ÐCED...Mythila Publishers, Puduvayal Download Question Bank From from the base of a lamp post at a...

16
Mythila Publishers, Puduvayal Download Question B 4.1. Similar Triangle 1. Show that ΔPST~ΔPQR ~ PST s SAS By com is P PQR ~ PST similarity SAS By common is P Thus, Thus, 3 2 2 4 4 3 2 1 2 2 a PST In ii) : PQR and PST In i) D D Ð D D Ð = = = = = + = = + = D D D PQ PS PR PT PQ PS PR PT PQ PS PR PT PQ PS 2. Is ΔABC ~ ΔPQR? AB PQ AB QR AB PQ ¹ = = D Thus, and PST In Corresponding sides are not propor D ABC is not similar to D PQR. 3. Check whether the which P Q CB C DB AD AC AE DB AD AC AE CB DE BC C ~ A s AA By co is QBA PQC 110 PQB 8 3 5 3 3 11 4 5 . 3 2 2 A In ii) : A and A In i) 0 D D Ð Ð = Ð = Ð ¹ = + = = + = D D D 5 5 3 3 ing Correspond similar not are A , A = Þ = = \ D D x x AB PQ QB CQ DE BC 4. Observe the figure and find Ð 73 X std Math Bank From https://winglishcoachings.weeb es PQR similarity mmon 5 2 3 2 2 35 2 3 2 2 : PQR and D = = + = = + = D PR PT AB QR ¹ = = = = D 5 2 10 4 2 1 6 3 PQR d rtional. triangles are x. P P Q similarity ommon 70 A : C and 0 = D 5 equal. sides g ÐP 5. In the figure Ð ΔCAB~ΔCED. Also c 6 8 2 10 9 s ing Correspond CED ~ CAB similarit AA By CED A common, is C CE and CAB In = Þ + = = \ D D \ Ð = Ð Ð D D x x CD CB DE AB 6. In figure, QA ,PB a If AO=10cm, BO=6c 9 6 10 Correspond B ~ OQ simil AA By op y (Verticall B AOQ 90 O and AOQ In Þ = = \ D D \ Ð = Ð = Ð = Ð D AQ AQ BP AQ BO AO OP OP OBP AQ 7. In figure, ΔACB PQ=4cm, BA=6.5cm and AQ. AQ AC AQ AB AP AC PQ BC APQ CB 5 . 6 8 . 2 4 8 Corresp ~ A = = = = \ D D . 6 4 8 cm 6 . 5 8 . 2 4 8 = = = = AQ A AC AC hematics Made Easy ebly.com/ QRP ~ BAC , similarity SSS By 2 1 3 6 3 3 2 1 12 6 2 1 6 3 : and In D D = = \ = = = = = = D D PR CA QP BC RQ AB PR CA QP BC RQ AB PRQ BAC ÐA=ÐCED prove that find the value of x. cm equal sides ty ED, are perpendicular to AB cm, PB=9cm. Find AQ. cm 15 equal sides ing d larity angles) pposite 0 , B d 0 = D OP ~ΔAPQ. If BC=8 cm, m and AP=2.8cm, find CA equal sides ing pond cm 15 5 . AQ

Transcript of ÐCED...Mythila Publishers, Puduvayal Download Question Bank From from the base of a lamp post at a...

Page 1: ÐCED...Mythila Publishers, Puduvayal Download Question Bank From from the base of a lamp post at a speed of 1.2m/sec. If the lamp post is 3.6m above the ground, find the length of

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4.1. Similar Triangles

1. Show that ΔPST~ΔPQR

~PST

similaritySASBy

commonisP

PQR~PST

similaritySASBy

commonisP

Thus,Thus,

3

2

24

4

3

2

12

2

andPSTInii):PQRandPSTIni)

DD

Ð

DD

Ð

=

=

=

=

=+

=

=+

=

DDD

PQ

PS

PR

PT

PQ

PS

PR

PT

PQ

PS

PR

PT

PQ

PS

2. Is ΔABC ~ ΔPQR?

AB

PQ

AB

QR

AB

PQ

¹

=

=

D

Thus,

andPSTIn

Corresponding sides are not proportional.

D ABC is not similar to D PQR.

3. Check whether the which

PQCB

C

DB

AD

AC

AE

DB

AD

AC

AE

CBDEBC

C~A

similarity AABy

commonis

QBAPQC

110PQB

8

3

53

3

11

4

5.32

2

AInii):AandAIni)

0

DD

Ð

Ð=Ð

¹

=+

=

=+

=

DDD

553

3

ing Correspond

similarnot

areA,A

=Þ=

=

\DD

xx

AB

PQ

QB

CQDEBC

4. Observe the figure and find Ð

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4.1. Similar Triangles

PQR

similarity

common

5

2

32

2

35

2

32

2

:PQRand

D

=

=+

=

=+

=

D

PR

PT

AB

QR¹

==

==

D

5

2

10

4

2

1

6

3

PQRand

Corresponding sides are not proportional.

triangles are similar and find the value of x.

P

PQ

similarity

common

70QBA

:Cand

0=

D

5

equal.sidesing

ÐP

5. In the figure ÐΔCAB~ΔCED. Also find the value of

cm68

2109

sidesing Correspond

CED~CAB

similarity AABy

CEDA

common,isC

CED,andCABIn

=Þ+

=

=

\

DD

\

Ð=Ð

Ð

DD

xx

CD

CB

DE

AB

6. In figure, QA ,PB are perpendicular to AB If AO=10cm, BO=6cm, PB=9cm. Find AQ.

96

10

Correspond

B~OQ

similarity AABy

oppositey(Verticall

BAOQ

90O

andAOQIn

Þ=

=

\

DD

\

Ð=Ð

=Ð=Ð

D

AQAQ

BP

AQ

BO

AO

OP

OP

OBPAQ

7. In figure, ΔACBPQ=4cm, BA=6.5cm and AP=2.8cm, find CA and AQ.

AQ

AC

AQ

AB

AP

AC

PQ

BC

APQCB

5.6

8.24

8

Correspond~A

==

==

\DD

.6

4

8

cm6.5

8.24

8

=

=

=

=

AQ

AQ

AC

AC

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QRP~BAC

,similaritySSSBy

2

1

36

33

2

1

12

6

2

1

6

3

:andIn

DD

==\

==

==

==

DD

PR

CA

QP

BC

RQ

AB

PR

CA

QP

BC

RQ

AB

PRQBAC

ÐA=ÐCED prove that CED. Also find the value of x.

cm

equalsides

similarity

CED,

PB are perpendicular to AB If AO=10cm, BO=6cm, PB=9cm. Find AQ.

cm15

equalsidesing Correspond

similarity

angles)opposite

90

,Band0

=

D OP

~ΔAPQ. If BC=8 cm, PQ=4cm, BA=6.5cm and AP=2.8cm, find CA

equalsidesing Correspond

cm15

5.

AQ

Page 2: ÐCED...Mythila Publishers, Puduvayal Download Question Bank From from the base of a lamp post at a speed of 1.2m/sec. If the lamp post is 3.6m above the ground, find the length of

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8. Two triangles QPR and QSR, right angled at P and S respectively are drawn on the same base QR and on the same side of QR. If PR and SQ intersect at T, prove that

T × TQPT × TR= S

T × TQPT × TR= S

ST

PT

TR

QT

SRTQT

STRTQ

SP

SRTQT

=

\

DD

\

Ð=Ð

=Ð=Ð

DD

equalsidesCorres.

~P

similarity AABy

angles)opp.Vertically

P

90

,andPIn0

9. If figure OPRQ is a square and MLN=90°. Prove that

(i) ΔLOP~ΔQMO

(ii) ΔLOP~ΔRPN

(iii) ΔQMO~ΔRPN

(iv) QR2 =MQ ×RN.

RNMQQR

RN

QO

´=

=

\

DD

DD

Ð=Ð

Ð=Ð

D

DD

Ð=Ð

=Ð=Ð

DD

2

o

RP

QM

aresidesing Correspond

RPN~QMO

(ii)&(i)From

~LOP

AABy

(Correspon

LPO

OLP

LOP,In

QMO~LOP

,similarity AABy

angles)ding (Correspon

OMQLOP

90OQMOLP

QMOLOP,In

10. In the given figure, Δ is right angled at C and DE^AB. Prove that ΔABChence find the lengths of AE and

ADE~ABC

,similarity AABy

commonisA

90ACBAED

ADE,&ABCIno

DD

Ð

=Ð=Ð

DD

13

36,

13

15

51213

3

AB

AD2

==

==

==

AB

AB

DEAE

AEDE

AC

AE

BC

ED

74 X std Mathematics Made Easy

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Two triangles QPR and QSR, right angled at P and S respectively are drawn on the same base QR and on the same side of QR. If PR and SQ intersect at T, prove that

If figure OPRQ is a square and MLN=90°.

