MT EDUCARE LTD.icse.maheshtutorials.com/images/ICSE_Step_Up_Papers/01Mathema… · 6 Angle at centre is twice the angle at circumference. AOC = 2 × ADC AOC = 2x But AOC = 160º [Given]

30

2003

1. In PQL and RPM,LPQ = MRP [Given]LQP = RPM [Given]

PQL ~ RPM [By AA axiom of similarity](i) Since, PQL ~ RPM

QLPM

=PLRM

QL × RM = PL × PM

(ii) In PQL and RQP, we haveQ = Q [Common angle]

QPL = PRM [Given] PQL ~ RQP [By AA axiom of similarity]

Then,PQQR =

QLPQ

PQ2 = QRQLHence proved

22. From the given figure,

ABD = ACD = 30º [Angle in same segment]Also, ADB = 90º [Angle in a semi-circle is 90º]In ABD, we have

DAB + ABD + ADB = 180º [Sum of all angles in a triangle is 180º]

x + 90º + 30º = 180º x + 120º = 180º x = 180º – 120º = 60º

Hence, the value of x is 60º

MT EDUCARE LTD.ICSE X

GeometrySTEP UP ANSWERSHEET

SUBJECT : MATHEMATICS

A

D

B

C

30º

xE

RQ L

P

M

MT EDUCARE LTD. X - ICSE (Maths)

31

2004

1. Given : In a figure, BAC = 30ºTo prove : BC = BDProof :

ACB = 90º [Angle in a semicircle]In ACB,ABC + ACB + BAC = 180º

ABC + 90º + 30º = 180º [ AB is a diameter of a circle and ACB is in semi-circle, so ACB = 90º]

ABC = 180º – 120º = 60ºBCD = BAC [Angles in alternate segment]

BCD = 30ºIn BCD,ABC is an exterior angle

ABC = BCD +BDC 60º = 30º + BDC BDC = 60 – 30

= 30º ... (ii)In BDC,

BDC = BCD = 30º [From equation (i) and (ii)] BC = BD [ sides opposite to equal angles

are equal]Hence proved

2. Given : DE BC

(i) To prove : ADE~ABCProof : In ADE and ABC, we have

DAE = BAC [Common angles]ADE = ABC [Corresponding angles as DE BC]

ADE ~ ABC [By AA axiom of similarity]Hence proved.

(ii) Since, ADE ~ ABC

ADAB

=DEBC

[If two triangles are similar, then

their corresponding sides areproportional]

ABAD

=BCDE

AD + BD

AD=

BCDE

[ AB = AD + BD]

30ºA B D

C

O

A

D E

B C

32

MT EDUCARE LTD.X - ICSE (Maths)

1 +BDAD

=BCDE

1 +BD

1 BD2

=BCDE

1AD = BD, given2

1 +112

=BCDE

1 + 2 =4.5DE

DE =4.53

= 1.5 cmADE ~ ABC

Ar(ΔADE)Ar(ΔABC) =

2

2

ADAB

=2AD

AB

=21

3

= 1 : 9Let Ar(ADE) = kand Ar(ABC) = 9k

Ar(ΔADE)Ar(trp. BCED) =

Ar(ΔADE)Ar(ΔABC) – Ar(ADE)

=k

9k – k =

k8 k

=18

= 1 : 8

2005

B = E = 90º [ AB BC and DE BC, given]and ACB = DCE [Common angle] ABC ~ DEC [By AA axiom of similarity]

ABDE

=ACDC

[ If two triangles are similar, then their corresponding sides are proportional]

B EC

D

A

1. Given, AB = 9 cm DE = 3 cmand AC = 24 cmIn ABC and DEC,


33

93

=24DC

DC =24 3

9

= 8 cm

Now, AD = AC – DCAD = 24 – 8AD = 16 cm

2. Let 0 be the centre and r be the radius of acircle. Join OM, ON and OS.We know that the tangent at any point of acircle and the radius through the point ofcontact are perpendicular to each other. OMQ = ONQ = 90ºAlso, NQ = QM [Tangents drawn from external

point to a circle are equal in length] NQ = QM= OM = ON = r [Say]Thus, OMQN is a square.

