Explanation:
Triangle ABC is similar to triangle FED
Using similar triangles theorem:
The ratio of corresponding sides are equal
AC corresponds to FD
AB corresponds to FE
BC corresponds to ED
AC/FD = AB/FE = BC/ED
AC/26 = 5/10 = 12/ED
solving for AC:
[tex]\begin{gathered} \frac{AC}{26}=\frac{5}{10} \\ \text{cross multiply:} \\ 10(AC)\text{ = 26(5)} \\ 10AC\text{ = 130} \\ AC\text{ = 130/10} \\ AC\text{ = 13} \end{gathered}[/tex]
5/10 = 12/ED
[tex]\begin{gathered} \frac{1}{2}=\frac{12}{ED} \\ 1(ED)\text{ = 2(12)} \\ ED\text{ = 24} \end{gathered}[/tex]
To check our answer we are asked to use another method
Using pythagoras' theorem:
Hypotenuse² = opposite² + adjacent²
For ABC:
AC² = 12² + 5²