D

Ð

Do

Equal,

RPN

,similarity

angles)ding (Correspon

PNR

90PRN

RPNLOP,

is right angled at C ABC~ΔADE and

hence find the lengths of AE and DE.

13

125 22

22

=

+=

+= BCAC

11. The perimeters of two similar triangles ABC and PQR are respectively 36 cm and 24cm . If PQ=10cm, find AB.

24

36

10

AB

24

36

PQ

AB

PQ

AB

=sidescorres.ofratio

~ABC

Þ=

=

===

D

AB

PR

AC

QR

BC

12. If ΔABC is similar to BC=3cm, EF=4cm and area of ΔABC=54cm2. Find the area of

DofArea

4

3

(sidescorres.ofratio

ABC

D

D

EF

EF

BC

13. If ΔABC~ΔDEF such that area of 9cm and the area of BC=2.1cm. Find the length of EF.

1.2(

(sidescorres.ofratio

ABCD

EF

EF

EF

BC

14. A vertical stick of length 6m casts a shadow 400cm long on same time a tower casts shadow 28m long. Using similarity, find height of the tower.

28

46

PQ

AB

PQR~ABC

similarity AABy

PR)||(ACRC

90QB

andABCIno

=Þ=

=

DD

Ð=Ð

=Ð=Ð

DD

PQPQ

QR

BC

15. A boy of height 90cm is walking away

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The perimeters of two similar triangles ABC and PQR are respectively 36 cm and 24cm . If PQ=10cm, find AB.

15

24

36

perimetersofratio=

PQR

=

=

D

ABC is similar to ΔDEF such that BC=3cm, EF=4cm and area of

. Find the area of ΔDEF.

2

2

2

2

2

2

cm969

1654

DofArea

54

4

3

DofArea

ABCofArea

Areaofratio=)

D~ABC

=

D=

D

D=

D

EF

EF

EFEF

BC

EF

DEF such that area of ΔABC is 9cm and the area of ΔDEF is16 cm 2and BC=2.1cm. Find the length of EF.

cm8.2

9

)1.2(16

16

9)1

DofArea

ABCofArea

Areaofratio=)

D~ABC

2

2

2

2

2

2

=

´=

=

D

D=

D

EF

EF

EFEF

BC

EF

A vertical stick of length 6m casts a shadow 400cm long on ground. At the same time a tower casts shadow 28m long. Using similarity, find height of the tower.

m42

,similarity

PR)

PQRD

A boy of height 90cm is walking away

Page 3: ÐCED...Mythila Publishers, Puduvayal Download Question Bank From from the base of a lamp post at a speed of 1.2m/sec. If the lamp post is 3.6m above the ground, find the length of

Mythila Publishers, Puduvayal

Download Question Bank From

from the base of a lamp post at a speed of 1.2m/sec. If the lamp post is 3.6m above the ground, find the length of his shadow cast after 4 seconds.

4.8BE

1.2

speeddistance

4time

m6.1

8.44

8.4

9.0

6.3

CD

AB

C~ABE

,similarity AABy

PR)||(ACEE

90DB

CandABEIno

=

=

=

=

=

+=

+=

=

DD

Ð=Ð

=Ð=Ð

DD

DE

DE

DE

DE

DE

DE

BE

DE

DE

16. A girl looks the reflection of the top of the lamp post on the mirror which is 66m away from the foot of the lamp post. The girl whose height is 12.5m is standing 2.5m away from the mirror.placed on the ground facing the sky and the girl, mirror and the lamppost are in a same line, find the height of the lamppost.

m328.5LPHeightpostLamp

6.87

4.05.1

LP

AB

LPC~ABC,similarity AABy

LCPACB

incidenceofangle

PB

:LPCABC,In

=

=

=

DD

Ð=Ð

=

Ð=Ð

DD

LP

CP

BC

17. Two poles of height ‘a’ meter and ‘are ‘p’ meter apart. Prove that the height of the point of intersection of the lines joining the top of each pole to the foot of

the opposite pole is given by a

)1.....(,

Let

p y x

y.LA

x , CL

=+

=

=

75 X std Mathematics Made Easy

Download Question Bank From https://winglishcoachings.weebly.com/

from the base of a lamp post at a speed of 1.2m/sec. If the lamp post is 3.6m above

length of his shadow

m4.8

41.2

timespeed

seconds4

´

´

A girl looks the reflection of the top of the lamp post on the mirror which is 66m away from the foot of the lamp post. The girl whose height is 12.5m is standing 2.5m away from the mirror. Mirror is placed on the ground facing the sky and the girl, mirror and the lamppost are in a same line, find the height of the lamppost.

m

reflectionof

90o

’ meter and ‘b’ meter ’ meter apart. Prove that the height

of the point of intersection of the lines joining the top of each pole to the foot of

ba

ab

+meter.

ba

abh

aphp

b

ph

a

php

b

ph

a

phyx

a

phx

h

a

x

p

+=

çè

æ+=

+=

+=++

=Þ=

=

Ð=Ð

=Ð=Ð

DD

1

(2),(1)

LO

AB

CL

CA

DCLO~DCAB

similarity AABy

common][ CC

90CLOCAB

LOC,andABCIno

18. Two vertical poles of are erected above a horizontal ground AC. Find the value of y.

2

6

3

6AC

AB

CA

CB(2),(1)

)1.....(6CA

CB

PA

QB

CA

CB

Q~P

similarity AABy

common][ CC

90QP

,QandPIno

=

=

çè

æ=

+

=++

=

=

DD

Ð=Ð

=Ð=Ð

DD

y

AC

AC

yAC

BCAB

y

y

BCAC

BCAC

BCAC

4.2. BPT & A

1. In ΔABC, D and E are points on the sides

AB and AC respectively such thatDE

(i) If 4

3=

DB

ADand AC = 15cm

6

454

154

3

theorem,BPTBy

»

=

=

=

AE

AE

EC

AE

DB

AD

ii) If AD = 8x – 7, DB = 5

X std Mathematics Made Easy

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b

b

ph

b

ph

b

phy

b

h

p

y

÷ø

ö

=Þ=

=

DD

Ð=Ð

=Ð=Ð

DD

1

DC

OL

AC

AL

ACD~ALO

similarity AABy

common][ AA

90ACDALO

ACD,andALOIno

Two vertical poles of heights 6m and 3m are erected above a horizontal ground AC. Find the value of y.

m2

6

3

1

6

1

36

)2....(3AC

AB

RC

BQ

AC

AB

Q~R

similarity AABy

common][ AA

90QR

ACD,andALOIno

÷ø

öçè

æ+

+

=

=

DD

Ð=Ð

=Ð=Ð

DD

y

yy

y

BACA

BACA

& ABT TheoremS

ABC, D and E are points on the sides

AB and AC respectively such thatDE P BC

and AC = 15cm find AE.

43.6

345

15

15

-

-

AE

AC

EC

AE

7, DB = 5x – 3, AE = 4x – 3 and

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EC = 3x – 1, find the value of x

1,2

1

012

0224

5)(34()13)(78(

13

34

35

78

theorem,BPTBy

2

2

-=

=--

=--

-=--

-

-=

-

-

=

x

xx

xx

xxxx

x

x

x

x

EC

AE

DB

AD

iii) AD = x, DB = x − 2, AE = x +2 and find the lengths of the sides AB

424

2

244

2

4

)(2()1(

1

2

2

theorem,BPTBy

22

-++=

-++=

+=

-+=

-+=

+=

-=-

+-=-

-

+=

-

=

xx

ECAEAC

xx

DBADAB

xxx

xxxx

x

x

x

x

EC

AE

DB

AD

2. In Δ

following casesshow that DE P

i) AB = 12cm, AD = 8cm, AE=12cm,

BC||DE

,BPTofConverseBy

3

2

18

12

3

2

12

8

:ABCIn

AC

AE

AB

AD

AC

AE

AB

AD

=

==

==

D

ii)AB=5.6cm,AD=1.4cm,AC=7.2cm,AE=

BC||DE

,BPTofConverseBy

2.7

8.1,

4

1

6.5

14

:ABCIn

AC

AE

AB

AD

AC

AE

AB

AD

=

===

D

3. In Fig.DEPAC,DCPAP. Prove that

76 X std Mathematics Made Easy

Download Question Bank From https://winglishcoachings.weebly.com/

x.

