Answer:
The distance between L and the line KH and HF is;
[tex]27[/tex]
The value of y is;
[tex]y=9[/tex]
Explanation:
Given the figure in the attached image;
[tex]\begin{gathered} KL=27 \\ KHL=(6y)^{\circ} \\ FHL=(4y+18)^{\circ} \end{gathered}[/tex]
From the image;
[tex]KL=FL=27[/tex]
So, the distance between L and the line KH and HF is;
[tex]27[/tex]
Also, the triangles KHL and FHL are congruent.
So, the angles KHL and FHL are congruent.
[tex]m\measuredangle KHL\cong m\measuredangle FHL[/tex]
Substituting the expressions;
[tex]\begin{gathered} (6y)^{\circ}=(4y+18)^{\circ} \\ 6y=4y+18 \\ 6y-4y=18 \\ 2y=18 \\ y=\frac{18}{2} \\ y=9 \end{gathered}[/tex]
Therefore, the value of y is;
[tex]y=9[/tex]
