Materi : Kesebangunan dan Kekonggruenan
Konsep garis bagi membuat
∆BCE kongruen ∆BFE
∆BCE { a¹ = 10 , t¹ = CE(x) , s¹ = BE }
∆BFE { a² = BF , t² = EF(x) , s² = BE }
s²/s¹ = a²/a¹
BE/BE = BF/10
BF = 1(10)
BF = 10 cm
BD² = AD² + AB²
BD² = 10² + 10²
BD² = 2.10²
BD = √(10²).2
BD = 10√2 cm
DF = BD - BF
DF = ( 10√2 - 10 ) cm
∆DFE { DF : EF : DE = 1 : 1 : √2 }
DF : DE = 1 : √2
( 10√2 - 10 )/DE = 1/√2
DE = √2( 10√2 - 10 )
DE = ( 20 - 10√2 )
CE = CD - DE
CE = 10 - ( 20 - 10√2 )
CE = ( 10√2 - 10 ) cm
Semoga bisa membantu
[tex] \boxed{ \colorbox{navy}{ \sf{ \color{lightblue}{ Answer\:by\: BLUEBRAXGEOMETRY}}}} [/tex]
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Materi : Kesebangunan dan Kekonggruenan
Konsep garis bagi membuat
∆BCE kongruen ∆BFE
∆BCE { a¹ = 10 , t¹ = CE(x) , s¹ = BE }
∆BFE { a² = BF , t² = EF(x) , s² = BE }
Panjang BF
s²/s¹ = a²/a¹
BE/BE = BF/10
BF = 1(10)
BF = 10 cm
Panjang BD
BD² = AD² + AB²
BD² = 10² + 10²
BD² = 2.10²
BD = √(10²).2
BD = 10√2 cm
Panjang DF
DF = BD - BF
DF = ( 10√2 - 10 ) cm
Panjang DE
∆DFE { DF : EF : DE = 1 : 1 : √2 }
DF : DE = 1 : √2
( 10√2 - 10 )/DE = 1/√2
DE = √2( 10√2 - 10 )
DE = ( 20 - 10√2 )
Panjang CE
CE = CD - DE
CE = 10 - ( 20 - 10√2 )
CE = ( 10√2 - 10 ) cm
Semoga bisa membantu
[tex] \boxed{ \colorbox{navy}{ \sf{ \color{lightblue}{ Answer\:by\: BLUEBRAXGEOMETRY}}}} [/tex]