)3-x

and EC = x − 1 AB and AC.

91

1

6

4

)2

+

AC

AB

x

ABC, D and E are points on the sides AB and AC respectively. For each of the

P BC.

12cm, AC = 18cm

AB=5.6cm,AD=1.4cm,AC=7.2cm,AE=1.8cm

4

1=

Prove that CP

BC

EC

BE=

CP

BC(2).and(1)From

EC

BE

:BPTBy

In

CP

BC

:BPTBy

In

=

=

=

ΔBCA, DE |

ΔBPA, DC |

4. ABCD is a trapezium in which AB

and P, Q are points on AD and BC, such thatPQ ||DC if PD = 18cm, BQ = 35cm and

QC = 15cm, find AD.

18

AP(2),&(1)From

15

35

QC

BQBPT,By

|RQABC,In

18

AP

PD

APBPT,By

|PRACD,In

RatPQmeet AC,Join

| AB ABCD,trapeziumIn

=

=

=

=

=

D

=

=

D

AD

5. In trapezium ABCD, AB

points on non-parallel sides AD and BC,

such that EFPAB. Show that

FC

BF

ED

AE

FC

BF

PC

AP

PC

AP

ED

AE

=

=

D

=

D

(2)&(1)From

....(BPT,By

AB||PRABC,In

....(BPT,By

DC||EPADC,In

PatEFmeetto ACJoin

EF||DC|| AB

ABCD,trapeziumIn

6. In fig, if PQ ||BC and PR

AQ

QB

AB

AQ

AD

AR= ii)i)

X std Mathematics Made Easy

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EC

BE

)2.....(DA

BD

:

.

)1.....(DA

BD

:

.

=

=

=

| ACΔBCA, DE |

| APΔBPA, DC |

ABCD is a trapezium in which AB|| DC

and P, Q are points on AD and BC, such DC if PD = 18cm, BQ = 35cm and

QC = 15cm, find AD.

m601842

4215

35

)2....(

AB||

)1....(

DC||

:

PQ||DC||

=+=

+=

=Þ=

=

=

=

=

PDAP

AP

RC

AR

RC

AR

RC

AR

RC

AR

ABCD, ABPDC, E and F are

parallel sides AD and BC,

. Show that FC

BF

ED

AE=

)2....(

)1....(

BC and PR || CD prove that

AR

DR

AQ

QB=

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AR

DR

AQ

QB

AR

ARAD

AQ

AQAB

AR

AD

AQ

AB

AR

AD

AQ

AB

AB

AQ

AD

AR

AD

AR

AC

AP

AB

AQ

=

-=

-

-=-

=

=

=

DD

11,1subtract

(1),ofreciprocalii)

(2),&(1)From

BPT,By

)1....(

BPT,By

PRADC,InBC||PQ ABC,Ini)

7. Rhombus PQRB is inscribed in that B is one of its angle. P,Q and R lie on AB, AC and BC. If AB = 12cm and BC = 6cm, find sides PQ, RB of the rhombus.

cm4RBPQ

7218

7218

6

12

PB

AP(2),(1)From

BPT,By

ABC,In

)1...(PB

APBPT,By

BC||PQABC,In

cm6BC,12APcm12AB

.rhombusofside

ABC.ininscribedisPQRSRhombus

22

==

=

=+-

-=

-

=

D

=

D

=-==

=

D

x

xxx

x

x

x

x

RC

BR

QC

AQ

QC

AQ

x

x

8. In figure DEPBC and CDPEF

= AB×AF AD2 .

= AB×AF AD

AB

AC

AE

AB

AC

AE

2

AD

AD

AF(2)(1),From

)2....(AD

BPT,By

BC||DEACD,In

)1....(AD

AFBPT,By

BC||DEACD,In

=

=

D

=

D

9. In fig, AD is the bisector of

4cm, DC = 3cm and AB = 6cm, find AC.

DC

BD=\

ÐD

AC

AB ABT,By

A.ofbisector theis ADABC,In

77 X std Mathematics Made Easy

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AC

AP= )2....(

DC||PR

Rhombus PQRB is inscribed in ΔABC such that B is one of its angle. P,Q and R lie on AB, AC and BC. If AB = 12cm and BC = 6cm, find sides PQ, RB of the rhombus.

)2..(

AB||QRABC,

6RC

ABC.

=

-=

RC

BR

QC

AQ

x

EF. Prove that

ÐA.If BD =

4cm, DC = 3cm and AB = 6cm, find AC.A.

cm5.4

4

18

184

6

3

4

=

=

=

=

AC

AC

AC

AC

10. In ΔABC, AD is the bisector of

side BC at D, if AB = 10cm, AC = 14cm and BC =6cm, find BD and DC.

2.5 – 6

x – 6DC

cm2.5BD

.22

5

7530

614

10

AC

AB ABT,By

bisector theis ADABC,In

=

=

=

==

=-

-=

=\

D

x

xx

x

x

DC

BD

11. Check if AD is a bisector of

i) AB = 5cm, AC = 10cm,BD =1.5cmii)AB =4cm, AC = 6cm, BD = 1.6cm

Aofbisector anotisAD

ABT,ofConverseBy

AC

AB

7

3

5.3

5.1

2

1

10

5

AC

AB

ABC,In

Ð

\

¹

==

==

D

DC

BD

DC

BD

12. In figure ÐQPR=90°, PS is its bisector. If

ST ^ PR, prove that

PT,PR=TR

TR......(2PQ=PRST

PR

TR=

PQ

ST

QPR~STR

similarity AABy

commonisR

QOR=STR

STRIn

PT........=ST

TSP=TPSPST,in

RPS=QPS

PS90°,=QPR

-

´´

DD

Ð

ÐÐ

D

ÐÐD

ÐÐ

Ð

X std Mathematics Made Easy

https://winglishcoachings.weebly.com/

ABC, AD is the bisector of ÐA meeting

side BC at D, if AB = 10cm, AC = 14cm and BC =6cm, find BD and DC.

cm3.52.5

cm

5.

A.ofbisector

=

Ð

DC

bisector of ÐA of ΔABC

i) AB = 5cm, AC = 10cm,BD =1.5cm,CD = 3.5cm 4cm, AC = 6cm, BD = 1.6cm, CD = 2.4cm

A Aofbisector aisAD

ABT,ofConverseBy

AC

AB

3

2

4.2

6.1

3

2

6

4

AC

AB

ABC,In

Ð

\

=

==

==

D

DC

BD

DC

BD

QPR=90°, PS is its bisector. If

PR, prove that =PQ×PRST×(PQ+PR)

PT,

)TR......(2

QPR

,similarity

common

90°=QOR

QPR&STR

(1)PT........

45°45°-90°-180°=

45°=90°/2TSP=

bisectorsit'isPS

D

=

Page 6: ÐCED...Mythila Publishers, Puduvayal Download Question Bank From from the base of a lamp post at a speed of 1.2m/sec. If the lamp post is 3.6m above the ground, find the length of

Mythila Publishers, Puduvayal

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PR.×PQ=PR)+(PQ×ST

PQPRPQ=PRST

ST)(PRPQ=PRST

in(2)valuethisusing

ST-PR=TRST,=PT

´-´´

-´´

13. ABCD is quadrilateral in which AB = AD,

the bisector of ÐBAC and ÐCAD intersect

the sides BCand CD at the points E and F

respectively. Prove that EF P BD.

BD.||EFHence

BPTofConverseBy

FD

CF=

BE

CE

(2)=(1)

-(2)---FD

CF=

AB

AC

FD

CF=

AD

AC

CADbisector theis AF ACD,In

-(1)---BE

CE=

AB

AC

BACofbisector theis AEABC,In

ÐD

ÐD

4.3. Pythagoras Theorem

1.

m30=

900=

576)+(324=

24+18=PN

pointstarting

positioncurrentofDistance=PN

22

2. There are two paths that one can choose to go from Sarah’s house to James house. One way is to take C street, and the other way requires to take A street and then B street. How much shorter is the direct path along C Street? (Using figure).

miles2.5=6.25=

2.25+4=

1.5+2=SJ

DistanceStreetCDirecta)

22

78 X std Mathematics Made Easy

Download Question Bank From https://winglishcoachings.weebly.com/

ST

in(2)

ST

ABCD is quadrilateral in which AB = AD,

CAD intersect

the sides BCand CD at the points E and F

BD.

CAD

BAC

Pythagoras Theorem

A man goes 18m due east and then 24m due north. Find the distance of his current position from the starting point?

fromposition

There are two paths that one can choose to go from Sarah’s house to James house.

take C street, and the other way requires to take A street and then B street. How much shorter is the direct path along C Street? (Using figure).