We know that, tangents drawn from an external point are equal in lenngths. PN = PS PQ – QN = PS 3 – r = PS ... (i)and MR = SR QR – QM = SR 4 – r = SR

On adding equation (i) and (ii), we get3 – r + 4 – r = PS + SR

7 – 2r = PR = 5 [In right PQR, PR = 2 2PQ + QR(Pythagoras theorem)

= 2 23 4 = 9 16

= 25 = 5 cm] 7 – 5 = 2r 2r = 2 r = 1 cm [Radius of incircle]

3. Given : In the given figure, PA = AMTo prove :(i) PMB is isosceles.(ii) PA . PB = MB2

Proof :(i) Since PA = AM [Given] AMP = APM ... (i)

[Angles opposite to equal sides are equal]

Since, angle between the tangent and chord is equal to the angle subtended

BAP

M

P

Q

N3 c

m

M R

S

Or

4 cm

34


by same chord in alternate segment. AMP = ABM = PBM APM = PBM

or BPM = PBM PM = MB

Hence, PMB is an isosceles triangle.(ii) Since, PM is tangent to the circle and PAB is a secant.

PAPB = PM2

PAPB = MB2 [PM = MB]Hence proved

4. To prove : BD = 2OCConstruction : Join CO and DB.

Proof : In OCA and BDA,OCA = BDA = 90º [ Angle in semi-circle is right

angle]and OAC = BAD [Common angle]

OAC ~ BAD [By AA axiom of similarity]

Then,OABA

=OCBD

[In similar triangles sides are

proportional]

OA2OA

=OCBD

[ BA = 2AO]

BD = 2OCHence proved

5.

On measuring, we get BCP = 30º

PA

C

3.5 cm

M

B

X

O120º

6 cm

N

A

C

B

D

O


35

6 Angle at centre is twice the angle at circumference.AOC = 2 × ADC

AOC = 2xBut AOC = 160º [Given]

160º = 2x

x =160º

2= 80º

Also, xº + yº = 180º 80º + y = 180º y = 180º – 80º yº = 100º

Now, 3y –2x = 3 × 100º – 2 × 80º= 300º – 160º= 140º

Hence proved

2006

1. Given, BAD = 65º, ABD = 70º and BDC = 45º(i) Since, ABCD is a cyclic quadrilateral.

So, the sum of opposite angles of a cyclicquadrilateral is 180º.

BCD + BAD = 180º 65º + BCD = 180º BCD = 180º – 65º = 115º(ii) In ABD,

DAB + ABD + ADB = 180º 65º + 70º + ADB = 180º ADB = 180º – 135º = 45º

ADC =BDA + BDC = 45º + 45º = 90ºi.e. ADC is in semi-circle.So AC is a diameter.

2. (i) Given, AB is a diameter of circle. ACB = 90º [ Angle in a semi-circle is 90º]

In ABC,ACB + CBA + BAC = 180º

90º + CBA + 34º = 180º CBA = 180º – 124º

CBA = 56º(ii) Since, CQ is a tangent to the circle. BCQ = BAC = 34º [Alternate segment property]

Since, AQ is a straight line. ABC + CBQ = 180º

AB

CD

65º 70º

45º

34ºA BQ

C

y

A C

x

160ºO

B

D

36


56º + CBQ = 180º CBQ = 180º – 56º = 124º

Now in CBQ,CBQ + CQB + BCQ = 180º

124º + CQB + 34º = 180º CQB = CQA = 22º

3. Since, PRT is a tangent at R and QR is a chord.

QRP = RSQ = yº [Alternate segment property]Since, SQ is a diameter of a circle. QRS = 90º [Angle in semi-circle]

In PRS,SPR + PRS + RSP = 180º [Sum of all angles of a triangle is

180º] xº + (yº + 90º) + yº = 180º xº + 2yº = 180º – 90º xº + 2yº = 90º

Hence proved

4. Given, PO = 6 cm, QO = 9 cm and ar(POB) = 120 cm2

In AOQ and BOP, we haveQAO = OBP = 90º [ PB AB and QA AB]

and AOQ = POB [Vertically opposite angles] AOQ ~ POB [By AA axiom of similarity]

Then,ar( AOQ)ar( BOP) =

2

2

(OQ)(PO)

ar(AOQ) =29

6

× 120

=81 120

36

= 270 cm2

Hence, the area of AOQ is 270 cm2

xº yº

O

Q

P

S

RT


37

5.

