Geometry And Mensuration Exp

Answers and Explanations 1 11 21 31 41 51 61 71

1. c

c a c c a b b c

2 12 22 32 42 52 62 72

c d b d b c b b

3 13 23 33 43 53 63 73

a a a c d c b a

4 14 24 34 44 54 64 74

d b a c b c c a

5 15 25 35 45 55 65 75

d a d d a a d b

Let the radius of the outer circle be x = OQ

6 16 26 36 46 56 66 76

a d b a c c a d

5. d

Since CD > DE, option (b) cannot be the answer. Similarly, since AB > AF, Option (c) cannot be the answer. We are not sure about the positions of points B and F. Hence, (a) cannot be the answer.

6. a

The gradient of the line AD is –1. Coordinates of B are (– 1, 0).

Hence, perimeter of the circle = 2πx But OQ = BC = x (diagonals of the square BQCO) Perimeter of ABCD = 4x Hence, ratio = 2. c

2 πx π = . 4x 2

So {L1(13) + 2 π } > 4 and hence

L2 (17 )

c d d b a c b d

8 18 28 38 48 58 68

a a c c c a d

9 19 29 39 49 59 69

d c b c b b d

10 20 30 40 50 60 70

b b d b a d b

A

Following rule should be used in this case: The perimeter of any polygon circumscribed about a circle is always greater than the circumference of the circle and the perimeter of any polygon inscribed in a circle is always less than the circumference of the circle. Since, the circles is of radius 1, its circumference will be 2 π . Hence, L1(13) > 2 π and L2(17) < π .

{L1(13 ) + 2π}

7 17 27 37 47 57 67 77

x+y=1

B

(0 , 0 )

D

(– 1, 0)

will (0 , – 1) C

be greater than 2.

Equation of line BC is x + y = –1. 3. a 5 00

10

7. c

0

Let the area of sector S1 be x units. Then the area of the corresponding sectors shall be 2x, 4x, 8x,16x, 32x and 64x. Since every successive sector has an angle that is twice the previous one, the total area

5 00

then shall be 127x units. This is

the circle. Hence, the total area of the circle will be 127x × 8

Area of shaded region =

1 100 100 × × = 2,500 sq m 2 2 2

Area of a ∆ is maximum when it is an isosceles ∆. 100 So perpendicular sides should be of length . 2

4. d

We have not been given the distances between any two points.

Geometry and Mensuration - Actual CAT Problems ‘99-’05

1 of the total area of 8

= 1016x units. Hence, angle of sector S1 is 8. a

π . 1016

We know that (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac = 3ab + 3bc + 3ac Now assume values of a, b, c and substitute in this equation to check the options. Short cut: (a – b)2 + (b – c)2 + (c – a)2 = 0. Hence, a = b = c.

Page 1

12. a

9. d E

(2 – 2x)

x

x

x

C

x

β G γ

α

A

D

F

B

2m E C G

A

α 2α

2α

3 α 2α F

B

α

3α

2α α

α

Let the length of the edge cut at each corner be x m. Since the resulting figure is a regular octagon,

α

∴

x 2 + x 2 = 2 – 2x ⇒ x 2 = 2 – 2x

⇒

2 x (1 +

13. b

A 3 km N C W

E

9 km

9 km

S

P

∆APS and ∆AOC are similar triangles. Where OC = r

∴

r = r+3

9 81 + (2r + 3)2

Now use the options. Hence, the diameter is 9 km.

B

14. a H

C

G

D

F E In order to reach E from A, it can walk clockwise as well as anticlockwise. In all cases, it will have to take odd number of jumps from one vertex to another. But the sum will be even. In simple case, if n = 4, then an = 2. For a2n–1 = 7 (odd), we cannot reach the point E.