2.5-3.5=

Differencec)

miles3.5=

1.5+2=Distance

StreetBb)

3. To get from point A to point B you must avoid walking throughwalk 34 m south and 41m east. To the nearest meter,how many meters would be saved if it were possible to make a way through the pond?

m754134

SB ASAB

East)&(South2 – Path

53.26=

2837=

1681+1156=

41+34=AB

pond)(Through1-Path

22

=+=

+=

4. The hypotenuse of a right triangle is 6m more than twice of the shortest side. Ifthird side is 2m less than the hypotenuse, find the sides of the triangle.

sidethirdoflength

hypotenuse

sideshorter ofLength

)2()10(

208

36244

6)+(2x

sidethirdoflength

hypotenuseoflength

sideshorter oflength

2

2

2

-

--

x + x

x x

x + + x

5. What length of ladder is needed to reach a height of 7 ft along the wall when the base of the ladder is 4 ft from the wall? Round off your answer to the

2

2

2

2

222

65.61<65<64=8

8betweenis65

65=65

1649

4272

BC+ AC=AB

ft.7=ACft,4=BC

ladder oflengthx

x. = x

+ = x

+ = x

Þ

=

X std Mathematics Made Easy

https://winglishcoachings.weebly.com/

mile1=

pathstwob/wDifference

miles

Street Athen&Street

To get from point A to point B you must avoid walking through a pond. You must walk 34 m south and 41m east. To the nearest meter,how many meters would be saved if it were possible to make a way

m21.74=

53.2675=

savedMetersEast)

pond)

-

The hypotenuse of a right triangle is 6m more than twice of the shortest side. If the third side is 2m less than the hypotenuse, find the sides of the triangle.

24=4+20=side

26=6+2(10)=hypotenuse

10side

2,100=)

020

1616436

)42(

42=

26+2=side

62=hypotenuse

x=side

22

222

=

-

x =

=

x + + x + = x

x + + = x

x +

x

x +

What length of ladder is needed to reach a height of 7 ft along the wall when the base of the ladder is 4 ft from the wall? Round off your answer to the next tenth place.

28.1=65.61

8.1.and

ladder

Page 7: ÐCED...Mythila Publishers, Puduvayal Download Question Bank From from the base of a lamp post at a speed of 1.2m/sec. If the lamp post is 3.6m above the ground, find the length of

Mythila Publishers, Puduvayal

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6. 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4m high. If the foot of the ladder is moved 1.6m towards wall, find the distance by which the top of the ladder would slide upwards on the wall.

0.8distancerequiredtheHence

0.8=EC

4.8=

EB=EC

=EB

25=EB

EB=5

EB=ED

In

1.4=

1.63=

AD AB=DB

3=AB

1625=AB

4+ AB=5

BC+ AB=AC

ABCtriangleIn

2

2

2

2

222

222

=

-

-

-

7. An insect 8m away initially from the foot of a lamp post which is 6 m tall, crawls towards it moving through a distance. If

to distance it has moved, how insect away from foot of the lamp post?

m7512568

8=BC

25610016

1664+36=x

)(8+6=x

BC+ AB=AC

8=CDBD=BC

x.=CD=AC

beinsectm,moving After

postlampof(heightm6=AB

lamp&insectb/w(Distancem8=BD

22

222

222

. = .=

x

.x = x =

x + x

x

x

x

-

-

Þ

-

-

--

Insect is 1.75 m away from foot of lamp post.8. An Aeroplane leaves an airport and flies

due north at a speed of 1000 km/hr. At the same time, another aeroplane leaves same airport and flies due west at a speed of 1200 km/hr. How far apart will be two planes after 1 hours? First plane from O , goes upto A towards north,

km1500

2

31000

time×Speed=Distance

=

´=OA

79 X std Mathematics Made Easy

Download Question Bank From https://winglishcoachings.weebly.com/

m.

m0.8

44.8

BCEB

4.8=23.04

1.9625

(1.4)+EB

DB+EB

DBEtriangle

22

22

-

-

-

An insect 8m away initially from the foot of a lamp post which is 6 m tall, crawls towards it moving through a distance. If its distance from top of lamp post is equal to distance it has moved, how far is the insect away from foot of the lamp post?

Cat

postlamp

foot of lamp post.An Aeroplane leaves an airport and flies due north at a speed of 1000 km/hr. At the same time, another aeroplane leaves same

d flies due west at a speed of 1200 km/hr. How far apart will be two

goes upto A towards north,

Second aeroplane starts from O at goes upto B towards west,

km1800

2

31200

=

´=OB

kms61300

61×3×100AB

61×9×100=

549×100=

18+(15100=

(1800)+(1500)=AB

OB+OA=AB

2

2

222

22

222

=

=

9. In the rectangle WXYZ,

and 26=YW+XZ cm. Calculate the length

and breadth of the rectangle?

=

=XY

=YZ

)5)(12(

6017

34289

)17(

XY

=YZLet,XYZIn

=XZXZ

=XZ

=YW+XZ

=YZ+XY

22

--

-

-

+-

=+

D

+

= PP

= P + P²

P + P² = P² +

P² = ² P

YZ

10.The perpendicular PS on the base QR of a ΔPQR intersects QR at S, such that

3SR=QS . Prove that

2

22

22

2

2

2PQ

PRPQ

PRPQ(2),(1)

PRPRS,In

PQPQS,In

QR

QR

QR

-

--

D

D

X std Mathematics Made Easy

https://winglishcoachings.weebly.com/

Second aeroplane starts from O at goes upto B

)

(1800)2

2

In the rectangle WXYZ, 17=YZ+XY cm,

cm. Calculate the length

and breadth of the rectangle?

cm5=cm12)-(17=

[breadth]P)-(17=

[length]cm12=P=

125,=P0

0

169

13

P).-(17=XY Then,,P=

cm13=26/2=

equal)are(DiagonalsYW=

cm26=

cm17=

2

Þ

=

=

=

P + P² =

²P² =

XZ

The perpendicular PS on the base QR of a PQR intersects QR at S, such that

. Prove that 222 QR+2PR=2PQ

22

22

2

2

22

222

22

22

QR+2PR=

2

QR

4

QR8

8SR

SR(3SR)

SRQS

(2)-----SR+PS=

(1)-----QS+PS=

4

QR=RS4RS=

RS+3RS=

RS+QS=

=

÷ø

öçè

æ=

=

-=

-=

Þ

Page 8: ÐCED...Mythila Publishers, Puduvayal Download Question Bank From from the base of a lamp post at a speed of 1.2m/sec. If the lamp post is 3.6m above the ground, find the length of

Mythila Publishers, Puduvayal

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11.In the adjacent figure, ABC is a right angled triangle with right angle at B and points D, E trisect BC. Prove that

222 5AD+3AC=8AE .

222

22

22

22

2222

22

22

222

22

22

222

22

222

5AD+3AC=8AE

)4x+(AB8=

32x+ AB8=

5AB+27x+ AB3=

5+)9x+(AB3=5AD+3AC

9

3

In

4

2

In

In

2

trisectEandDGiven

x + = AB

x) + (= AB

+ BC = ABABC, AC

x + = AB

x) + (= AB

+ BE = ABABE, AE

+ x= AB

+ BD = ABABD, AD

x, BC = E = BD = x, B

EC = xBD = DE =

D

D

D

12.P and Q are mid-points of the sides CA and CB of a Δ

Prove that 22 5AB=)BP+( AQ4

222

22

222

22

2222

2222

222

222

5AB=)BP+( AQ4

()BC(AC5

4BCBC4AC

4BC(2QC)4AC

4QC4AC)BP(AQ4

QCACBP AQ

(2),(1)

(2).....CPBCBP,BPCIn

(1).....QC ACAQAQC,In

BC+=

++=

++=

++=+

++=+

+

+=D

+=D

4.4. Circles and Tangents

1. Find the length of the tangent drawn from a point whose distance from the centre of a circle is 5cm and radius of circle is 3cm.

4cmPT

tangent,ofLength

16925PT

PT35

PTOTOP,

theoremPythagorasBy

:OTPIn

cm5OPcm,3OTradius

2

222

222

=

=-=

+=

+=

D

==

80 X std Mathematics Made Easy

Download Question Bank From https://winglishcoachings.weebly.com/

In the adjacent figure, ABC is a right angled triangle with right angle at B and points D, E trisect BC. Prove that

22

22

5x+5AB

)x+(AB5

3

BC.trisect

xx, BC =

points of the sides CA ABC, right angled at C.

