The trapezoids LMNK and LMAB are simillar, which means that their sides are related by the same ratio. With this we can form the following equation:
[tex]\frac{2x-2}{-15+6x}=\frac{-15+6x}{32}[/tex]
We need to solve for x.
[tex]\begin{gathered} 32\cdot(2x-2)=(-15+6x)^2 \\ 64x-64=225-180x+36x^2 \\ 36x^2-180x-64x+225+64=0 \\ 36x^2-180x-64x+225+64=0 \\ 36x^2-244x+289=0 \end{gathered}[/tex]
Now we have to find the roots of the quadratic equation.
[tex]\begin{gathered} x=\frac{-(-244)\pm\sqrt[]{(-244)^2-4(36)(289)}}{2\cdot36} \\ x=\frac{244\pm\sqrt[]{17920}}{64} \\ x=\frac{244\pm133.87}{64} \\ x=\frac{244+133.87}{64}=5.9 \\ x=\frac{244-133.87}{64}=1.72 \end{gathered}[/tex]
Therefore the midsegment is equal to:
[tex]\begin{gathered} -15+6\cdot(5.9)=20.4_{} \\ or \\ -15+6\cdot(1.72)=-4.68 \end{gathered}[/tex]
Since the length can't be negative, the correct result is 20.4.