Page 2

r

O

10. a

11. d

2 2 +1

D

Let ∠EAD = α. Then ∠AFG = α and also ∠ACB = α. Therefore, ∠CBD = 2α (exterior angle to ∆ABC). Also ∠CDB = 2α (since CB = CD). Further, ∠FGC = 2α (exterior angle to ∆AFG). Since GF = EF, ∠FEG = 2α. Now ∠DCE = ∠DEC = β (say). Then ∠DEF = β – 2α. Note that ∠DCB = 180 – (α + β). Therefore, in ∆DCB, 180 – (α + β) + 2α + 2α = 180 or β = 3α. Further ∠EFD = ∠EDF = γ (say). Then ∠EDC = γ – 2α. If CD and EF meet at P, then ∠FPD = 180 – 5α (because β = 3α). Now in ∆PFD, 180 – 5α + γ + 2α = 180 or γ = 3α. Therefore, in ∆EFD, α + 2γ = 180 or α + 6α = 180 or α = 26 or approximately 25.

A

2) = 2 ⇒ x =

Let BC = y and AB = x. Then area of ∆CEF = Area(∆CEB) – Area(∆CFB) =

xy 1 2x 1 x . .y– . .y = 6 2 3 2 3

Area of ABCD = xy ∴ Ratio of area of ∆CEF and area of ABCD is

xy 1 : xy = 6 6

Work with options. Length of wire must be a multiple of 6 and 8. Number of poles should be one more than the multiple.

Geometry and Mensuration - Actual CAT Problems ‘99-’05

15. d Since area of ∆ABC = 80 =

1 AB × CD 2

80 × 2 = 8 . In ∆ACD; AD = 20 Hence DB = 20 – 6 = 14. So CD =

W all La dde r x

8m

142 + 82 = 196 + 64 = 260 unit

So CB = 18. c

y

D

2m

G roun d Let the length of the ladder be x feet. We have 82 + y2 = x2 and (y + 2) = x Hence, 64 + (x – 2)2 = x2 ⇒ 64 + x2 – 4x + 4 = x2 ⇒ 68 = 4x ⇒ x = 17

40 ° A x

B

E

16. d

25

D

40

y y

x

F

C

Here ∠ACE=180 – 2x , ∠BCF = 180 – 2y and x + y + 40° = 180° (In ∆DEF) So x + y = 140° So ∠ACB= 180° – ∠ACE – ∠BCF = 180° – (180° – 2x) – (180° – 2y) = 2(x + y) – 180° = 2 × 140 – 180 = 100°

C

20 24

102 – 82 = 6

25 20

19. b A

B

32

z

CE =

252 – 202 = 15

(Since DBC is isosceles triangle.) Assume ABCD is a quadrilateral where AB = 32 m, AD = 24 m, DC = 25 m, CB = 25 m and ∠DAB is right angle. Then DB = 40 m because ∆ADB is a right-angled triangle and DBC is an isosceles triangle.

x–3

x–3

We can find the value of x, using the answer choices given in the question. We put (a), (b), (c) and (d) individually in the figure and find out the consistency of the figure. Only (b), i.e. 11 is consistent with the figure.

1 × 15 × 20 = 300 sq. m 2 Hence area of ABCD = 384 + 300 = 684 sq. m Area of ∆ BCD = 2 ×

xm

20. c

C

x+4

y

1 So area of ∆ ADB = × 32 × 24 = 384 sq. m 2

17. a

x

xm

10

20

P ath 60

A

D 20

B

Let’s assume AB be the longest side of 20 unit and another side AC is 10 unit. Here CD ⊥ AB.

Geometry and Mensuration - Actual CAT Problems ‘99-’05

Let width of the path be x metres. Then area of the path = 516 sq. m ⇒ (60 + 2x)(20 + 2x) – 60 × 20 = 516 ⇒ 1200 + 120x + 40x + 4x2 – 1200 = 516 ⇒ 4x2 + 160x – 516 = 0 ⇒ x2 + 40x – 129 = 0 Using the answer choices, we get x = 3.