2

22

22

22

2

)2

AC

(2CP)4BC

4CP4BC

CP

QCBC =

+

+

+

+

4.4. Circles and Tangents

Find the length of the tangent drawn from a point whose distance from the centre of

is 5cm and radius of circle is 3cm.

theorem

2. Length of tangent to a circle from a point P, which is 25cm away from the centre is 24cm. What is the radius of circle?

4cmPTLength

49OT

765625OT

OT2425

PTOT

PythagorasBy

:OTPIn

OTcm,25OP

2

22

22

=

=

-=

+=

+=

D

=

OP

3. PQ is a chord of length 8cm to a circle of radius 5cm. The tangents at P and Q intersect at a point T. Find the length of the tangent TP.

3

16=TP(2from

3

16y

32,6y

4+y=y)+(3(1),from

+(TR=OT(2),from

OPTP=OTOPT,In

+TR=TPPRT,In

..........y+3=

RT+OR=OT

cm3=OR

45=OR

+OR=OPORP,In

4=QR=PR

OTy,=TRLet

2

2

222

22

22

22

222

22

÷ø

öçè

æ

=

=

-D

D

-

D

^

4. O is the centre of the circle with radius 5cm. T is a point such that OT=13cm and OT intersects circletangent to circle at

=24x x144

x)(12

ATAET,In

= APLet

=ET

13

OT

OP

2

2

2

2

2

=

-+

=-

=D

=

=

=

AB

X std Mathematics Made Easy

https://winglishcoachings.weebly.com/

ength of tangent to a circle from a point P, which is 25cm away from the centre is 24cm. What is the radius of circle?

76

OT

PT

theoremPythagoras

cm24OT

2

2

=

PQ is a chord of length 8cm to a circle of 5cm. The tangents at P and Q

intersect at a point T. Find the length of

3

204+

3

16

5+

OP+)PR+

OP

(2)...PR

(1).............

RT

PR

cm4

PQ.ofbisector r

2

22

22

2

2

2

2

^

TP

TR

O is the centre of the circle with radius 5cm. T is a point such that OT=13cm and OT intersects circle at E, if AB is the tangent to circle at E, find length of AB.

3

20

3

102

3

10x24x144=

8+x

AE

x12=TAx = AE=

8=513=

125

13OT5,OE

22

22

22

22

=Þ´=

=Þ-

=

+=

-

=Þ+=

+=

===

AB

ET

PTPT

PTOP

Page 9: ÐCED...Mythila Publishers, Puduvayal Download Question Bank From from the base of a lamp post at a speed of 1.2m/sec. If the lamp post is 3.6m above the ground, find the length of

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5. i) In fig, ΔABC is circumscribing a circle. Find the length of BC.

cm10=6+4=

CL+BL=BC

cm6=3-9=

AM- AC=

CM=CL

cm4=BL=BN

cm3= AM=AN

pointexternal samefromTangents

ii) A circle is inscribed in Δsides 8cm, 10cm, 12cm find AD, BE and CF

cm, CF = cm, BE = AD

x = = x +

y = = y +

z = + z =

= x + y + z

= x + y + z

.....z + x =

.....y + z =

....x + y =

57is

7125(3)from

83(2)from

1512(1),from

15

30)(2Adding

(3)10

(2)8

.(1)12

Þ

Þ

Þ

6. ΔLMN is a right angled triangle with L=90°. A circle is inscribed in it.

6cm, 8cm. Find the radius of the circle.

24824

106868

2

1

2

1

NOLofArea

of AreaMLN ofArea

cm10

68

MNLMN,In22

222

=Þ=

++=´

´+´=´´

D

D=D

=

+=

+=D

rr

r

NLrLMLNML

MN

LNML

7. If radii of two concentric circles are 4cm and 5cm, find length of the chord of one circle which is a tangent to other circle.

cm632BC

2ABBC

cm3 AB

AB45

ABOAOB

BC.OAOAB,In

cm;5OBcm,4OA

222

222

=´=

=

=

+=

+=

^D

==

8. In two concentric circles, a chord of length 16cm of larger circle becomes a tangent to the smaller circle whose radius is 6cm. find the radius of the larger circle.

81 X std Mathematics Made Easy

Download Question Bank From https://winglishcoachings.weebly.com/

ABC is circumscribing a circle.

equalpoint

ΔABC having 12cm find AD, BE and CF

cm cm, CF = 3

7

5

3

LMN is a right angled triangle with L=90°. A circle is inscribed in it. Lengths of the sides containing the right angle are

8cm. Find the radius of the circle.

2

MONofArea

´+´

D

rMNr

centric circles are 4cm he chord of one

circle which is a tangent to other circle.

In two concentric circles, a chord of length 16cm of larger circle becomes a tangent to the smaller circle whose radius is 6cm. find the radius of the larger circle.

10Radiuscirclelarger

10OB

6O

OTOB

AB.OTOBT,In

TBAT

22

2

=

=

=

=

^D

=

B

9. Two circles with centres O and O of radii 3cm and 4cm, intersect at points P and Q, such that OP and O P are tangents to two circles. Find length of common chord PQ.

(PR)2=PQ

16=PR

32=x 10

-(5-9=x-16

-(5-9=PRRP,O'In

x-16=PRORP,In

OQR~OPR,

OQO'~OPO'

x =OR

beRLet

(tangentPO'OP

(radiuscm3=PO'

(radiuscm4=OP

2

2

2

22

çè

æ-

Þ

D

D

DD

DD

Þ

^

10. In fig O is the centre of a circle. PQ is a chord and the tangent PR at P makes an

angle of 50° with PQ. Find

o

oo

o

o

oo

100=POQ

40180=

OPQ180=POQ

40=OQP=OPQ

of(Radii OQ=OP

40=5090=OPQ

Ð

--

Ð-Ð

ÐÐ

11. PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that ÐPOR=120°. Find Ð

oo

o

o

o

6090OPQ

(Radius90OQP

pair)(linear 60POQ

120PORGiven

=-=Ð

12. A tangent ST to a circle touches it at B. AB is a chord such that 65°.Find ÐAOB, where “O” is the centre of circle.

X std Mathematics Made Easy

https://winglishcoachings.weebly.com/

cm10

cm10

8

BTOT

AB.

cm6OTcm,8TB

22

22

+

+

==

Two circles with centres O and O of radii 3cm and 4cm, intersect at points P and Q, such that OP and O P are tangents to two circles. Find length of common chord PQ.

cm4.8=(2.4)2=(PR)

4.25

16

5

16x

x)-

x)-

90=ORPOQR

OQO'

x-5=RO'

thatsuchPQofpointabe

?)areradius&(tangent

circle)2ndof(radius

circle)1stof(radius

2

2

2

2

O

=÷ø

ö

è

æ

ÐÞ

In fig O is the centre of a circle. PQ is a chord and the tangent PR at P makes an

angle of 50° with PQ. Find ÐPOQ.

o

o

40

OQPOPQ

isosceles)isOPQ(

equal)arecircleaof

tangent)(Radius40

Ð-

D

^

PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that

ÐOPQ.

o30

tangent)

pair)

^

A tangent ST to a circle touches it at B. AB is a

ÐABT = AOB, where “O”

is the centre of circle.

Page 10: ÐCED...Mythila Publishers, Puduvayal Download Question Bank From from the base of a lamp post at a speed of 1.2m/sec. If the lamp post is 3.6m above the ground, find the length of

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130=)2(65=

APB2=AOB

ncecircumfereon

ndedanglesubte2

centreat

subtendedAngle

65=APB=TBA

altemateinAngles

oo

o

ÐÐ

îíì

=þýü

ÐÐ

13. Show that in a triangle medians are concurrent Or Verify CEVA’S theoem

Ceva�s Theorem :

Let ABC be a triangle and let D,E,F be points on lines BC, CA, AB. Then the cevians AD, BE, CF are concurrent if and only if

1FB

AF

EA

CE

DC

BD=´´

Medians :

Line segments joining each vertex toof the corresponding opposite sides.Thus medians are the cevians where D, E, F are midpoints of BC, CA and AB.

1FB

AF

EA

CE

DC

BD

(3)and(2)(1),g Multiplyin

(3)..............1FB

AF

FBAF

ABofmidpointaofpointmidaisF

(2)..............1EA

CE

EACE

CAofmidpointaofpointmidaisE

(1)..............1DC

BD

DCBD

BCofmidpointaofpointmidaisD

=´´

=

=

=

=

=

=

Ceva’s theorem is satisfied.

