Cir3

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5/25/13 Exam Reports

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1.3 Marks Consider a circle with unit radius. There are 7 adjacent sectors, S1, S2, S3,...., S7in the circle such that their total area is (1/8)

thof the area

of the circle. Further, the area of the j thsector is twice that of the (j 1)thsector, for j = 2, ..., 7. What the angle, in radians, subtended bythe arc of S1at the centre of the circle?

[CAT 2000]

1) /508

2) /2040

3) /1016

4) /1524

Solution:

Let the area of sector S1bexunits.

The area of the sectors S2, S3, S4, S5, S6, S7will be 2x, 4x, 8x,16x, 32xand 64x

The total area of 7 sectors = 127xunits = (1/8) total area of circle = (1/8)

127x= /8units

A circ le subtends an angle of 2 at the centre.

Hence, (1/8)th of the circle will subtend an angle of /4 at the centre.

i.e. area of the seven sectors i.e. 127xwill subtend an angle of /4 at the centre.

Sector S1, whose area is x will subtend an angle of /(127 4) at the centre.

The required angle = /508 radians

Hence, option 1.

2.3 Marks

A certain c ity has a c ircular wall around it , and this wall has four gates pointing north, south, east and west. A house stands outs ide thecity, three km north of the north gate, and it can just be seen from a point nine km east of the south gate. What is the diameter of the wallthat surrounds the city?

[CAT 2001]

1) 6 km

2) 9 km

3) 12 km

4) None of these

Solution:

In the figure given below, E is the north gate, A is the south gate and C is the house which can be seen from the point B. AE is the diameterof the wall that surrounds the city.

Collapse All

Section I

Reports for

Circles - Past

CAT Questions

1

Overview Solution Key Unattempted Questions Report

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testfunda.com/LMS/Student/NewReports.aspx 2

DB and AB are the tangents to the circle from B.

DB = AB = 9 km

Also, CD2= CE CA

x2= 3 (3 + AE) = 9 + 3AE ...(i)

Now in ABC, (x+ 9)2= (3 + AE)2 + 92

x2+ 92+ 18x= (3 + AE)2 + 92

x2 + 18x= (3 + AE)2 ...(ii)

Now, start substituting AE from the options.

AE = 9, satis fies the equations

Hence, option 2.

Alternatively,

We have 9, AE + 3 and 9 +x, as the sides of a right-angled triangle. The most common Pythagorean triplet is 9, 12, 15. Using this, AE = 9andx= 6. Substit ute these values in the equations and verify.

Hence, option 2.

3.3 Marks

There is a common chord of 2 circles with radius 15 and 20. The distance between the two centres is 25. The length of the chord is

[CAT 2002]

1) 48

2) 24

3) 36

4) 28

Solution:

In ABC,

AB2+ AC2= BC2

BAC = 90Let BP =x

PC = 25 x

202=x2+ AP2 ...(i)

In APC,

225 = AP2+ (25 x)2

225 = AP2+ 625 +x2 50x ...(ii)

Equating (i) and (ii), we get,

225 = 625 + 400 50x

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3 Marks

1)

2)

3)

4) None of the above

Solution:

Consider the diagram below as per the conditions given in the question,

Speed on outer ring road = 30 km/hr

Speed on inner ring road = 20 km/hr

Let Rand rbe radii of the Outer Ring road and the Inner Ring road respectively.

Outer Ring road is twice as long as Inner Ring road, we have,

2R= 2 (2r)

R= 2r

Length of inner ring road = 2rkm and outer ring road = 4rkm

In the figure above, in OW2N

1,

Hence, option 3.

6.3 Marks

Amit wants to reach N2from S1. It would take him 90 minutes if he goes on minor arc S1- E1on OR, and then on the chord road E 1- N2. What is the radius of the outer ring road in km?

1) 60

2) 40

3) 30

4) 20

Solution:

Total distance to be travelled =

S1E1arc of the Outer Ring road + Chord E1N2

Given the speeds,

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r= 15 km

Radius of the Outer Ring road R= 2r= 30 km

Hence, option 3.

