The length of segment CF is 15
Step-by-step explanation:
Let [tex]A_{1}[/tex] = ABEC = 133
Let [tex]A_{2}[/tex] = ABFD = 140
Let A = ABFC = 161
Let [tex]A_{rectangle}[/tex] = A = 8 AD
Let x = CF
[tex]A_{1}[/tex]+ [tex]A_{2}[/tex] = A + [tex]A_{0}[/tex]
133 + 140 = 161 + [tex]A_{0}[/tex]
[tex]A_{0}[/tex] = 133 + 140 - 161
[tex]A_{0}[/tex] = 122 [tex]A_{0}[/tex] = 8AD
8AD = 122
AD= [tex]\frac{122}{8}[/tex]
AD=14
A=[tex]\frac{(8+x)}{2}[/tex] × AD
161 =[tex]\frac{(8+x)}{2}[/tex] ×14
161 = (8 + x )×7
23 = 8 + x
x = 23 - 8
x= 15