2007

1.

Radius = 1.5 cm

2. Given, AT = 16 cm and AB = 12 cmWe know that, chord AB and tangent at P meet at point T, then

TA × TB = PT2

16 × (AT – AB) = PT2

16 × (16 – 12) = PT2

PT2 = 16 × 4 PT2 = 64 PT = 8 cm

3. Given PBA = 45ºCalculate the value of PQBSince, AB is a diameter of circle,

A

B60º

X

D C

O

2 cm

5 cm

2 cm

4.2 cm

5.8 cm

C

O

E

A

B

X

6.4 cm

60º

P

T

BA

38


So APB is in semi-circle. APB = 90º

In right APB,APB + PBA + BAP = 180º

90º + 45º + BAP = 180º 135º + BAP = 180º BAP = 180º – 135º = 45º PQB = BAP = 45º [Angles in the same segment are

equal]Hence, the value of PQB is 45º

4. In ACE,AEC + EAC + ACE = 180º [ Sum of all the angles of a

triangle is 180º] AEC + 62º + 43º = 180º AEC = 180º – 105º = 75ºSince, points A, B, D and E are the pointson a circle, so quadrilateral ABDE is acyclic quadrilateral.We know that, sum of opposite angles is 180º

ABD + AED = 180º a + 75º = 180º [ AEC = AED = 75º] a = 105º

EDF = BAE [Exterior angle property] c = 62ºIn ABF,

ABF + AFB + BAF = 180º [By angle sum property of a triangle]

105º + b + 62º = 180º b = 180º – (105º + 62º)

= 180º – (167º) = 13ºHence, a = 105º, b = 13º and c = 62º

5. Given : DE BC andADDB

=32

(i) Now,ADDB

=32

AD

AD DB=

33 2

ADAB =

35

In ADE and ABC, we getDAE = BAC [Common angle]

A E F

BD

C

b

ac

62º

43º

D E

B C

A

F

A B

Q

P

O45º


39

and ADE = ABC [Corresponding angles as DE BC] ADE ~ ABC [By AA axiom of similarity]

Then,DEBC

=ADAB

[ If two triangles are similar, then

their corresponding sides areproportional]

DEBC =

35 ... (i)

(ii) To prove : DEF ~ CBFProof : In DEF and CBF, we have

FED = FBC [Alternate interior angles as DE BC]and DFE = BFC [Vertically opposite angles] DEF ~ CBF [By AA axiom of similarity]

Then,DEBC

=FEFB

[ If two triangles are similar, then

FEFB

=35

[From equation (i)]

(iii)Since, DEF ~ CBF

ar( DEF)ar( CFB) =

2

2

DEBC

=23

5

=925

[From equation (i)]

Hence, ar(DEF) : ar(CFB) = 9 : 25

6.

(ii) 9.4 cm

AQ

D

P

C

R X

B

4 cm

45º

M

5 cm

their corresponding sides are proportional]

40


2008

1. (i) area APO : ABC = 4 : 25(ii) area APO : CQO = 4 : 9

2. (i) QOP = 112º(ii) QCP = 68º

3. 12 cm

4.

AXBY is a square

2009

1. (i) 75º(ii) 15º(iii)105º

2. (i) 24 m(ii) 1000 cm3

C DX

P

Q

BA

E FY

8 cm


41

3.

