The iconographical subject ”Christ the wine” in Byzantine and Post-byzantine Art
ÐCED...Mythila Publishers, Puduvayal Download Question Bank From from the base of a lamp post at a...
Transcript of ÐCED...Mythila Publishers, Puduvayal Download Question Bank From from the base of a lamp post at a...
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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
=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
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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
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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
=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
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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
=+
=
=
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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
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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
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x.
)3-x
and EC = x − 1 AB and AC.
91
1
6
4
)2
=Þ
=Þ
=Þ
+
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)
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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
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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
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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
=
=Ð
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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
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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
=
-Þ
-
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
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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
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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
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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
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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
==
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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
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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
=Þ
=Þ
^
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
-
=Þ+=
+=
===
AB
ET
PTPT
PTOP
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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
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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
-Ð
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
=Ð
=Ð
=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
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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
=Þ
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.
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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
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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
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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
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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
Mythila Publishers, Puduvayal
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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
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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
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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
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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)
¥
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
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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
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
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)
Mythila Publishers, Puduvayal 88 X std Mathematics Made Easy
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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