Step 1

Two triangles are similar if the lenght of corrspanding sides are proportional.

Step 2

\(\displaystyle\triangle{P}{M}{L}\sim\triangle{T}{R}{Q}\)

\(\displaystyle\Rightarrow\frac{{{P}{M}}}{{{T}{R}}}=\frac{{{M}{L}}}{{{R}{Q}}}\)

\(\displaystyle\Rightarrow\frac{{{15}}}{{{14}}}=\frac{{{7}{x}-{9}}}{{{2}{x}+{2}}}\)

\(\displaystyle\Rightarrow{35}{\left({2}{x}+{2}\right)}={14}{\left({7}{x}-{9}\right)}\)

\(\displaystyle\Rightarrow{70}{x}+{70}={98}{x}-{126}\)

\(\displaystyle\Rightarrow{70}+{126}={98}{x}-{70}{x}\)

\(\displaystyle\Rightarrow{196}={28}{x}\)

\(\displaystyle\Rightarrow{x}=\frac{{196}}{{28}}={7}\)

Step 3

Now

\(\displaystyle{L}{M}={7}{x}-{9}\)

\(\displaystyle{L}{M}={7}{\left({7}\right)}-{9}\)

\(\displaystyle{L}{M}={40}\)

Two triangles are similar if the lenght of corrspanding sides are proportional.

Step 2

\(\displaystyle\triangle{P}{M}{L}\sim\triangle{T}{R}{Q}\)

\(\displaystyle\Rightarrow\frac{{{P}{M}}}{{{T}{R}}}=\frac{{{M}{L}}}{{{R}{Q}}}\)

\(\displaystyle\Rightarrow\frac{{{15}}}{{{14}}}=\frac{{{7}{x}-{9}}}{{{2}{x}+{2}}}\)

\(\displaystyle\Rightarrow{35}{\left({2}{x}+{2}\right)}={14}{\left({7}{x}-{9}\right)}\)

\(\displaystyle\Rightarrow{70}{x}+{70}={98}{x}-{126}\)

\(\displaystyle\Rightarrow{70}+{126}={98}{x}-{70}{x}\)

\(\displaystyle\Rightarrow{196}={28}{x}\)

\(\displaystyle\Rightarrow{x}=\frac{{196}}{{28}}={7}\)

Step 3

Now

\(\displaystyle{L}{M}={7}{x}-{9}\)

\(\displaystyle{L}{M}={7}{\left({7}\right)}-{9}\)

\(\displaystyle{L}{M}={40}\)