Math, asked by Prakhar2908, 1 year ago

Pls answer Q.51

Need Qualitative Answers with good explanation.

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Answered by Anonymous
11

To prove Δ AEF and Δ ABC similar :


∠EAF = ∠BAC [ Common ]

∠AEF = ∠ABC [ corresponding ∠s ]

Δ AEF ≈ Δ ABC [ A.A criteria ]


So : AE : AB = EF : BC [ corresponding ∠s of Δ ]


Finding EF


So : AE / AB = EF / BC

==> AE / AB  = EF / 15 cm

==> AB / AE = 15 cm / EF [ Invertendo ]

==> AB / AE - 1 = 15 / EF - 1

==> ( AB - AE ) / AE = ( 15 - EF ) / EF

==> EB / AE = ( 15 - EF ) / EF

==> 3 / 2 = ( 15 - EF ) / EF

==> 3 EF = 30 - 2 EF

==> 5 EF = 30

==> EF = 30 / 5

==> EF = 6


EF = 6 cm ....................(1)


Similarly : Δ CFG ≈ Δ ACD


Finding FG


FG : AD = CG : CD

Given : AE : EB = 2 : 3

So : DG : CF = 2 : 3 ..............................(2)


==> We know that : -

AD / FG = DC / CG [ corresponding parts of similar Δ ]

==> AD / FG - 1 = DC / CG - 1

==> ( AD - FG ) / FG = ( DC - CG ) / CG

AD = BC = 15 cm

==> ( 15 - FG ) / FG = DG / CG

==> ( 15 - FG ) / FG = 2 / 3 [ See (2) ]

==> 45 - 3 FG = 2 FG

==> 5 FG = 45

==> FG = 45/5

==> FG = 9


FG = 9 cm


Difference between EF and FG


                  FG = 9 cm

       ( - )      EF =  6 cm

               -----------------------

          FG - EF = 3 cm


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Hope it helps !

______________________________________________________________________

Answered by Alfreddegreat
6

Answer:


Step-by-step explanation:To prove Δ AEF and Δ ABC similar :



∠EAF = ∠BAC [ Common ]


∠AEF = ∠ABC [ corresponding ∠s ]


Δ AEF ≈ Δ ABC [ A.A criteria ]



So : AE : AB = EF : BC [ corresponding ∠s of Δ ]



Finding EF



So : AE / AB = EF / BC


==> AE / AB  = EF / 15 cm


==> AB / AE = 15 cm / EF [ Invertendo ]


==> AB / AE - 1 = 15 / EF - 1


==> ( AB - AE ) / AE = ( 15 - EF ) / EF


==> EB / AE = ( 15 - EF ) / EF


==> 3 / 2 = ( 15 - EF ) / EF


==> 3 EF = 30 - 2 EF


==> 5 EF = 30


==> EF = 30 / 5


==> EF = 6



EF = 6 cm ....................(1)



Similarly : Δ CFG ≈ Δ ACD



Finding FG



FG : AD = CG : CD


Given : AE : EB = 2 : 3


So : DG : CF = 2 : 3 ..............................(2)



==> We know that : -


AD / FG = DC / CG [ corresponding parts of similar Δ ]


==> AD / FG - 1 = DC / CG - 1


==> ( AD - FG ) / FG = ( DC - CG ) / CG


AD = BC = 15 cm


==> ( 15 - FG ) / FG = DG / CG


==> ( 15 - FG ) / FG = 2 / 3 [ See (2) ]


==> 45 - 3 FG = 2 FG


==> 5 FG = 45


==> FG = 45/5


==> FG = 9



FG = 9 cm



Difference between EF and FG



                  FG = 9 cm


       ( - )      EF =  6 cm


               -----------------------


           FG - EF = 3 cm









Hope it helps !


______________________________________________________________________


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