7.3 Marks

Amit wants to reach E2from N1using first the chord N1 W2and then the inner ring road. What will be his travel time in minutes on

the basis of information given in the above question?

1) 60

2) 45

3) 90

4) 105

Solution:

Total distance to be travelled, chord N1 W2+ W2 E2arc of the Inner Ring road

Given the speeds,

= 105 minutes

Hence, option 4.

8.3 Marks There are two concentric circles such that the area of the outer circle is four times the area of the inner circle. Let A, B and C be threedistinct points on the perimeter of the outer circle such that AB and AC are tangents to the inner circle. If the area of the outer circle is 12square centimetres then the area (in square centimetres) of the triangle ABC would be

[CAT 2003 Leaked Test]

1)

2)

3)

4)

Solution:

Consider the diagram below, as per the given conditions.

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Let rand Rbe the radii of the inner ci rcle and outer circle respect ively.

As area of the outer circle is 4 times the area of the inner circle, we have, R= 2r

OAM = = 30

Similarly,

OAM = OBM = OAC = OCA = 30

OBC = OCB = 30

BAC = ACB = CBA

ABC is an equilateral triangle.

But, area of the outer circle = (2r)2= 4r2= 12

Hence, option 3.

9.3 Marks

Three horses are grazing within a semi-circular field. In the diagram given below, AB is the diameter of the semi-circular field with centre atO. Horses are tied up at P, R and S such that PO and RO are the radii of semi-circles with centres at P and R respectively, and S is thecentre of the circle touching the two semi-circles with diameters AO and OB. The horses tied at P and R can graze within the respective

semi-circles and the horse tied at S can graze within the circle centred at S. The percentage of the area of the semi-circ le with diameter ABthat cannot be grazed by the horses is nearest to


1) 20

2) 28

3) 36

4) 40

Solution:

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Consider the diagram below.

Let rbe the radius of the smaller semi-circles and sbe the radius of the smaller circle.

OS = 2r s

PS = r+ s

PO = r

But, PSO is a right-angled triangle.

PS2= PO2+ SO2

(r+ s)2= r2+ (2r s)2

r2+ s2+ 2rs= r2+ 4r2+ s2 4rs

28%

Hence, option 2.

10.3 Marks

In the figure given below, AB is the chord of a circle with centre O. AB is extended to C such that BC = OB. The straight line CO isproduced to meet the circle at D. If ACD = y and AOD =x such thatx= ky, then the value of kis


1) 3

2) 2

3) 1


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Solution:

Consider the figure shown above.

In OBC, BC = OB

BOC = BCO = y (i)

Also, OB = OA = Radius of the circle

OBA is an exterior angle of OBC.

OBA = BOC + BCO = 2y

OAB = OBA = 2y (ii)

In AOB, AOB = 180 4y

Now, AOD + AOB + BOC = 180

x+ 180 4y+ y= 180

x= 3y

The value of k= 3

Hence, option 1.

Alternatively,

mACD = 1/2 [m(arc AD) m(arc BM)]

2y=xy

x= 3y

Hence, option 1.

11.3 Marks

In the figure below, the rectangle at the corner measures 10 cm 20 cm. The corner A of the rectangle is also a point on the circumferenceof the circle. What is the radius of the circle in cm?


1) 10 cm

2) 40 cm

3) 50 cm


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Solution:

Consider the figure shown above.

(r 20)2+ (r 10)2= r2

r2+ 400 40r+ r2+ 100 20r= r2

r2 60r+ 500 = 0

r2 50r 10r+ 500 = 0

r (r 50) 10 (r 50) = 0

r= 50 or r= 10

But, rcannot be 10.

r= 50

Hence, option 3.

Group Question

Answer the following questions based on the information given below.

Consider three circular parks of equal size with centres at A 1, A2and A3respectively. The parks touch each other at the edge as shown in the

figure (not drawn to scale). There are three paths formed by the triangles A1A2A3, B1B2B3and C1C2C3, as shown. Three sprinters A, B, and C

begin running from points A1, B1and C1respectively. Each sprinter traverses her respective triangular path clockwise and returns to her starting

point.

