Answer:
Perimeter of ΔABC is 9.5 in.
Step-by-step explanation:
Given:
ΔABC [tex]\sim[/tex] ΔDEF
DE = 6 in.
EF = 5.25 in.
DF = 3 in.
AB = 4 in.
We need to find the Perimeter of ΔABC.
Solution:
First we will find the sides of ΔABC.
Now By Triangle similarity property which states that:
"When two triangles are similar the the ratio of their corresponding sides are equal."
From Above property we can say that;
[tex]\frac{AB}{DE} =\frac{BC}{EF}=\frac{AC}{DF}\\\\\frac{4}{6}=\frac{BC}{5.25}=\frac{AC}{3}[/tex]
Now we will find BC and AC
[tex]\frac{4}{6}=\frac{BC}{5.25}\\\\BC = \frac{4\times5.25}{6}= 3.5 \ in[/tex]
Also;
[tex]\frac{4}{6}=\frac{AC}{3}\\\\AC = \frac{4\times3}{6} = 2 \ in[/tex]
Now In ΔABC
AB = 4 in
BC = 3.5 in
AC =2 in.
Now Perimeter of ΔABC can be calculated as sum of all sides.
Perimeter of ΔABC = AB +BC +AC = 4 + 3.5 + 2 = 9.5 in
Hence Perimeter of ΔABC is 9.5 in.
