angle ABC is such that AB is equal to 3cm, bc is equal to 2cm and ca i equal to 2.5cm, if angle DEF angle ABC and EF is equal to 44cm, then perimeter of angle DEF is.
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Given :
AB = 3cm, BC = 2cm, and CA = 2.5cm, and EF = 4cm.
Also, [tex]\rm ΔABC ≈ ΔDEF ΔABC≈ΔDEF[/tex]
Thus, [tex]\large\rm\frac{ AB}{DE}[/tex]
= [tex]\large\rm\frac{BC}{EF}[/tex]
= [tex]\large\rm\frac{AC}{DF}\times[/tex][tex]\large\rm\frac{3}{DE}[/tex]=[tex]\large\rm\frac{2}{4}[/tex]=[tex]\large\rm\frac{2.5}{DF}[/tex]
Hence, DE = 6cm and DF = 5cm
Perimeter of [tex]\text{ ΔDEF = DE+EF+EF }[/tex]
[tex]\text{ =4+5+6 }[/tex]
[tex]\text{ =15cm } [/tex]
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Therefore, Perimeter of [tex]\rm{ ΔDEF \: = \: 15cm }[/tex]
Answer:
Given: AB=3cm, BC=2cm and CA=2.5cm and EF=4cm
Given: AB=3cm, BC=2cm and CA=2.5cm and EF=4cm Also, △ABC∼△DEF
Given: AB=3cm, BC=2cm and CA=2.5cm and EF=4cm Also, △ABC∼△DEFThus, DEAB=EFBC=DFAC
Given: AB=3cm, BC=2cm and CA=2.5cm and EF=4cm Also, △ABC∼△DEFThus, DEAB=EFBC=DFACDE3=42=DF2.5
Given: AB=3cm, BC=2cm and CA=2.5cm and EF=4cm Also, △ABC∼△DEFThus, DEAB=EFBC=DFACDE3=42=DF2.5Hence, DE=6cm and DF=5cm
Given: AB=3cm, BC=2cm and CA=2.5cm and EF=4cm Also, △ABC∼△DEFThus, DEAB=EFBC=DFACDE3=42=DF2.5Hence, DE=6cm and DF=5cmPerimeter of Δ DEF = DE+EF+EF
Given: AB=3cm, BC=2cm and CA=2.5cm and EF=4cm Also, △ABC∼△DEFThus, DEAB=EFBC=DFACDE3=42=DF2.5Hence, DE=6cm and DF=5cmPerimeter of Δ DEF = DE+EF+EFPerimeter of Δ DEF = 4+5+6
Given: AB=3cm, BC=2cm and CA=2.5cm and EF=4cm Also, △ABC∼△DEFThus, DEAB=EFBC=DFACDE3=42=DF2.5Hence, DE=6cm and DF=5cmPerimeter of Δ DEF = DE+EF+EFPerimeter of Δ DEF = 4+5+6Perimeter of Δ DEF = 15 cm