Given:
KM=12 cm.
KO=1+KL
[tex]KL=\frac{1}{3}LM[/tex]
Since KM=12 cm, we can write
[tex]\begin{gathered} KM=KL+LM \\ KM=\frac{1}{3}LM+LM \\ 12\text{ =}\frac{4}{3}LM \\ LM=\frac{12\times3}{4} \\ LM=9 \end{gathered}[/tex]
Therefore, KL can be calculated as,
[tex]\begin{gathered} KL=\frac{1}{3}LM \\ =\frac{1}{3}\times9 \\ =3 \end{gathered}[/tex]
Now, KO can be calculated as,
[tex]\begin{gathered} KO=1+KL \\ =1+3 \\ =4 \end{gathered}[/tex]
Now, using geometric property,
[tex]KM\times KL=KN\times KO[/tex]
Putting the values in the above equation, KN can be calculated as,
[tex]\begin{gathered} 12\times3=KN\times4 \\ KN=\frac{12\times3}{4} \\ KN=9 \end{gathered}[/tex]
Now, ON can be calculated as,
[tex]\begin{gathered} ON=KN-KO \\ =9-4 \\ =5 \end{gathered}[/tex]
Since LM=9 is a chord longer than MN in the given circle, the length of MN is less than 9.
Therefore, the segments with length 9 are LM and KN.