Page 3

21. b 24. d

A 3 0°

A

F

D

B

E

C

3 0° y

4 B

3

D

C

Area of ∆ABE = 7 cm2 Area of ∆ABEF = 14 cm2 Area of ∆ABCD = 14 × 4 = 56 cm2

Let BC = x and AD = y. As per bisector theorem, Hence, BD =

25. b

4x 3x ; DC = 7 7

In ∆ABD, cos30° = ⇒ 2× 4× y×

4 BD AB = = 3 DC AC

(1 , 1 ) 2

16x 49 2× 4× y

(4)2 + y 2 –

(0 , 0 )

Let a = 0

3 16x 2 = 16 + y 2 – 2 49

⇒ 4 3y = 16 + y 2 –

16x 2 49

Similarly, from ∆ADC, cos30° = 9x 2 49 Now (i) × 9 – 16 × (ii), we get ⇒ 3 3y = 9 + y 2 –

1 (2) (1) = 1 2 Note: Answer should be independent of a and area of the triangle does not have square root. Hence, area =

... (i)

9x 2 49 2×3× y

9 + y2 –

26. d

... (ii)

36 3y – 48 3y = 9y 2 – 16y 2 ⇒ y =

12 3 7

27. c

22. a C 15

(2 , 0 )

1 ⇒ Diagonal = 5 2 Distance saved = 3 – 5 ≈ 0.75 ≠ Half the larger side. Hence, incorrect. 3 ⇒ Diagonal = 5 4 Distance saved = (4 + 3) – 5 = 2 = Half the larger side. Check choices, e.g.

Area = 40 × 20 = 800 If 3 rounds are done, area = 34 × 14 = 476 ⇒ Area > 3 rounds If 4 rounds ⇒ Area left = 32 × 12 = 347 Hence, area should be slightly less than 4 rounds.

20

A

B

28. b

B

25

20

15

Let the chord = x cm

1 1 x ⇒ x = 24 cm ∴ (15 × 20) = × 25 × 2 2 2

P A x

23. a

C

D

25 – x

Total area = 14 × 14 = 196 m2  π × r2  2 Grazed area =  4  × 4 = πr = 22 × 7(r = 7)  

= 154 m2 Ungrazed area is less than (196 – 154) = 42 m2, for which there is only one option.

Page 4

Q M

(15)2 – x 2 = (20)2 – (25 – x)2 ⇒x=9 ⇒ BD = 12

Area of ∆ABD =

1 × 12 × 9 = 54 2

s=

1 (15 + 12 + 9) = 18 2

r1 =

Area ⇒ r1 = 3 s

Geometry and Mensuration - Actual CAT Problems ‘99-’05

Area of ∆BCD = s =

The corresponding speeds are

1 × 16 ×12 = 96 2

20π, 30π and 15 5 kmph. Thus time taken to travel one circumference of

1 (16 + 20 + 12) = 24 2

IR =

Area ⇒ r2 = 4 s In ∆PQM, PM = r1 + r2 = 7 cm QM = r2 – r1 = 1 cm r2 =

Hence, PQ = 29. d

and one length of the chord road =

32. c

50 cm

r r hr , one circumference of OR = hr hr. 10 7.5

If KL = 1, then IG = 1 and FI = 2

r hr 15

Sum of the length of the chord roads = 4r 5 and the length of OR = 4π r. Thus the required ratio =

5:π

2 Hence, tan θ = = 2 1 Thus, θ none of 30, 45 and 60°. 30. c

Area of quadrilateral ABCD =

1 (2x + 4 x ) × 4 x = 12x 2

Area of quadrilateral DEFG =

1 (5 x + 2 x ) × 2 x = 7 x 2

33. c

The total time taken by the route given =

(i.e. 90 min.) Thus, r = 15 km. The radius of OR = 2r = 30 kms

34. d

The total time taken =

r r 7r + = 20 15 60

Hence, ratio = 12 : 7 31. d

Since r = 15, total time taken =

The surface area of a sphere is proportional to the square of the radius. Thus,

SB 4 = SA 1

(S. A. of B is 300% higher than A)

A

VB 8 = VA 1

N2

36. b

W2

B

r

NOTE: You will realize that such a circle is not possible (if r = 3.125 how can CE be 5). However we need to check data sufficiency and not data consistency. Since we are able to find the value of r uniquely using second statement the answer is (a).