[CAT 2003 Re-Test]

12.3 Marks

Let the radius of each circular park be r, and the distances to be traversed by the sprinters A, B and C be a, band c, respectively.Which of the following is true?

1)

2)

3)

4)

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Solution:

The radius of each circular park = r

A1B1= A1E = r

Distance travelled by A = a= 3 2r= 6r

A1B1D is a right angle triangle with A1B1D = 30 and B1A1D = 60

Similarly, A1C1E is a right angle triangle with A1C1E = 30 and C1A1E = 60

Hence, option 1.

13.3 Marks

Sprinter A traverses distances A1A2, A2A3, and A3A1at average speeds of 20, 30 and 15 respectively. B traverses her entire path at

a uniform speed of ( ). C traverses distances C1C2, C2C3, and C3C1at average speeds of and

120 respectively. All speeds are in the same unit. Where would B and C be respectively when A finishes her sprint?

1) B1, C1

2) B3, C3

3) B1, C3

4) B1, Somewhere between C3and C1

Solution:

Time taken by A to travel through distance a(A1A2+ A2A3+ A3A1)

Distance travelled by B in the same time,

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B travels a full round in the same time as A. Thus, B will be at B 1.

For C, time taken to traverse through C1C2

C will be at C3.

Hence, option 3.

14.3 Marks

Sprinters A, B and C traverse their respective paths at uniform speeds u, vand wrespectively. It is known that u2 : v2: w2is equal toArea A : Area B : Area C, where Area A, Area B and Area C are the areas of triangles A1A2A3, B1B2B3, and C1C2C3respectively.

Where would A and C be when B reaches point B 3?

1) A2, C3

2) A3, C3

3) A3, C2

4) Somewhere between A2and A3, Somewhere between C3and C1

Solution:

Given that u2: v2: w2= Area A : Area B : Area C (i)

The area of an equilateral triangle is proportional to the square of its side,

Area A : Area B : Area C = (A1A2)

2

: (B1B2)

2

: (C1C2)

2

(ii)

From (i) and (ii),

u: v: w= A1A2: B1B2: C1C2

2u: 2v: 2w= 2(A1A2) : 2(B1B2) : 2(C1C2) = (A1to A3) : (B1to B3): (C1to C3)

Thus, C and A would be at C3and A3respectively when B reaches B3.

Hence, option 2.

15.3 Marks

Consider two different cloth-cutting processes. In the first one, ncircular cloth pieces are cut from a square cloth piece of side a in thefollowing steps: the original square of side a is divided into nsmaller squares, not necessarily of the same size; then a circle of maximumpossible area is cut from each of the smaller squares. In the second process, only one circle of maximum possible area is cut from thesquare of side aand the process ends there. The cloth pieces remaining after cutting the circles are scrapped in both the processes. Theratio of the total area of scrap cloth generated in the former to that in the latter is:

[CAT 2003 Re-Test]

1) 1 : 1

2)

3)

4)

Solution:

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A1,A2,A3,A4,A5, ... ,An

Also, letA1+A2+A3+A4+ A5+ ... +An=A ... (i)

Where,Awill be the Area of the original square.

Now, ifA= a2 (where ais a side of the square), then the area of the largest circle which can be drawn in it will have an area of (a/2)2= /4

a2= /4 A

Case 1: When the cloth is cut using the 2ndprocess

The area of the scrap material will be:

Case 2: When the cloth is cut using the 1st process

The sum of the areas of the maximum circles that can be cut out from the nsquares

Also, the sum of the areas of the nsquares = Area of the original square =A

Area of the sc rap material will be:

From Cases 1 and 2, it is clear that the ratio of scrap left in the 1 stprocess to the 2ndprocess is 1 : 1.

Hence, option 1.