4. Given : AB = 7 cm and BC = 9 cm

(i) To prove : ACD ~DCBProof : In ACD and DCB,

C = C [Common angles]CDB = BAD [Anlges in alternate segment]

ACD ~ DCB [By AA axiom of similarity](ii) Since, chord AB and tangent at D intersect each other at point C. AC × BC = CD2

16 × 9 = CD2

CD2 = 144 = (12)2

CD = 12 cm

5. Given : BA CE and AF : AC = 5 : 8

(i) To prove : ADF ~ CEFProof : In ADF and CEF,

AFD = CFE [Vertically opposite angles]DAF = FCE [Alternate angles, since BA CE]

ADF ~ CEF [By AA axiom of similarity](ii) Given : CE = 6 cm

Since, ADF ~ CEF

ADCE

=AFCF

AF 5 AC 8 AC 8–1 –1AC 8 AF 5 AF 5

AC – AF 8 – 5 CF 3 AF 5AF 5 AF 5 CF 3

A B

D

C7 cm 9 cm

CB

ED F

A

5.5cm

3.4

cm4.9 cm

A

OY

C

B

X

42


AD6

=53

AD =5 6

3

AD = 5 × 2AD = 10 cm

(iii)Given, DF BC ADF = ABC [Corresponding angles]

and DAF = BAC [Common angle] ADF ~ ABC

Then,ar( ADF)ar( ABC) =

2

2

(AF)(AC) =

258

AF 5AC 8

=2564

ar(ADF) : ar(ABC) = 25 : 64

2010

1. BAO = 40º

2. (i) 12 cm(ii) Area of ABC = 36 cm2

3.

(iv) 5.2 cm

CD

P

60ºB

A

6 cm

3.5

cm


43

4.

2011

1. (i) In ADB and CAB,ADB = CAB [both 90º]ABD = CBA [common angle]

ABC DBA [AA similarity criterion]In ADC and BAC,ADC = BAC [both 90º]ACD = ACB [common angle]

DAC ABC [AA similarity criterion]If two triangles are similar to one triangle, then the two triangles aresimilar to each other.

DAC DBA or CDA ADB

(ii) Since the corresponding sides of similar triangles are proportional.

CDAD

=DADB

AD2 = DB x CD AD2 = 18 x 8

AD = 12 cm

(iii) The ratio of the areas of two similar triangles is equal to the ratio of thesquares of their corresponding sides.

So,Ar( ADB)Ar( CDA)

=2

2

ADCD

=14464 =

94

Thus, the required ratio is 9:4 .

C

A B

D8 cm

18 cm

4 cm

44


2.

PQ = 4.8 cm

3. In AOC, ACO = 30º, [Given]OAC = 90º [radius is prependicular to the

tangent at the point of contact]

By angle sum property,ACO + OAC + AOC = 1800

AOC = 1800 – (900 + 300) = 600

Consider AOC and BOCAO = BO [radii]AC = BC [tangents to a circle from an

external point are equal in length]OC = OC [Common]

AOC BOC.(i) BCO = ACO = 30º(ii) AOC = BOC = 60º

AOB = AOC + BOC = 120º(iii) AOB = 2APB

So, APB =12

AOB

=12

× 120º

= 60º

4. Let the radii of the circles with A, B, and Cas centres be r1, r2 and r3 respectively.

According to the given information,AB = 10 cm = r1 + r2 ...(1)

OP

Q

R

M6 cm 3.5 cm

C

A

B

30º

PO


45

B P A

Q

C

R

BC = 8 cm = r2 + r3 ...(2)CA = 6 cm = r1 + r3 ...(3)

Adding equations (1), (2) and (3),2(r1 + r2 + r3) = 24

r1 + r2 + r3 = 12 ...(4)Subtracting (1) from (4),

r3 = 12 –10 = 2 cmSubtracting (2) from (4),

r1 = 12 – 8 = 4 cmSubtracting (3) from (4),

r2 = 12 – 6 = 6 cmThus, the radii of the three circles with centres A, B and C are 4 cm, 6 cmand 2 cm respectively.

2012

1. 11.25 cm.

