

DCE= ECB= 30

Therefore, CED= 60. Option (b)

In the given figure, EADF is a rectangle and ABC is a triangle whose vertices lie on the sides of EADF. AE = 22, BE =16, CF = 16, CF = 16 and BF = 2. Find the length of the line joining the
mid-points of the sides AB and BC.
(CAT 1997)

A. 44
B. 5
C. 3.5
D. none of these

#### SOLUTION

Solution : B

Option (b)

CD = (22 – 16) = 6

So in the right angled triangle ADC, AD = 8 and CD = 6.

Therefore AC = 10

Therefore length of the line joining the mid – points of

AB and BC = 12 (10) = 5

(length of the line joining the mid – point of two sides of a triangle is half the 3$$^{rd}$$

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 triangle ΔCBE to that of the triangle ΔADE ? (CAT 1997)

A. 1:4
B. 1:2
C. 1:3
D. data insufficient

#### SOLUTION

Solution : A

CBA = CDA

(a chord of a circle subtends equal angle an all its circumference)

Similarily BCD = BAD and BEC = AED

(opp. angles)

Therefore ΔCBE ΔADE (AAA similarity rule)

Now BCDA=1224=12 Thus, ratios of areas = 1:4,

since Ratio of Areas of 2 similar triangles = Ratio of squares of the corresponding sides

A. 120
B. 66
C. 93
D. 87

#### SOLUTION

Solution : C

Answer the questions based on the following information:
A rectangle PRSU is divided into two smaller rectangles PQTU, and QRST by the line TQ.PQ = 10cm. QR = 5 cm and Rs = 10cm. Points A, B, F are within rectangle PQTU, and points C, D, E are within the rectangle QRST. The closest pair of points among the pairs (A,C), (A,D), (A,E) (F,C), (F,E), (B,C), (B,D), (B,E) are 10√3 cm apart.
Which of the following statements is necessarily true?

A. The closest pair of points among the six given points cannot be (F, C).
B. Distance between A and B is greater than that between F and C.
C. The closest pair of points among the six given points is (C, D), (D, E) or (C, E).
D. None of the above

#### SOLUTION

Solution : A

The diagonal length of a rectancle PUSR = 513 = 18 (approx)

Among given eight pairs the shortest distance = 103

So, the six points A, B, F, C, D and E are near corner of rectangle PUSR.

So, (F,C) cannot be the shortest distance.

AB > AF > BF; CD > DE > CE; and BF = 65 cm. which is the closest pair of points among all the six given points?            (CAT 1999)

A. B ,F
B. C ,D
C. A ,B
D. None of these

#### SOLUTION

Solution : D

(d) Cannot be determined

ABCD is a rhombus with the diagonals AC and BD intersecting at the origin on the x y plane. The equation of the straight line AD is x + y =1. What is the equation of BC?  (CAT 2000)

A. x + y = -1
B. x – y = -1
C. x + y = 1
D. None of these

#### SOLUTION

Solution : A

Hence, eqn of BC which passes through (0,-1) and (-1,0) and is parallel to x+y=1 is

(y-0)=-1(x+1)

Y=-x-1  x+y= -1. Option (a)

In triangle DEF shown below, points A, B and C are taken on DE, DF and EF respectively such that EC = AC and CF = BC. D = 40 then what is ACB in degrees?(CAT 2001)

A. 140
B. 70
C. 100
D. None of these

#### SOLUTION

Solution : C

Option (c)

The angle in question is p

p+180-2x+180-2y=180

thus p-2(x+y)= 180……………..(1)

also, BDX = x+y (exterior angle= sum of 2 interior angles not adjacent to it)

x+y= 140………(2)

using the 2 statements, p= 100

A ladder leans against a vertical wall. The top of the ladder is 8m above the ground. When the bottom of the ladder is moved 2m farther away from the wall, the top of the ladder rests against the foot of the wall. What is the length of the ladder ?
(CAT 2001)

A. 10m
B. 15m
C. 20m
D. 17m

#### SOLUTION

Solution : D

Option (d)

Length of the ladder = x+2

Thus, 82+x2=(x+2)2

Solving for x, x=15