16.3 Marks

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 Tsuch that CT becomes a tangent to the circle at point C. If ATC = 30 andACT = 50, then the angleBOA is:

[CAT 2003 Re-Test]

1) 100

2) 150

3) 80

4) Cannot be determined

Solution:

In ACT,

ACT = 50 andATC = 30,

CAT = 100

Applying the Alternate Segment theorem,

ABC = 50

Since CAT is the external angle of ABC, the sum of ABC and BCA is 100,

BCA = 50

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BOA = 100

Hence, option 1.

17.3 Marks

The length of the circumference of a circle equals the perimeter of a triangle of equal sides, and also the perimeter of a square. The areascovered by the circle, triangle, and square are c, t, and s, respectively. Then,

[CAT 2003 Re-Test]

1) s> t> c

2) c> t> s

3) c> s> t

4) s> c> t

Solution:

Let radius of the circle be r, a side of the equilateral triangle be a, and a side of the square be x.

The circumference/perimeter of the circle, triangle and square are equal. Hence,

2r= 3a= 4x= k

The areas of the circle, triangle and square are c, t, srespectively. Hence,

c> s> t

Hence, option 3.

Group Question


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 thattouches 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 ratio4 : 3. It is also known that the length of PO is 28 cm.

[CAT 2004]

18.3 Marks

What is the ratio of the length of PQ to that of QO?

1) 1 : 4

2) 1 : 3

3) 3 : 8

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4) 3 : 4

Solution:

SQO ~ RPO

42 = 56 7x

x= 2

PQ = 7 cm, OQ = 21 cm

PQ : OQ = 1 : 3

Hence, option 2.

19.3 Marks

What is the radius of the circle II?

1) 2 cm

2) 3 cm

3) 4 cm

4) 5 cm

Solution:

Radius of circle II = 1.5 2 = 3 cm.

Hence, option 2.

Alternatively,

We can solve this question and the previous question together using options as follows:

Consider each of the options in this question.

1. If the radius of circle II is 2, the radius of I is 8/3. PQ is then 14/3 and OQ is 70/3.

PQ : OQ = 1 : 5, which is not there in the options of the previous question.

2. If the radius of circle II is 3, the radius of I is 4. PQ is then 7 and OQ is 21.

PQ : OQ = 1 : 3, which is there in the options of the previous question. This is possible.





The radius of circle II is 3 and PQ : OQ = 1 : 3

Hence, option 2.

20.3 Marks

The length of SO is ________.

1)

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2)

3)

4)

Solution:

SO2= OQ2 SQ2

Hence, option 3.

21.3 Marks

In the adjoining figure, chord ED is parallel to the diameter AC of the c ircle. If CBE = 65, then what is the value of DEC?

[CAT 2004]

1) 35

2) 55

3) 45

4) 25

Solution:

mEBC = 65

mEOC = 130 (angle subtended by an arc at the centre is twice the angle subtended by the same arc on the circumference)

As AC || ED, mOCE = mDEC

OEC is an isosceles triangle.

mOCE = mOEC = (180 130)/2 = 25

mDEC = 25

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Hence, option 4.

22.3 Marks

On a semicircle with diameter AD, chord BC is parallel to the diameter. Further, each of the chords AB and CD has length 2, while AD has

length 8. What is the length of BC?

[CAT 2004]

1) 7.5

2) 7

3) 7.75


Solution:

Quadrilateral ABCD is an isosceles trapezium. Let AM = DN =x

AD = 8, AO = DO = BO = 4

OM = ON = 4 x

BMO and BMA are right triangles.

OB2= OM2+ MB2

16 = (4 x)2+ h2 (i)

Also,

AB2= MB2+ AM2

4 =x2+ h2 (ii)

From (i) and (ii),x= 0.5

BC = 2(4 x)

BC = 7

Hence, option 2.

23.3 Marks

Let C be a circle with centre P0and AB be a diameter of C. Suppose P1is the mid-point of the line segment P0B, P2is the mid-point of the

line segment P1B and so on. Let C1, C2, C3, ... be circles with diameters P0P1, P1P2, P2P3 ... respectively. Suppose the circles C1,C2,

C3, ... are all shaded. The ratio of the area of the unshaded portion of C to that of the original circle C is:

[CAT 2004]

1) 8 : 9

2) 9 : 10

3) 10 : 11

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:

Solution:

Let the radius of C be r.

and so on.

