Question

1) Construct ∆LMN, in which ∠M = 60degree

, ∠N = 80degree

and LM + MN + NL = 11cm
​

Answer 1

Answer:

MS = LM and NT = LN …..(i)  MS + MN + NT = ST [S-M-N, M-N-T]  ∴ LM + MN + LN = ST …..

(ii)  Also,  LM + MN + LN = 11 cm ….

(iii)  ∴ ST = 11 cm [From (ii) and (iii)]  ii. In ∆LSM  LM = MS ∴ ∠MLS = ∠MSL = x° …..(iv) [isosceles triangle theorem]  In ∆LMS, ∠LMN is the exterior angle.  ∴ ∠MLS + ∠MSL = ∠LMN [Remote interior angles theorem]  ∴ x + x = 60° [From (iv)]  ∴ 2x = 60° 

∴ x = 30° 

∴ ∠LSM = 30°  ∴ ∠S = 30°  Similarly, ∠T = 40° 

iii. Now, in ∆LST  ∠S = 30°, ∠T = 40° and ST = 11 cm  Hence, ALST can be drawn. iv. Since, LM = MS 

∴ Point M lies on perpendicular bisector of seg LS.  Also LN = NT

∴ Point N lies on perpendicular bisector of seg LT. 

∴ Points M and N can be located by drawing the perpendicular bisector of LS and LT respectively. 

∴ ∆LMN can be drawn. Steps of construction: i. Draw seg ST of length 11 cm.  ii. From point S draw ray making angle of 30°. 

iii. From point T draw ray making angle of 40°. 

iv. Name the point of intersection of two rays as L. 

v. Draw the perpendicular bisector of seg LS and seg LT intersecting seg ST in M and N respectively. 

vi. Join LM and LN. Hence, ∆LMN is the required triangle.