Answer:
The length of sides KM, LM and KN are [tex]\sqrt{109}, \frac{48}{5}[/tex] and 16 respectively. The area of KLMN is [tex]\frac{192\sqrt{5}}{5}[/tex] square unit.
Step-by-step explanation:
According to given information: KLMN is a trapezoid, ∠N= ∠KML, [tex]\frac{LM}{KN}=\frac{3}{5}[/tex], ME ⊥ KN, KE=8 [tex]ME=3\sqrt{5}[/tex].
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]
Let angle N and angle KML be θ.
Since angle KML and angle MKE are alternate interior angles, therefore angle MKE is θ.
In triangle KME,
[tex]\tan\theta=\frac{3\sqrt{5} }{8}[/tex] .... (1)
In triangle KNE,
[tex]\tan\theta=\frac{3\sqrt{5} }{EN}[/tex] .... (2)
Equate (1) and (2),
[tex]\frac{3\sqrt{5} }{8}=\frac{3\sqrt{5} }{EN}[/tex]
[tex]EN=8[/tex]
The length of side KN is
[tex]KN=KE+EN=8+8=16[/tex]
Sides LM:KN are in the ratio 3:5. Let the side lengths are 3x and 5x respectively.
[tex]5x=16[/tex]
[tex]x=\frac{16}{5}[/tex]
[tex]3x=3\times \frac{16}{5}=\frac{48}{5}[/tex]
The area of KLMN is
[tex]Area=\frac{1}{2}(b_1+b_2)h[/tex]
[tex]A=\frac{1}{2}\times (LM+KN)\times ME[/tex]
[tex]A=\frac{1}{2}\times (\frac{48}{5}+16)\times 3\sqrt{5}[/tex]
[tex]A=\frac{192\sqrt{5}}{5}[/tex]
Therefore length of sides KM, LM and KN are [tex]\sqrt{109}, \frac{48}{5}[/tex] and 16 respectively. The area of KLMN is [tex]\frac{192\sqrt{5}}{5}[/tex] square unit.