Thus the areas of circles with diameters P0P1, P1P2, P2P3, are

Required ratio = 11 : 12

Hence, option 4.

24.3 Marks

A circle with radius 2 is placed against a right angle. Another smaller circle is also placed as shown in the adjoining figure. What is the

radius of the smaller circle?

[CAT 2004]

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2)

3)

4)

Solution:

AB = AK = 2

Quadrilateral ABCD is a square.

Let the centre of the smaller circle be O and let its radius be x.

Hence, option 4.

Alternatively,

By visual inspection, we can see that the radius of the smaller circle is less than half the radius of the larger circle. After finding theapproximate values of the four options, we can eliminate options 2 and 3.

Now, the approximate values of options 1 and 4 are 0.17 and 0.343 respectively. Also AC = 2.82

The distance between the smaller circle and point C is less than the radius of the small circle. This distance is given by

AC AK 2(KO) = 2.82 2 2(KO)

Considering option 1, 2.82 2 2(0.17) = 0.48, which is more than 0.17.

Option 1 is eliminated.

Considering option 4.

2.82 2 2(0.343) = 0.134, which is less than 0.343.

Option 4 is the answer.

Hence, option 4.

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25.3 Marks

Two identical circles intersect so that their centres, and the points at which they intersect, form a square of side 1 cm. The area in sq. cmof the portion that is common to the two circles is

[CAT 2005]

1)

2)

3)

4)

Solution:

Let the two circles with centres P and Q intersect at M and N.

Quadrilateral PMQN is a square.

m MPN = m MQN = 90

The area common to both the circles = 2(Area of sector P-MN Area of PMN)

Hence, option 2.

26.3 Marks

A jogging park has two identical circular tracks touching each other, and a rectangular track enclosing the two circ les. The edges of therectangles are tangential to the circles. Two friends, A and B, start jogging simultaneously from the point where one of the circular trackstouches the smaller side of the rectangular track. A jogs along the rectangular track, while B jogs along the two circular tracks in a figure ofeight. Approximately, how much faster than A does B have to run, so that they take the same time to return to their starting point?

[CAT 2005]

1) 3.88%

2) 4.22%

3) 4.44%

4) 4.72%

Solution:

Let r be the radius of the circular tracks.

Length and breadth of the rectangular track are 4rand 2rrespectively.

Length (perimeter) of the rectangular track = 12r

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Length of the two circular tracks (figure of eight) = 4r

If A and B have to reach their starting points at the same time,

(where aand bare the speeds of A and B respectively)

= 4.7%

Hence, option 4.

27.3 Marks

What is the distance in cm between two parallel chords of lengths 32 cm and 24 cm in a circle of radius 20 cm?

[CAT 2005]

1) 1 or 7

2) 2 or 14

3) 3 or 21

4) 4 or 28

Solution:

The two chords AB and CD can be on the same side or the opposite sides of the centre O.

Let M and N be the midpoints of AB and CD.

MN is the distance between the two chords.

MB = 12 cm and ND = 16 cm

OM and ON are perpendicular to AB and CD respectively.

ON2= 202 162(By Pythagoras theorem)

ON = 12 cm

Similarly, OM = 16 cm

Case 1: AB and BC are on the same side of the centre.

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MN = OM ON = 4 cm

Case 2: MN = OM + ON = 28 cm

Hence, option 4.

28.3 Marks

Four points A, B, C and D lie on a straight line in the X-Y plane, such that AB = BC = CD, and the length of AB is 1 metre. An ant at Awants to reach a sugar particle at D. But there are insect repellents kept at points B and C. The ant would not go within one metre of anyinsect repellent. The minimum distance in metres the ant must traverse to reach the sugar particle is

[CAT 2005]

1)

2) 1 +

3)

4) 5

Solution:

The ant will not go into the circles with centers B and C and radius = 1 m

The minimum distance that the ant has to traverse = the distance of the path A-H-G-D

HG = 1 m

AH + GD = m

The ant must traverse (1 + ) m.

