Explain why the triangles are similar and then find the value of x.
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1 ) From figure ( i )
In ∆PQR , ∆LST
<Q = <T ( given )
<R = <S = 90°
Therefore ,
∆PRQ ~ ∆LST ( A.A similarity )
Then ,
QR/ST = PQ/TL
=> 3/4.5 = 5/x
=> 3x = 5 × 4.5
=> x = ( 5 × 4.5 )/3
x = 7.5
_______________________
2 ) from figure ( ii ) ,
In ∆ABC and ∆PQR
<ABC = <PQC = 70°
( corresponding angles )
<C = <C ( common angle )
Therefore ,
∆ABC ~ ∆PQC ( A.A similarity )
then
AB/PQ = BC/QC
=> 5/x = ( BQ + QC )/QC
=> 5/x = ( 3 + 3 )/3
=> 5/x = 2
x = 5/2
x = 2.5
_____________________
3 ) from figure ( iii ) ,
In ∆ABC and ∆DEC
<A = <E ( given )
<BCA = <ECD
( vertically opposite angles )
Therefore ,
∆ABC ~ ∆DEC ( A.A similarity )
then ,
BC/CD = AB/DE
=> 22/x = 24/14
=> 22/x = 12/7
=> x/22 = 7/12
=> x = ( 7 × 22 )/12
x = 77/6 = 12 5/6
______________________
4 ) From figure ( iv )
Given AB//ST
then ,
RA/RS = RB/RT----( 1 )
( Thales theorem )
<RAB = <RST-----( 2 )
( corresponding angles )
From ( 1 ) and ( 2 ),
∆RAB ~ ∆RST
[ SAS similarity ]
RA/RS = RB/RT = AB/ST
RA/RS = AB/ST
6/8 = 9/x
=> x = 12
______________________
5 ) FROM figure ( v ),
In ∆PQR , MN // QR
PM/PQ = PN/PR ( Thales theorem )
<PNM = <PRQ ( corresponding angles )
∆PQR ~ ∆PMN
[ SAS similarity ]
PM/PQ = PN/PR = MN/QR
=> PN/PR = MN/QR
4/( 4 + x ) = 5/15
=> 4/( 4 + x ) = 1/3
x = 8
_____________________
6 ) from figure ( vi )
In ∆XYZ , AB // ZY
XA/XZ = XB/XY ---( 1 )
[ Thales theorem ]
<XAB = <XZY ----( 2 )
[ corresponding angles ]
∆XZY ~ ∆XAB
[ SAS similarity ]
XA/XZ = AB/ZY
=> XA/( XA + AZ ) = AB/ZY
=> x/( x + 7.5 ) = 12/18
=> x/( x + 7.5 ) = 2/3
after simplification ,
x = 15
_________________
7) From figure ( vii )
In ∆ABC , ∆EDC
<ABC = <EDC = 90°
<BCA = <DCE
[ angle of incidence and angle of
reflection are same ]
∆ABC ~ ∆EDC
[ A.A similarity ]
AB/ED = BC/CD
=> 1.6/x = 1.5/15
=> after simplification
x = 16
______________________
8 ) From figure viii
Data not sufficient .
______________________
☺️
In ∆PQR , ∆LST
<Q = <T ( given )
<R = <S = 90°
Therefore ,
∆PRQ ~ ∆LST ( A.A similarity )
Then ,
QR/ST = PQ/TL
=> 3/4.5 = 5/x
=> 3x = 5 × 4.5
=> x = ( 5 × 4.5 )/3
x = 7.5
_______________________
2 ) from figure ( ii ) ,
In ∆ABC and ∆PQR
<ABC = <PQC = 70°
( corresponding angles )
<C = <C ( common angle )
Therefore ,
∆ABC ~ ∆PQC ( A.A similarity )
then
AB/PQ = BC/QC
=> 5/x = ( BQ + QC )/QC
=> 5/x = ( 3 + 3 )/3
=> 5/x = 2
x = 5/2
x = 2.5
_____________________
3 ) from figure ( iii ) ,
In ∆ABC and ∆DEC
<A = <E ( given )
<BCA = <ECD
( vertically opposite angles )
Therefore ,
∆ABC ~ ∆DEC ( A.A similarity )
then ,
BC/CD = AB/DE
=> 22/x = 24/14
=> 22/x = 12/7
=> x/22 = 7/12
=> x = ( 7 × 22 )/12
x = 77/6 = 12 5/6
______________________
4 ) From figure ( iv )
Given AB//ST
then ,
RA/RS = RB/RT----( 1 )
( Thales theorem )
<RAB = <RST-----( 2 )
( corresponding angles )
From ( 1 ) and ( 2 ),
∆RAB ~ ∆RST
[ SAS similarity ]
RA/RS = RB/RT = AB/ST
RA/RS = AB/ST
6/8 = 9/x
=> x = 12
______________________
5 ) FROM figure ( v ),
In ∆PQR , MN // QR
PM/PQ = PN/PR ( Thales theorem )
<PNM = <PRQ ( corresponding angles )
∆PQR ~ ∆PMN
[ SAS similarity ]
PM/PQ = PN/PR = MN/QR
=> PN/PR = MN/QR
4/( 4 + x ) = 5/15
=> 4/( 4 + x ) = 1/3
x = 8
_____________________
6 ) from figure ( vi )
In ∆XYZ , AB // ZY
XA/XZ = XB/XY ---( 1 )
[ Thales theorem ]
<XAB = <XZY ----( 2 )
[ corresponding angles ]
∆XZY ~ ∆XAB
[ SAS similarity ]
XA/XZ = AB/ZY
=> XA/( XA + AZ ) = AB/ZY
=> x/( x + 7.5 ) = 12/18
=> x/( x + 7.5 ) = 2/3
after simplification ,
x = 15
_________________
7) From figure ( vii )
In ∆ABC , ∆EDC
<ABC = <EDC = 90°
<BCA = <DCE
[ angle of incidence and angle of
reflection are same ]
∆ABC ~ ∆EDC
[ A.A similarity ]
AB/ED = BC/CD
=> 1.6/x = 1.5/15
=> after simplification
x = 16
______________________
8 ) From figure viii
Data not sufficient .
______________________
☺️
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