Hence the Medians are concurrent

14. Suppose AB, AC ,BC have lengths 13, 14

and 15. If 8

5,

3

2==

EA

CE

FB

AFFind BD and DC.

12BD3,DC

DCDC415

1.....(DC4

18

5

3

2

DC

BD

,1

:theoremsCeva'Using

==

+=

+=

=

=´´

=´´

DCBDBC

BD

FB

AF

EA

CE

DC

BD

82 X std Mathematics Made Easy

Download Question Bank From https://winglishcoachings.weebly.com/

nce

nded

segmentaltemate

Show that in a triangle medians are Or Verify CEVA’S theoem

and let D,E,F be points on lines BC, CA, AB. Then the cevians AD, BE,

ine segments joining each vertex to midpoint of the corresponding opposite sides.Thus medians are the cevians where D, E, F

AB

CA

BC

Suppose AB, AC ,BC have lengths 13, 14

Find BD and DC.

12

DC

)1

15. Show that the angle bisectors of a triangle are concurrent.

ofbisectorsAngle

(4)Subs.in

theorem, ABTBy

bisectorsareCDBF, AE,If

FA

CF

EC

BE

DB

AD

FA

CF

EC

BE

DB

AD

(3)(2),(1),Muliplying

.......(3)FA

CF

OA

OC

bisector angleCOA,In

.......(2)EC

BE

OC

OB

bisector angleBOC,In

.......(DB

AD

OB

OA

bisector angleAOB,In

D\

´

=

=

´´

´´

=

=

=

BC

AB

AC

AB

CB

CA

DB

AD

FA

CF,

AC

AB

EC

BE

Δ

Δ

Δ

16. An artist has created a triangular stained glass window and has one strip of small length left before completing the window. She needs to figure out out portion based on the lengths of the other sides as shown in the figure.By Ceva's theorem, Cevians AD, BEintersect at exactly one point if and only if

cm2=FB

×30=60

×10=5×4×3

×DC= AF×CE×BD

1FB

AF

EA

CE

DC

BD=´´

17. In figure,ABCB=90°,BC=3cm and AC such that AD =1cm and E is the midpoint of AB. Join D and E and extend DE to meet CB at F. Find BF.

AC34

BC ABAC

theorem,PythagorasBy

BFBCFBFC

DA2,EBAE

:assumptionBy

22

222

=Þ+=

+=

=+=

==

X std Mathematics Made Easy

https://winglishcoachings.weebly.com/

Show that the angle bisectors of a triangle

.concurrentare

1

,,ofbisectors

......(4)1FA

CF

OA

OC

OC

OB

OB

OA

FA

CF

sidesbothon(3)

.......(3)

OFisCOAofbisector

.......(2)

OEisBOCofbisector

)1.......(

ODisAOBofbisector

D

=

ÐÐÐ

=

´´=

Ð

Ð

Ð

CB

CA

,BC

AB

FA

CF

CBA

n artist has created a triangular stained glass window and has one strip of small length left before completing the window. She needs to figure out the length of left out portion based on the lengths of the other sides as shown in the figure.By Ceva's theorem, Cevians AD, BE,CF intersect at exactly one point if and only if

cm

FB×

FB×3×

FB×EA×

is a triangle with B=90°,BC=3cm and AB=4cm. D is point on AC such that AD =1cm and E is the mid-point of AB. Join D and E and extend DE to meet CB at F. Find BF.

.5

theorem,

3BF

1

=

+

=

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41 – 5

– AD ACCD

=Þ=

=

CD

BF3BF4BF

11

4

32

2

1RB

AR

QA

CQ

PC

BP

:theoremMenelaus'By

Þ+=

=´+

´

=´´

BF

BF

18. In ΔABC, with B=90°, BC = 6cm and AB = 8cm, D is a point on AC such that AD = 2cm and E is the mid-point of AB. Join D to E and extend it to meet at F. find BF.

8.2 – 10

– AD ACCD

10AC68

BC ABAC

theorem,PythagorasBy

6BFBCFBFC

2DA4,EBAE

:assumptionBy

22

222

=Þ=

=

=Þ+=

+=

+=+=

===

CD

2BF12

8

64

4

:theoremMenelaus'By

=Þ=´+

´BF

BF

19.

the following way, BP=2m, C

Menelaus Theorem :

A necessary and sufficient condition for points P, Q, R of sides BC, CA, AB of

collinear is ......(1RB

AR

QA

CQ

PC

BP=´´

straightsameaonlieRQ,P,trees

1

2

10

5

3

6

2

(1)invaluestheseng Substituti

RBandm5QAm,6PC

10RAm,3CQm,2BP

\

=

´´=

==

===

83 X std Mathematics Made Easy

Download Question Bank From https://winglishcoachings.weebly.com/

1BF =

ABC, with B=90°, BC = 6cm and AB = AC such that AD = point of AB. Join D

to E and extend it to meet at F. find BF.

In a garden containing several trees, three particular trees P,Q,R are located in

Q=3m, RA=10 m, PC=6 m, QA=5 m, RB=2 m, where A,B,C are points such that P lies on BC, Q lies on AC and R lies on AB. Check whether the trees P,Q,R lie on a same straight line.

A necessary and sufficient condition for points sides BC, CA, AB of DABC to be

)1......(

line.straight

m2RB

m,10

=

5.1. Area of a Triangle

1. Find area of triangle formed by the points

=D2

1triangleofArea

i) 5) –(–3,&6)(–4,1), –(1,

Taking A, B, C in anti clockwise direction

unitssq.24

19][292

1

(4 – 3)20{(62

1

61

41

2

1

=

+=

++=

îíì

-

-=D

ii) (–3,&1) –(–8,4), – (–10,

sq.units11.5

62]85[2

1

(1232)3(50[2

1

54

310

2

1

=

-=

-++=

îíì

--

--=D

iii) (5,-2).&(5,6),(-3,5)

A(–3, 5) , B(5, –2), C(5,6)

unitssq.32

}361{2

1

(2525)30{(62

1

25

53

2

1

=

+=

-++=

îíì

-

-=D

2. Vertices of given triangles taken in order and areas are provided Find value of ‘

i) 2)(6,8),,(0),(0, p Area

2

48[0 – 0]2p[0

280

60

2

1

p –

p

+++

îíì

ii) 2) –(5,6),(5,),( p, p Area

(3p – 10) – (11p

30(5p – 5p)10 – (6p

26

55

2

1

p

p

++

îíì

-

iii) 4)(7,,2) –,(,2)(–1, p Area

X std Mathematics Made Easy

https://winglishcoachings.weebly.com/

5.1. Area of a Triangle

Find area of triangle formed by the points

þýü

îíì

1321

1321

yyyy

xxxx

5)

Taking A, B, C in anti clockwise direction

5)] – 18 – (4

15

13

þýü

--

-

5) –(–3,

10)]40(12

41

108

++

þýü

--

--

2), C(5,6)

18)}10(25

56

35

--

þýü-

Vertices of given triangles taken in order and areas are provided Find value of ‘p’.

Area =20Sq.units

44882

4048

400]48

200

0

20Sq.units

p p

p –

=Þ=

=

=+

=þýü

=D

Area =32Sq.units

131048

6430)(3p

642p) – 30

232

Sq.units23

p p

p

p

=Þ=

=+

=

=þýü

=D

Area =22Sq.units

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5

44342

444)}14(214)4{(2

222722

171

2

1

22Sq.units

=

=+

=---++

=þýü

îíì

-

--

=D

k

k

kk

k

3. Determine if the points are collinear

ofAreaCondition,tyCollineari D

i) 8)(–8,,6)(–5,,,2

÷ø

öçè

æ3

1-

collinear.arepoints3

0

)]67(67[2

1

481524403[( [2

1

3863

85

2

1 21

21

\

=

---=

------=

þýü

îíì --

=D--

( )

ii) )(,)(),( bc, a ab, c ca, b +++

collineararepoints3

0

-bc-b-=

2

1

c+cb+b+ba+a+ac

c(c-c)+b(b-c)+c(b+b)+b(a+a)+(c

2222

\

=

þýü

îíì

++++=D

= a

cbbaaccb

acba

iii) P( 1.5, 3), Q(6, 2), R( 3,4)