N1

W1

2 .5

We can get the answer using the second statement only. Let the radius be r. AC = CB = 2.5 and using statement B, CE = 5, thus OC = (r – 5). Using Pythagoras theorem, (r – 5)2 + (2.5)2 = r2 We get r = 3.125

7 7  th less than B i.e.  × 100  87.5% 8 8 

For questions 32 to 34:

2 .5 C O

The volume of a sphere is proportional to the cube of the radius.

Or, VA is

7 hr. = 105 min. 4

E

35. a

r 2 ∴ B = rA 1

Thus,

r r 3 + = 30 15 2

E2

E1

1 th the area of triangle ABC. Thus by knowing 4 either of the statements, we get the area of the triangle DEF. be

S2 S1

If the radius of the inner ring road is r, then the radius of the outer ring road will be 2r (since the circumference is double). The length of IR = 2π r, that of OR = 4π r and that of the chord roads are r 5 (Pythagoras theorem)

Geometry and Mensuration - Actual CAT Problems ‘99-’05

The question tells us that the area of triangle DEF will

37. c

In this kind of polygon, the number of convex angles will always be exactly 4 more than the number of concave angles. NOTE : The number of vertices have to be even. Hence the number of concave and convex corners should add up to an even number. This is true only for the answer choice (c).

Page 5

A

38. c

=

10 36 = 28% (approx.)

r B

 Area that can be grazed    Area of the field  

26 36

The fraction that cannot be grazed = C

2r

40. a

Since the area of the outer circle is 4 times the area of the inner circle, the radius of the outer circle should be 2 times that of the inner circle. Since AB and AC are the tangents to the inner circle, they should be equal. Also, BC should be a tangent to inner circle. In other words, triangle ABC should be equilateral. The area of the outer circle is 12. Hence the area of inner circle is 3 or the radius is

It is very clear, that a regular hexagon can be divided into six equilateral triangles. And triangle AOF is half of an equilateral triangle. Hence the required ratio = 1 : 12

P

41. b

6 0°

3 . The area of π

equilateral triangle = 3 3 r2, where r is the inradius. Hence the answer is 39. b

A

9 3 π

Given ∠APB = 60° and AB = b.

If the radius of the field is r, then the total area of the

∴ PQ =

2 πr . field = 2 The radius of the semi-circles with centre's P and

R=

∴

πr 2 4 Let the radius if the circle with centre S be x.

b2 3b2 + h2 = 4 4

∴ 2h2 = b2

Hence, their total area =

r r  and RS =  + x  . 2 2 

42. d

C Q

Applying Pythagoras theorem, we get

r (r – x)2 +   2

2

r  =  + x 2 

Solving this, we get x =

P A

B 6

r . 3 πr 2 . 9

2 1 1 The total area that can be grazed = πr  +  4 9

13πr 2 36 Thus the fraction of the field that can be grazed

Page 6

10 D

8

2

Thus the area of the circle with centre S =

=

b × 3 2

b , h and PQ form a right angle triangle. 2

Next,

r . 2

Thus, OS = (r – x), OR =

B

Q

Triangle ABC is a right angled triangle.

1 1 × BC × AB = × BD × AC 2 2 Or, 6 × 8 = BD × 10. Thus BD = 4.8. Therefore, BP = BQ = 4.8. So, AP = AB – BP = 6 – 4.8 = 1.2 and CQ = BC – BQ = 8 – 4.8 = 3.2. Thus, AP : CQ = 1.2 : 3.2 = 3 : 8 Thus

Geometry and Mensuration - Actual CAT Problems ‘99-’05

43. b

Using the Basic Proportionality Theorem,

(

AB BD = PQ QD

(

AB BQ = = 3 : 1. CD QD Thus CD : PQ = BD : BQ = 4 : 3 = 1 : 0.75

3 +1

C2C3 C C , 3 1 40 3 + 1 120 3

(

)

