D and E are points on the sides AB and AC respectively of a ∆ABC suc

1.	D and E are points on the sides AB and AC respectively of a ∆ABC such that DE║BC. Find the value of x, when(i) AD = x cm, DB = (x – 2) cm, AE = (x + 2) cm and EC = (x – 1) cm. (ii) AD = 4cm, DB = (x – 4) cm, AE = 8cm and EC = (3x – 19) cm. (iii) AD = (7x – 4) cm, AE = (5x – 2) cm, DB = (3x + 4) cm and EC = 3x cm.
Answer» (i) In ∆ ABC, it is given that DE \|\| BC. Applying Thales’ theorem, we have : AD/DB = AE/EC ⇒ x/x−2 = x+2/x−1 ⇒X(x-1) = (x-2) (x+2) ⇒x² – x = x² − 4 ⇒x= 4 cm (ii) In ∆ ABC, it is given that DE ‖ BC. Applying Thales’ theorem, we have : AD/DB = AE/EC ⇒ 4/x−4 = 8/3x−19 ⇒ 4 (3x-19) = 8 (x-4) ⇒12x – 76 = 8x – 32 ⇒4x = 44 ⇒ x = 11 cm (iii) In ∆ ABC, it is given that DE \|\| BC. Applying Thales’ theorem, we have : AD/DB = AE/EC ⇒ 7x−4/3x+4 = 5x−2/3x ⇒3x (7x-4) = (5x-2) (3x+4) ⇒21x² – 12x = 15x² + 14x-8 ⇒6x² – 26x + 8 = 0 ⇒(x-4) (6x-2) =0 ⇒ x = 4, 1/3 ∵ x ≠ 1/3 (as if x = 1/3 then AE will be negative) ∴ x = 4 cm

D and E are points on the sides AB and AC respectively of a ∆ABC such that DE║BC. Find the value of x, when(i) AD = x cm, DB = (x – 2) cm, AE = (x + 2) cm and EC = (x – 1) cm. (ii) AD = 4cm, DB = (x – 4) cm, AE = 8cm and EC = (3x – 19) cm. (iii) AD = (7x – 4) cm, AE = (5x – 2) cm, DB = (3x + 4) cm and EC = 3x cm.

Answer»

(i) In ∆ ABC, it is given that DE || BC.

Applying Thales’ theorem, we have :

AD/DB = AE/EC

⇒ x/x−2 = x+2/x−1

⇒X(x-1) = (x-2) (x+2)

⇒x² – x = x² − 4

⇒x= 4 cm

(ii) In ∆ ABC, it is given that DE ‖ BC.

Applying Thales’ theorem, we have :

AD/DB = AE/EC

⇒ 4/x−4 = 8/3x−19

⇒ 4 (3x-19) = 8 (x-4)

⇒12x – 76 = 8x – 32

⇒4x = 44

⇒ x = 11 cm

(iii) In ∆ ABC, it is given that DE || BC.

Applying Thales’ theorem, we have :

AD/DB = AE/EC

⇒ 7x−4/3x+4 = 5x−2/3x

⇒3x (7x-4) = (5x-2) (3x+4)

⇒21x² – 12x = 15x² + 14x-8

⇒6x² – 26x + 8 = 0

⇒(x-4) (6x-2) =0

⇒ x = 4, 1/3

∵ x ≠ 1/3 (as if x = 1/3 then AE will be negative)

∴ x = 4 cm