collineararepoints3

0

18018[2

1

)}6618()9243{(2

1

3423

5.13651

2

1

\

=

-=

-+--+=

ü

îíì -

=D

4. Find the value of ‘a’ for which given points are collinear. (i) 3) –(6,,a)(4,3),(2,

Since given points are collinear

0=a

0=4a-

0=6)+(6a-6)+(2a

0=6)-6a+(12-18)+12-(2a

0333

2642

2

1

0

=þýü

îíì

-

-

=D

a

(ii) a, – 4(– ,2a)1,(–a2a), – 2(a, +

84 X std Mathematics Made Easy

Download Question Bank From https://winglishcoachings.weebly.com/

44

22Sq.units

the points are collinear

0=D

)]4-

c ab-a-ac-

b)+a(a-a)+c(c

22

’ for which given

iven points are collinear,

)2 – 6a, a

+12-(2a

6222

41

2

1

îíì

-

-+-

aa

aa

5. i) If (Q4), – P(–1, b, c

and if 42b+c = , then find thevalues of

and(1)Solving

n,informatiogiven

4()20(

14

51

2

1

+-----

îíì

--

-

bbc

c

b

ii) If the points A(-3,

collinear and if a+b =

solving,By

that,Given

4b+(9a-36)+5a-3b(-

59

43

2

1

îíì

-

-

b

a

6. Find area of quadrilateral w

îíì

=21

21

2

1Area

yy

xx

2)(2,4), –(–8,2), –(–9,(i)

unitssq.35

)]12(58[2

1

)422436([2

1

42

89

2

1

–(–8,B2), –(–9,A

=

--=

-++=

îíì

--

--=D

2) –(–1,6),(–8,0),(ii)(–9,

unitssq.34

)]35(33[2

1

612270[[2

1

06

98

2

1

C0),(–9,B6),(–8,A

=

--=

-++=

îíì --

=D

(

X std Mathematics Made Easy

https://winglishcoachings.weebly.com/

0=a

0=4a-

0=6a-6-6+2a

0=6a)+(6-6)+(2a

0=6)-6a+(12-18)+

02226

4

0

=þýü

--

-

=D

aa

aa

1) –R(5,&)b, c are collinear

, then find thevalues of b,c.

.2c,3b,(2)and

......(2)4c2b

)1.....(7

0)15

04

1

0

==

=+

=-

=++

=þýü

-

-

=D

cb

c

C(4,-5),b)(a,B,9)A(-3, are

1a+b = , then find a and b.

1-=b2,=asolving,

....(2)1ba

....(1)3b+2a-

0=15)+4b

09

3

=+

=

=þýü-

Find area of quadrilateral with vertices

þýü

143

143

yyy

xxx

3) –(1,,2)

)186416()

223

921

(2,2)D3), –(1,C4),

---

þýü

--

-

3) –(–6,,

)]163054()6

623

813

2) –(–1,D3), –(–6,C

+++--

þýü

--

---

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iii) 3)4,(– &(–5,12)11),(5,6),(8,

unitssq.79

]49109[2

1

5530()24156088([2

1

312116

4558

2

1

=

+=

----+=

îíì --

=D

7. Find the value of k, if the area of a quadrilateral is 28 sq. units, whose vertices are –(3,),(–3,2), –(–4, k

21

– (3k – 4k) – (11

– 4 – 3k(6 – 4) – 96(–4k

322

42334

2

1

k

+++

îíì

---

---

8. 4)R(9.5,,4)Q(13.5,,P(11,7) are

sides AB, BC and AC of ΔABC. Find the coordinates of the vertices A, B and C. of ΔABC, compare this with area of ΔPQR.

),A(

vertex APQR,4ogramofparallel

are),(),,(),,P(If

321321

th

332211

yyyxxx

yxRyxQyx

-+-+

1)(12,C=

411,9.5+(13.5=CVertex

7)(15,B=

4+79.5,13.5+(11=BVertex

7)(7, A=

4+713.5,9.5+(11Vertex A

-

-

-=

unitssquare24=ABC Area

196][1482

1=

2

1=

77

157

2

1

84)+15+[(49

D

-

îíì

=D

-

(12,C7),(15,B7),(7, AtriangleofArea

6=176.5][164.52

1=

=

447

5.913.511

2

1

(94.566.5)+54+[(442

1

-

îíì

=

-

4)R(9.5,,4)Q(13.5,,P(11,7)ΔofArea

triangleof(Area4= ABCtriangleofArea

85 X std Mathematics Made Easy

Download Question Bank From https://winglishcoachings.weebly.com/

)244855

6

8

+-

þýü

, if the area of a quadrilateral is 28 sq. units, whose

3)(2,and2) – .

5

567

5610) –

5612) –

282

4

28

– k

k

=

=

=

=

=þýü

-

=D

midpoints of

sides AB, BC and AC of ΔABC. Find the coordinates of the vertices A, B and C. , area

re this with area of ΔPQR.

isvertex

vertices,3

7)

4)

4)

-

-

-

71

712

7)]+84+(105

þýü

1)(12,

sq.units6

7

115

44)]+38+(94.5

þýü

PQR)triangle

9. Quadrilateral swimming pool is surrounded by concrete patio. Find area of patio.

units.square122=

90-212=

ABCDofArea

unitssquare90

(-90)]-[(90)2

1=

+12+42+[(62

1=

25

63

2

1Area

unitssq.212=

212][2122

1=

(-212)]-[(212)2

1=

36+80+[(162

1=

48

84

2

1Area

-=

=

îíì

--

-=

+

îíì

--

-=

patioofArea

EFGHralquadrilateofArea

ABCDralquadrilateofArea

10. In the figure, find the area of

units.sq.3=

(-0.5)]-[(5.5)2

1=

3)+4.5+[(-22

1=

1)(1.5,E3)(-2,F

units.square3.75=

29.5)+(-222

1=

(-29.5)]-[(-22)2

1=

6)-13.5-[(-2.52

1=

D

FEDtriangleii)

AGFtrianglei)

units.square13.875=

12)+(15.752

1=

(-12)]-[(15.75)2

1=

0.75+2+[(42

1=

BCEGralquadrilateiii)

X std Mathematics Made Easy

https://winglishcoachings.weebly.com/

uadrilateral swimming pool is surrounded by concrete patio. Find area of patio.

units.

EFGHofArea

units

12)]-42-6-(-30-30)+

547

363

(-212)]

24)]-100-24-(-64-80)+36

8610

4106

-

þýü

-

--

þýü

-

-

EFGH

ABCD

In the figure, find the area of

6)]-1+(4.5-3)

3)(1,Dand

units.

(-29.5)]

15)]-1-(-13.5-6)

units.

(-12)]

2)]-4.5-1.5-(-4-9)+

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11. Floor of a hall is covered with identical titles in shapes of triangles. One triangle has vertices at 1) –(–1,,2)3,(–

floor of the hall is completely covered by 110 titles, find the area of the floor.

sq.units660

6110floor ofArea

unitssq.6

)]9(3[2

1

612()223[(2

1

212

113

2

1 Areatile

=

´=

=

--=

----+-=

îíì

-

--=

12. Given diagram is plan for constructing a new parking lot at a camus. It is estimated that it would cost Rs1300 per sq. ft. What will be total cost for making parking lot?

Rs.20800

130016costtotal

Rs.1300ftsq.rate/

units.sq.16

53} – {852

1

(10 – 2)2845{(102

1

952

452

2

1Area

D(1,7),C(4,9),B(5,5) A(2,2),

=

´=

=

=

=

+++=

îíì

=

verticeswithralquadrilateislot Parking

13.4)C(7,,B(1,6),4)5,A( --- has to be painted.

If one bucket of paint covers 6 sq. ft., how many buckets of paint will be required to paint the whole glass

buckets10=6

60ftsq.60For

1bucketftsq.6For

feetsquare60=

(120)2

1=

58]-[-622

1=

(58)]-[-622

1=

+(-4-28)-4-[(-302

1=

464

715

2

1

=

=

îíì

--

-=D

86 X std Mathematics Made Easy

Download Question Bank From https://winglishcoachings.weebly.com/

loor of a hall is covered with identical titles in shapes of triangles. One triangle

(1,2),1) . If the

floor of the hall is completely covered by 110 titles, find the area of the floor.

)6

2

1

þýü-

plan for constructing a new parking lot at a camus. It is estimated

would cost Rs1300 per sq. ft. What will be total cost for making parking lot?

14)}920(10

27

21

+++

þýü

vertices

A triangular shaped glass with vertices at has to be painted.