) )

(

) ( )

)

(

(

)

1+ 3 i.e. 3 × 2r, 3 × 2r, r 40 40 60

20

r – 10

)

,

(

∠AOD = 30° (external angle of ∆AOC ) Thus k = 3

A

3 r 10

2+2 3 r 3 2+2 3 r 2+2 3 r 3 × × , , 40 40 120 3 +1 3 +1

∠OAB = 20 (opposite equal sides)

45. c

(

(

∠BOC = 10° (opposite equal sides) ∠OBA = 20° (external angle of ∆BOC )

r – 20

)

3 + 2 × 10 ×

)

C1C2 40 3

If y = 10°,

10

(

∴ B will be at B1. Now time taken by for each distance are

Multiplying the two we get,

44. a

3 r = 10

= 2r + 3r × 3 = B1B2 + B2B3 + B3B1

PQ BQ = . CD BD

and

)

= 10 3 + 20 ×

i.e.

C

r

(

)

1+ 3 3 3 r, r, r 20 20 60

We can observe that time taken for C1C2 and C2C3

B

3 3 3 r+ r= r , which is same as time 20 20 10 taken by A. Therefore, C will be at C3. combined is

Let the radius be r. Thus by Pythagoras’ theorem for ∆ABC we have (r – 10)2 + (r – 20)2 = r2 i.e. r2 – 60r + 500 = 0. Thus r = 10 or 50. It would be 10, if the corner of the rectangle had been lying on the inner circumference. But as per the given diagram, the radius of the circle should be 50 cm.

48. b

In similar triangles, ratio of Area = Ratio of squares of corresponding sides. Hence, A and C reach A3 and C3 respectively.

49. a

The whole height h will be divided into n equal parts. Therefore, spacing between two consecutive turns

For questions 46 to 48: A1A2 = 2r, B1B2 = 2r + r 3 , C1C2

=

h . n

= 2r + 2r 3 50. b

Hence, a = 3 × 2r

The four faces through which string is passing can be shown as

b = 3 × (2r + r 3 )

(

c = 3 × 2r + 2r 3

46. a

)

Difference between (1) and (2) is 3 3r and that between (2) and (3) is 3 3r . Hence, (1) is the correct choice.

47. c

Time taken by A =

2r 2r 2r  2r × 9  3 + + = r  = 20 30 15  60  10

3 r. Therefore, B and C will also travel for time 10

(

Now speed of B = 10 3 + 20

)

Therefore, the distance covered

Geometry and Mensuration - Actual CAT Problems ‘99-’05

n 4

n

n

n 4

n

n 4 n 4

n

Therefore, length of string in each face

=

n n2 +   4

= n2 +

2

17n n2 = 4 16

Therefore, length of string through four faces =

17n × 4 = 17n 4

Page 7

51. c

As h/n = number of turns = 1 (as given). Hence h = n.

Area of circle

=

Area of rec tan gle

52. c

PQ || AC

r2

CQ AP 4 ∴ = = QB PB 3

lb

QD || PC

d2 4 = 1 lb 3

∴

PD CQ 4 = = DB QB 3

As

PD 4 = DB 3 4 PB 7

∴ PD = ∴

AP AP = PD 4 PB 7

7 AP × 4 PB

=

7 4 × 4 3 =7:3

C

96°

d2 1 = 4lb 3

∴

l2 + b2 1 = 4lb 3

∴

l2 + b2 4 = lb 3

∴

l b 4 + = b l 3

180 – 2y

∴

AE BC = AD DC

∴

AE b = AD l

Let B

Using exterior angle theorem

∠A + ∠B = 96 i.e. x + y = 96 … (i) Also x + (180 – 2y) + 96 = 180°

∴ x – 2y + 96 = 0 ∴ x – 2y = –96 Solving (i) and (ii), y = 64° and x = 32°

… (ii)