2. (i) 50º(ii) 40º

3. In ABC and AMP(i) ABC = AMP [each 900]

BAC = PAM [common] ABC AMP [By AA similarity]

Since the triangles are similar, we have

(ii)ABAM

=BCMP

=ACAP

ABAM

=BC12

=1015

Taking,BC12

=1015

BC =12 10

8cm15

Now using Pythangoras theorem in triangle ABC,AB2 + BC2 – AC2

AB2 = 102 – 82 = 36AB = 6 cm

Hence, AB = 6 cm and BC = 8 cm

BA P

C

M

46


4.

5. 3 cm

2013

1. (i) In ABC,DAB + ABD + ADB = 1800

650 + 700 + ADB = 1800

ADB = 1800 – 700 – 650

= 450

Now, ADC = ADB + BDC= 450 + 450

= 900

ADC is the angle of semi-circle so AC is a diameter of the circle.(ii) ACB = ADB (angle subtended by the same segment) ACB = 450

2. (i) ADC = 80º(ii) DCT = 60º

A B

D

C450

650

700

C

B

A

D40º

100º O

T


47

3.

On measuring BCP = 30º

4. (i) In ABC and DEC,ABC = DEC = 90º (perpendiculars to BC)ACB = DCE (Common)ABC DEC (AA criterion)

(ii) Since ABC DEC,

ABDE

=ACCD

64 =

15CD

6 × CD = 60

CD =606 = 10 cm

(iii) ABC ~ DECar(ABC) : ar(DEC) = AB2 : DE2

= 62 : 42

= 36 : 16= 9 : 4

2014

1. (i) 32º(ii) 148º(iii) 32º

2. Consider the given triangleA

B D C

A

D

CB E

6 cm120º

3.5 cm

X

A

B

P

C

48


Given that ABC DAC

(i) Consider the triangles ABC and DAC

ABC = DAC [given]

C = C [common]

By AA Similarity, ABC DAC

(ii) Hence the corresponding sides are proportional

ABDA

=ACDC

=BCAC

85

=4

DC

DC =4 5

8

DC =52

cm

= 2.5 cm

ABDA

=BCAC

85

=4

BC

BC =8 4

5

BC =325

cm

= 6.4 cm(iii)We need to find the ratios of the area of the triangles ABC and DAC.

Since the triangles ABC and DAC are similar triangles, we have

Thus,Area(ΔACD)Area(ΔABC) =

2

2

DAAB

So,Area( ACD)Area(ΔABC)

=

2

2

58

Area(ΔACD)Area(ΔABC) =

2564

= 25 : 64

3. (i) 12 cm(ii) 8 cm


49

4.

Radius = 1.5 cm

2015

1. (i) x = 25º(ii) y = 50º(iii) z = 40º

2.

3. Construction : Join AD and CB.

In APD and CPBA = C [Angles in the same segment]

C

A

5 cm

B N

O5.5

cm

6.5 cm

A

xF O

5 cm

B

E D

C

x

A C

P

BD

50


D = B [Angles in the same segment] APD CPB [By AA Postulate]

APCP

=PDPB

[Corresponding sides of similar

triangles] AP × PB = CP × PD

4. (i) Consider ADE and ACBA = A [Common]

mB = mE = 900

ACB ADE. [AA postulate]

(ii) ConsiderADE ~ ACB

AEAB =

ADAC =

DEBC ... (i)[Corresponding sides of similar

triangles]

Consider ABCBy applying Pythagoras Theorem, we have

AB2 + BC2 = AC2

AB2 + 52 = 132

AB = 12 cmFrom equation (1), we have

412 =

AD13 =

DE5

13 =

AD13

AD =133 cm

Also,4

12 =DE5

DE =20 512 3 cm

(iii) We need to find the area of ADE and quadritateral BCED.

Area of ADE =1

× AE × DE2

A

B C

D

E

C

4 cm

D

B

A

E

9 cm13 cm

12 cm

5 cm


51

=1

× 4 ×2

53

= 210cm

=1 10

×BC × AB2

–3

=1 10

×5 ×122

-3

=10

303

=90 10

3

= 2803

cm

Thus ratio of areas of ADE to quadrilateral BCED =

1013

803

8

5.

