Given: KLMN is a trapezoid m∠N = m∠KML ME ⊥ KN , ME = [tex] 3\sqrt{5} [/tex], KE = 8, LM:KN = 3:5 Find: KM, LM, KN, Area of KLMNI know that KM is [tex] \sqrt{109} [/tex]

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.