I

1 + x2 4 = x 3

∴ 3x 2 − 4x + 3 = 0

B b

D

1 4 +x= x 3

3 + 3x 2 = 4x

∴ ∠DBC = y = 64o

A

∆ CBD

b =x l Therefore, from (i), we get

y

y D

E

:

… (i)

∴ We have to find AE , i.e. b . AD l

x

54. a

3

3

∴

53. c

x

π

=

AE AD = ∴ CB DC E

A

lb

1

=

Now ∆ AEB

=

πr 2

∴x =

=

C

=

−( −4) ± 16 − 4

( 3)

3

2 3

4 ± 16 − 12 2 3 4±2 2 3

BD = 2r

Page 8

Geometry and Mensuration - Actual CAT Problems ‘99-’05

=

6

58. b

B

a

3 0°

3 a

C

2 3

a

a 2

OR =

3 1 OR 1 3

E 1 3

, i.e. 1: 3 .

Q ∆ ACE is equilateral triangle with side

P

56. c

1 + L∞ 1 − 2 2 = = A 2A A + + L∞ 2

=

=

P

(

P 2

2P ( 2 + 1)

=

2A

2 ×2

(

)

2 +1

a

)

2 −1

=

(

2 2+ 2

×

3 a.

3 2 a ×6 4

Area of hexagon =

It’s standard property among circle, square and triangle, for a given parameter, area of circle is the highest and area of the triangle is least whereas area of the square is in-between, i.e. c > s > t.

P+

a

a

From options, the answer is

55. c

D

A

2 3

Area as ∆ACE =

3 ( 3a)2 4

Therefore, ratio =

1 2

1 59. d

2A

2 × 4a

(

2×a

)

2 +1 2

The required answer is 34 × 0.65 × 0.65 = 14.365 Because we get two similar triangles and area is proportional to square of its side.

60. b

x

)

a 2

57. a

o

∠BAC = ∠ACT + ∠ATC = 50 + 30 = 80

And ∠ACT = ∠ABC (Angle in alternate segment) o So ∠ABC = 50

x/2

∠BCA = 180 − (∠ABC + ∠BAC )

= 180 − (50 + 80 ) = 50o

In original rectangle ratio =

o Since ∠BOA = 2∠BCA = 2 × 50 = 100

In Smaller rectangle ratio = Alternative Method: Given

Join OC

∠OCT = 90° (TC is tangent to OC)

x 2 = ⇒x=2 2 2 x 2

∠OCA = 90° – 50° = 40° ∠OAC = 40° (OA = OC being the radius)

Area of smaller rectangle =

∠BAC = 50° + 30° = 80° ∠OAB = 80° – 40° = 40° = ∠OBA (OA = OB being the radius) ∠BOA = 180° – (∠OBA + ∠OAB) = 100°

61. b

x 2 2 x 2  

x × 2 = x = 2 2 sq. units 2

OP PR 4 = = OQ QS 3 OP = 28 OQ = 21 PQ = OP – OQ = 7 PQ 7 1 = = OQ 21 3

Geometry and Mensuration - Actual CAT Problems ‘99-’05

Page 9

62. b

PR + QS = PQ = 7

=

PR 4 = QS 3

a3

a3

⇒ QS = 3 63. c

SO = OQ2 – QS2

a3

= 212 – 32

Circum radius for equilateral triangle

= 24 × 18 = 12 3

=

side 3

64. d Therefore

66. b

Circle C

r 4

C2

r 8

C3

r 16

M

M

C

B

Radius r

C1

a 3 =a 3

2

A

2

E

D

F

8

1 1 × AB × BD = × AD × BE 2 2 2 82 − 22 = 8 × BE

BE =

Area of unshaded portion of C ⇒ either Area of C

60 15 = 4 2 2

= 1−

= 1−

Area of shaded portion Area of C

 15  15 1 AE = 22 −  =  = 4 − 2 4 2  

  r 2  r 2  π    +   + …  4   8    

 1 1 BC = EF = 8 −  +  = 7 2 2

πr 2

1  1  1 16 = 1−  + + … = 1 − 1  42 82  1− 4

=

11 12

67. d

O 2 A

65. a

C D

O B

DF, AG and CE are body diagonals of cube. Let the side of cube = a Therefore body diagonal is a 3

