Question

In exercises 1 and 2 , identify the bisector of ST then find ST

Answer 1

Given: The line segment ST as shown in the image

To Determine: The bisector of ST and the value of ST

Solution

It can be observed from the first image, the bisector of ST is line MW

`\begin{gathered} ST=SM+MT \\ SM=MT(given) \\ MT=19(given) \\ Therefore \\ ST=19+19 \\ ST=38 \end{gathered}`

For the second image, the bisector of ST is line LM

`\begin{gathered} ST=SM+MT \\ SM=3x-6 \\ MT=x+8 \\ SM=MT(given) \\ Therefore \\ 3x-6=x+8 \\ 3x-x=8+6 \\ 2x=14 \\ x=(14)/(2) \\ x=7 \end{gathered}`
`\begin{gathered} SM=3(7)-6=21-6=15 \\ MT=7+8=15 \\ ST=SM+MT \\ ST=15+15 \\ ST=30 \end{gathered}`

For first exercise, the bisector is MW, ST = 38

For the second exercise, the bisector is LM, ST = 30r