Answer:
The length of KM is [tex]\sqrt{109}[/tex] units.
Step-by-step explanation:
Given information: KLMN is a trapezoid, ∠N= ∠KML, FD=8, [tex]\frac{LM}{KN}=\frac{3}{5}[/tex], F∈ KL, D∈ MN , ME ⊥ KN KF=FL, MD=DN, [tex]ME=3\sqrt{5}[/tex].
From the given information it is noticed that the point F and D are midpoints of KL and MN respectively. The height of the trapezoid is [tex]3\sqrt{5}[/tex].
Midsegment is a line segment which connects the midpoints of not parallel sides. The length of midsegment of average of parallel lines.
Since [tex]\frac{LM}{KN}=\frac{3}{5}[/tex], therefore LM is 3x and KN is 5x.
[tex]\frac{3x+5x}{2}=8[/tex]
[tex]\frac{8x}{2}=8[/tex]
[tex]x=2[/tex]
Therefore the length of LM is 6 and length of KN is 10.
Draw perpendicular on KN form L and M.
[tex]KN=KA+AE+EN[/tex]
[tex]10=6+2(EN)[/tex] (KA=EN, isosceles trapezoid)
[tex]EN=2[/tex]
[tex]KE=KN-EN=10-2=8[/tex]
Therefore the length of KE is 8.
Use pythagoras theorem is triangle EKM.
[tex]Hypotenuse^2=base^2+perpendicular^2[/tex]
[tex](KM)^2=(KE)^2+(ME)^2[/tex]
[tex](KM)^2=(8)^2+(3\sqrt{5})^2[/tex]
[tex]KM^2=64+9(5)[/tex]
[tex]KM=\sqrt{109}[/tex]
Therefore the length of KM is [tex]\sqrt{109}[/tex] units.