If one bucket of paint covers 6 sq. ft., how many buckets of paint will be required to

20)]+42+

44

5

þýü

-

-

5.2.Slope of a Straight line

1. What is the slope of line whose inclination

an=

ithlinetheofSlope

qtm

w

(i) 90° (ii) 0°

m

m

ii)

=m

90°tan=

?tan=m

90°=i)

¥

qq

2. What is inclination of line whose slope is

(i) 0 (ii) 1

tan

ii)

0°.=nInclinatio

0=tan

0=mi)

q

3. Find the slope of a line joining the points

=

joining linetheofSlope

2

2

x

ym

-

-

2

1

)6(3

12m

),(),( 2211

=

---

--=

m

yxyx

2)3,(– 1)6,(– i)

26

3

m

),(),(

31

72

21

73

2211

-=

+

--=

÷ø

öçè

æ÷ø

öçè

æ

m

yxyx

7

3,

7

2,

2

1,

3

1-iii)

q

q

q

sin2

cos2

sinsinθ

cos(cos

(),(

sin(&cosθv)(sinθ

211

Þ-

=

--

--=

--

m

xyx

),

4. The line through the points

has slope2

1-. Find the value of

1126

2

1

29

3

2

1

Þ-=-

-=

+

-

-=

a

a

m

X std Mathematics Made Easy

https://winglishcoachings.weebly.com/

5.2.Slope of a Straight line

What is the slope of line whose inclination ninclinatioith

q

q

(ii) 0° (iii) 30°

3/1

30tan

tanm

30iii)

0=m

0°tan=

?tan=m

0°=

o

o

=

=

=

=

q

qq

What is inclination of line whose slope is

(ii) 1 (iii) 3

°60

3tan

3miii)

45°=

1=tan

1=mii)

=

=

=

q

q

q

q

Find the slope of a line joining the points

),(),(joining

1

1

2211

x

y

yxyx

-

-

undefined,0

16

1414

106m

),(),() 2211

-=

-

---=

m

yxyx

6)–(14,10)(14,ii)

5

1

50

50m

),(),( 2211

=

-

--=

m

yxyx

0)(0,)5(5,ii)

q

q

cot

sinθ

)cos

),

cosθsinθ

22

-=Þ m

y

),

The line through the points 3)(9,,)(-2, a

. Find the value of a.

2

17=Þa

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Mythila Publishers, Puduvayal 87 X std Mathematics Made Easy

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1.larPerpendicuarelinestwoIf

Parallel,arelinestwoIf

LinesParallel &lar Perpendicu

21

21

-=

=

mm

mm

5. The line r passes through 2)(–2, and (5,8)

and line s passes through (–8,7) and 0)(–2,

Is the line perpendicular to s?

other,eachtor arelinesThese

1=

6

7

7

6mm

6

7=m

)8(2

70=mslope

2,0)7)((-8,b)

7

6=m

2)(5

28=mslope

8)2)(5,2,a)(

21

2

2

1

1

^

-

÷ø

öçè

æ -÷ø

öçè

æ=

-

---

-

--

-

-

-

6. The line p passes through the points (12,4)2), –(3, and the line passes through

the points .2)(12,2), –(6, Is parallel to q?

othereachto||arelinesThese

mm

3

2m

6

4=

612

)2(2=mslope

.2)(12,2), –(6,b)

3

2=m

9

6=

312

)2(4=mslope

(12,4)2), –a)(3,

21

2

2

1

1

=

=

-

--

-

--

7. What is the slope of a line perpendicular to line joining A(5,1) and P where P is mid point of the segment joining 6,4)( –(4,2)

.3m

1m3

1

1m.mrlinesFor

3

1

5-1-

1-3=m

.

3)1,P(=

2

4+2,

2

(-6)+4=

22=midpoint

2

2

21

1

2121

=

^

-=÷ø

öçè

æ-

-=^

-=

-

÷ø

öçè

æ

÷ø

öçè

æ

3)P(-1,and1) A(5,tor ofslope

3)P(-1,and1) A(5,ofslope

4)(-6,and(4,2)point -Mid

+ yy ,

+ xx

8. The line through the points 8)(4,&6)(-2, is

perpendicular to the line through the points 24),(&12)(8, x . Find the value of x.

4=x

8)(=4

1-=8

12

3

1

1mm

other,eachtor arelinesThese

8

12=

8

1224=mslope

24),12)((8,b)

3

1=m

2+4

2=

2)(4

68=mslope

8)6)(4,2,a)(

21

2

1

1

--

÷ø

öçè

æ

-÷ø

öçè

æ

-=

^

-

--

-

-

x

x

x

x -

-

x

9. Show that the given points are collinear. a) 5).(12,2),(7,4), – 3,(–

collineararepointsiven

BCofSlope ABofSlope

5

3=

7-12

2-5=BCofSlope

5

3=

10

6=

(-3)-7

(-4)-2= ABofSlope

G\

=

b) 2)(2,1), –(6,5),(–2,

collineararepointsiven

BCofSlope ABofSlope

2

1=

6-2

12=BCofSlope

2

1=

8

4=

26

51= ABofSlope

G\

=

-

+

--

+

--

10. If the three points (1,–3)3),,(1), –(3, a are

collinear, find the value of a.

713

4

ACofSlope ABofSlope

collineararepointsiven

12

2=

31

13= ACofSlope

3

4=

3a

(-1)-3= ABofSlope

=Þ=-

=\

=

-

+-

-

-

aa

G

a

11. Without using Pythagoras theorem, show that the points 7)–(4, ,3)–(2, ,(1,-4) form a

right angled triangle.

2=

2-4

(-3-7-=BCofSlope

1=

1-2

(-4-3-= ABofSlope

7) –C(4,,3) –B(2,,A(1,-4)

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1-=

(-1)1=CAofSlope ABofSlope

1-=

1-4

(-4-7-=CAofSlope

´

Given points are vertices of right triangle To check this with pythagoras theorem

2

22

222

222

222

AC20

182BC+ AB

18=1)(4+4)+7(=CA

20=2)(4+3)+7(=BC

2=1)(2+4)+3(=AB

==

+=

--

--

--

12. Show that 14)(3,N &12)(9,M5),(0,L form a

right angled triangle and check whether they satisfies Pythagoras theorem

1-=

(3)3

1-=NLofSlopeMN ofSlope

3=

0-3

5-14=NLofSlope

3

1-=

9-3

12-14=MN ofSlope

9

7=

0-9

5-12=LMofSlope

÷ø

öçè

æ´

Given points are vertices of right triangle To check this with pythagoras theorem

222

22

222

222

222

LM=NL+MN

130=90+40=NL+MN

90=9+3=NL

40=2+6=MN

130=7+9=LM

13. 1)(2,D1),C(5,2), –(6,B2), –A(1, be four point

i) Find slope of line segment a) AB b) CD

0

52

11CDofSlope

0

16

22 ABofSlope

=

-

-=

=

-

+-=

ii) Find slope of line segment a) BC b)AD

3

12

21 ADofSlope

3

65

21BCofSlope

=

-

+=

-=

-

+=

iii) What do you deduce from your answer

CD||AB

CDofslope ABofslope

\

=

parallelnotareCD,AB

ADofslopeBCofslope

\

¹

Quadrilateral ABCD is a trapezium

14. Show that the points form parallelogram 5)5,(– D&2.5) – (2.5,C4), – (10,B3.5),(2.5,A

1

2.5-5-

2.5+5CDofSlope

1

2.5-10

3.5-4- ABofSlope

-=

=

-=

=

5

1

3-7

2.5-5- ADofSlope

5

1

10-2.5

4+2.5-BCofSlope

-=

=

-=

=

CD||AB

CDofslope ABofslope

\

=

parallelareCD,AB

ADofslopeBCofslope

\

=

Quadrilateral ABCD is a parallelogram

15. If )(D&3) –(1,C3), –(–2,B2),(2,A x, y form a

parallelogram, find the value of x and y.

1-x

3+y

2.5-5-

2.5+5CDofSlope

4

5

2-2-

2-3- ABofSlope

=

=

-=

=

2-x

2-y ADofSlope

0

2+1

3+3-BCofSlope =

=

=

Since ABCD is a parallelogram

5x17=4(2)5x

(1)in2ysub

2y0=2y

02x

2y

ADofslopeBCofslope

(1)-----17=4y5x

4

5

1x

3+y

CDofslope ABofslope

=Þ-

=

=Þ-

=-

-

=

-

-=

-

=

16. Let 7) –D(7,&7) –(5,C4), – (9,B4), – (3,A

Show that ABCD is a trapezium.

0

57

7+7-CDofSlope

0

3-9

4+4- ABofSlope

=

-=

=

=

4

3

3-7

4+7- ADofSlope

4

3

95

4+7-BCofSlope

=

=

-=

-=

CD||AB

CDofslope ABofslope

\

=

parallelnotareCD,AB

ADofslopeBCofslope

\

¹

Quadrilateral ABCD is a trapezium