2016

1. (i) 32º(ii) 64º(iii) 58º

C

M

AB

D

5.5 cm

PT

6 cm

NQ

R105º

3Area of quad.BCED = Area of ABC Area of ADE

52


2. (i)

(ii) The figure is an irregular hexagon (or arrow head).

3.

4. (i) Since PQRS is a cyclic quadrilateral,

RSP = RQT ... (i)[Exterior angle property]In TPS and TRQ,

PTS RTQ [Common angle]RSP = RQT [From (i)]

TPS ~ TRQ [AA similarity criterion](ii) Since TPS ~ TRQ,

SPQR =

TPTR

A(–4, 4)

C(0,–3)

B(–3, 0)

S

R

Q TP

C

B

O

A Y F

M

E

N

DX

5 cm

L


53

SP4

=186

SP =18 4

6

SP = 12 m(iii) Since TPS ~ TRQ,

ar( TPS)ar( TRQ) =

2

2

SPRQ

27

ar( TRQ) =212

4

ar(TRQ) =279

ar(TRQ) = 3 cm2

ar(PQRS) = ar(TPS) – ar(TRQ) ar(PQRS) = 27 – 3 ar(PQRS) = 24 cm2

5.

6. (i) 600 m(ii) 2 m2

(iii) 175500000 m3

7.

(iv) XY = 5 cm

P

O

A

Q

C B

M2.5

cm 2.5 cm

3 cm 5 cm

54


2017

1. Steps of construction :(i) Draw line AB = 7 cm and

CAB = 60º. Cut off AC = 5 cm.Join BC. ABC is the requiredtriangle.

(ii) Draw angle bisectors of Aand B.

(iii) Bisector of B meets AC at Mand bisector of A meets BCat N.

(iv) P is the point which isequidistant from AB, BC and AC.

(v) Draw a perpendicular frompoint P to AB and let itintersect at point D

(vi) With DP as the radius, draw a circle touching the three sides of thetriangle (incircle.)

2. (i) BAQ = 30ºSince AB is the bisector of CAQ

CAB = BAQ = 30ºAD is the bisector of CAP and P – A – Q

DAP + CAD + CAQ = 180º CAD + CAD + 60º = 180º 2CAD = 180º – 60º CAD = 60º

So, CAD + CAB = 60º + 30º= 90º

So, BD is the diameter of the circle.

(ii) BAQ = ACB (Alternate segment theorem) ACB = 30ºIn ABC,

CAB = 30º (AB is the bisector of CAQ) ACB = CAB AB = BC (Sides opposite to equal angles

are equal) ABC is an isosceles triangle.

3. (i) DAE = 70ºSince ABCD is cyclic quadrilateral

BCD = DAE[Exterior angle property]

BCD = 70º

C

D

M

A

N

B

P

60º

5 cm

7 cm

A

B A

D

E

O

P QA

C

B

D


55

(ii) BOD = 2BCD [Angle at centre is twice theangle at the circumference]

BOD = 2(70º) BOD = 140º

(iii) In OBD, OB = OD OBD = ODBBy Angle Sum property,OBD + ODB + BOD = 180º

2OBD + 140º = 180º 2OBD = 40º OBD = 20º

4. (i) In PQR and SPR, we haveQPR = PSR (Given)PRQ = PRS (Common)

PQR ~ SPR (A.A. Axiom)

PQSP

=QRPR

=PRSR

... (i)

(ii) Now,QRPR

=PRSR

... [From (i)]

QR6

=63

QR =6 6

3

= 12 cm

Also,PQSP

=PRSR

...[From (i)]

8

SP=

63

8

SP= 2

SP =82

= 4 cm

(iii)Area of PQRArea of SPR

=2

2

PQSP

=2

2

84

=6416

= 4

Q

P

RS 3 cm

8 cm 6 cm

MT EDUCARE LTD.icse.maheshtutorials.com/images/ICSE_Step_Up_Papers/01Mathema… · 6 Angle at centre is twice the angle at circumference. AOC = 2 × ADC AOC = 2x But AOC = 160º [Given]

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