Question

KLMN is an isosceles trapezium whose diagonals cut at X and KL is parallel to NM. If ∠KNL = 25°, ∠KMN = 30°, find 

(a)∠KXN 

(b) ∠MLN.

Answer 1

In Δs KLN and KLM 

KN = LM            (Isos. trap.) 

∠NKL = ∠KLM     (Prop. of isos. trap.) 

KL = KL              (Common) 

∴ ΔKLN ≅ ΔKLM         (SAS) 

⇒ ∠KML = ∠KNL =25°        (c.p.c.t.) 

∴ ∠LMN = 25° + 30° = 55° 

Now ∠KNM =∠LMN 

⇒ ∠KNL + ∠LNM = 55° ⇒ ∠LNM = 55° – 25° = 35° 

In ΔNXM, ∠NXM = 180° – (∠XNM + ∠XMN) 

= 180° – (30° + 30°) = 180° – 60° = 120° 

(i) Now, ∠KXN = 180°– ∠NXM = 180° – 120° = 60°    (Linear pair) 

Also, ∠LXM = ∠KXN = 60° 

(ii) Now, in ΔMLX, ∠MLX (∠MLN) = 180° – (60° + 25°) 

= 180° – 85° = 95°