Hence, option 2.

29.3 Marks

In the following figure, the diameter of the circle is 3 cm. AB and MN are two diameters such that MN is perpendicular to AB. In addition,CG is perpendicular to AB such that AE:EB = 1:2, and DF is perpendicular to MN such that NL:LM = 1:2. The length of DH in cm is

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[CAT 2005]

1)

2)

3)

4)

Solution:

AO = OD = 1.5 cm

AE + EB = 3 cm and AE:EB = 1:2

AE = 1 cm and EB = 2 cm

OE = AO AE = 1.5 1 = 0.5 cm

Similarly, NL = 1 cm, LM = 2 cm, and OL = 0.5 cm

OEHL is a square as all its angles are right angles and OE = OL

EH = HL = 0.5 cm

1.5 = 0.5 + (0.5 + DH)

2.25 = 0.25 + 0.25 + DH + DH

DH + DH 1.75 = 0

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Hence, option 2.

30.3 Marks

P, Q, S, and R are points on the circumference of a circle of radius r, such that PQR is an equilateral triangle and PS is a diameter of thecircle. What is the perimeter of the quadrilateral PQSR?

[CAT 2005]

1)

2)

3)

4)

Solution:

PQR is an equilateral triangle and PS is the diameter.

m PQS = m PRS = 90 (angles subtended in a semi-circle)

and m PQM = m PRM = m QPR = 60 (each angle in an equilateral triangle = 60)

PS bisects QPS as it is the median of PQR.

m PMQ = m PMR = 90

m QPS = m RPS = 30

m PSQ = m PSR = 60

Radius = r

PS = 2r

As PQS, PQM, MQS are 30-60-90 triangles,

QS = r, PQ = 3r

Similarly, RS = r and PR = 3r

= =

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Hence, option 1.

Group Question


A punching machine is used to punch a circular hole of diameter two units from a square sheet of aluminium of width 2 units , as shown below.The hole is punched such that the circular hole touches one corner P of the square sheet and the diameter of the hole originating at P is in linewith a diagonal of the square.

[CAT 2006]

31.3 Marks

The proportion of the sheet area that remains after punching is:

1)

2)

3)

4)

5)

Solution:

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1) 32

2) 50

3) 40.5

4) 81

5) undeterminable

Solution:

Let CB =xcm

ACD and ADB are similar triangles.

AD/AB= AC/AD

AD2= AC AB

(AC2+ CD2) = 2 (2 +x)

40 = 2 (2 +x)

x= 18

Diameter AB = 20 cm

Radius = 10 cm

Area = 50 sq. cm.

Hence, option 2.

34.3 Marks

Two circles with centres P and Q cut each other at two distinct points A and B. The circles have the same radii and neither P nor Q fallswithin the intersection of the circles. What is the smallest range that includes all possible values of the angle AQP in degrees?

[CAT 2007]

1) Between 0 and 90

2) Between 0 and 30

3) Between 0 and 60

4) Between 0 and 75

5) Between 0 and 45

Solution:

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P and Q do not lie within the intersection of the two circles.

So they lie on the circumferences or outside the circumferences. If they lie on the circumferences, then APQ forms an equilateral triangle.So, m AQP = 60

From the diagram, if they lie outside the circumferences,

m AQ'P' < 60

Also, m AQP would be 0if A, Q and P were collinear.

But as P and Q cut each other in two distinct points, A, Q and P cannot be collinear.

m AQP > 0

The value, m

AQP lies between 0

and 60

Hence, option 3.

35.3 Marks

Two circles, both of radii 1 cm, intersect such that the circumference of each one passes through the centre of the circle of the other. Whatis the area (in sq cm) of the intersecting region?

[CAT 2008]

1)

2)

3)

4)

5)

Solution:

Let O and P be the centres of the circles.

OR = OP = PR = 1cm

PRO is an equilateral triangle.

m ROP = 60

m ROS = 120

Now, area of the intersecting region = 2(area of sector O-RPS) 2(area of PRO)

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Hence, option 5.

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