Page 10

Let the radius of smaller circle = r

Geometry and Mensuration - Actual CAT Problems ‘99-’05

∴ O′B = r 2

M

71. b

∴ OB = O′B + O′D + OD

C

=r 2 +r +2

Also OB = 2 2 ⇒r 2 +r +2= 2 2

A

⇒r =6–4 2

D

68. d

B

E

O

H

L

B F

G 6 5°

N

A

C

O

1 2 DL = DH + HL HL = OE =

D

E

In ∆ABC,

∠B = 90° (Angles in semicircle)

1 2 OB = AO = radius = 1.5 DL = DH +

Therefore ∠ABE = 90 – 65 = 25° Also ∠ABE = ∠ACE ( angle subtended by same arc AE)

DO2 = OL2 + DL2

Also ∠ACE = ∠CED [AC ED]

2

2

1 3  1    =   +  DH +  2 2 2 

Therefore ∠CED = 25°

2

69. b 2

B

A

9 0°

1 cm C 6 0° 6 0° 9 0° 6 0°

P

Hence option (b)

Q

Drawn figure since it have not to be within distance of 1 cm so it will go along APQD. AP =

1 1  ⇒  DH +  = 2 ⇒ DH = 2 – 2 2 

D

72. a

A

90 π × 2π × 1 = 360 2

D

π 2 So the minimum distance = AP + PQ + QD =

9

Also AP = QD =

70

30

C

So here we can say that triangle BCD and triangle ABC will be similar. Hence from the property of similarity

30

70

10

AB 12 = Hence AB = 16 12 9 AC 12 = Hence AC = 8 6 9 Hence AD = 7 AC = 8

30

70 10 30 70

Geometry and Mensuration - Actual CAT Problems ‘99-’05

12

Here ∠ACB = θ + 180 – (2θ + α ) = 180 – (θ + α )

30

30

θ

B

π π + 1+ = 1+ π 2 2

70. c

6 α

SADC = 8 + 7 + 6 = 21 SBDC = 27 Hence r =

21 7 = 27 9

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73. a

75. d

P

A

B

r

r

R 3 0° Q

r

r

r

r

R

a

x

A covers 2r + 2r + 4r + 4r = 12 r

1 20 °

B covers 2πr + 2πr = 4πr dis tance

x S

4πr 12r π = ⇒ SB = S A SB S A 3

a Here cos30° = 2r

SB – S A π–3 × 100 = × 100 = 4.72% SA 3

a=r 3

Here the side of equilateral triangle is r 3 From the diagram cos120° =

2

2

x +x –a

Hence Option (d)

2

2x 2

76. d

a2 = 3x2 x=r

(

Hence the circumference will be 2r 1 + 3

)

D 12 C 4 B 16 20 A 12 20 O

Hence answer is (a). 74. b

Let the rectangle has m and n tiles along its length and breadth respectively. The number of white tiles W = 2m + 2(n – 2) = 2 (m + n – 2) And the number of Red tiles = R = mn – 2 (m + n – 2) Given W = R ⇒ 4 (m + n – 2) = mn ⇒ mn – 4m – 4n = – 8 ⇒ (m – 4) (n – 4) = 8 ⇒ m – 4 = 8 or 4 ⇒ m = 12 or 8 ∴ 12 suits the options.

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OB2 = OA2 – AB2 = 202 – 162 = 144 OB = 12 OD2 = 202 – 122 = 400 – 144 = 256 OD = 16 BD = 4 Only one option contains 4 hence other will be 28. Hence option (d)

Geometry and Mensuration - Actual CAT Problems ‘99-’05

Geometry and Mensuration - Actual CAT Problems ‘99-’05

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