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.(i) If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC. (ii) If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD. (iii) If AD/DB = 4/7 and AC = 6.6cm, find AE. (iv) If AD/AB = 8/15 and EC = 3.5cm, find AE.
Answer» (i) In ∆ ABC, it is given that DE ∥ BC. Applying Thales’ theorem, we get: AD/DB = AE/EC ∵ AD = 3.6 cm , AB = 10 cm, AE = 4.5cm ∴ DB = 10 − 3.6 = 6.4cm Or, 3.6/6.4 = 4.5/EC Or, EC = 6.4×4.5/3.6 Or, EC= 8 cm Thus, AC = AE + EC = 4.5 + 8 = 12.5 cm (ii) In ∆ ABC, it is given that DE \|\| BC. Applying Thales’ Theorem, we get : AD/DB = AE/EC Adding 1 to both sides, we get : AD/DB + 1 = AE/ EC +1 ⇒ AB/DB = AC/EC ⇒ 13.3/DB = 11.9/5.1 ⇒ DB = 13.3×5.1/11.9 = 5.7 cm Therefore, AD=AB-DB=13.5-5.7=7.6 cm (iii) In ∆ ABC, it is given that DE \|\| BC. Applying Thales’ theorem, we get : AD/DB = AE/EC ⇒ 4/7 = AE/EC Adding 1 to both the sides, we get : 11/7 = AC/EC ⇒ EC = 6.6×7/11 = 4.2 cm Therefore, AE = AC – EC = 6.6 – 4.2 = 2.4 cm (iv) In ∆ ABC, it is given that DE ‖ BC. Applying Thales’ theorem, we get: AD/AB = AE/AC ⇒ 8/15 = AE/ AE+EC ⇒ 8/15 = AE/ AE+3.5 ⇒ 8AE + 28 = 15AE ⇒ 7AE = 28 ⇒ AE = 4cm

D and E are points on the sides AB and AC respectively of a ∆ABC such that DE║BC.(i) If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC. (ii) If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD. (iii) If AD/DB = 4/7 and AC = 6.6cm, find AE. (iv) If AD/AB = 8/15 and EC = 3.5cm, find AE.

Answer»

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

Applying Thales’ theorem, we get:

AD/DB = AE/EC

∵ AD = 3.6 cm , AB = 10 cm, AE = 4.5cm

∴ DB = 10 − 3.6 = 6.4cm

Or, 3.6/6.4 = 4.5/EC

Or, EC = 6.4×4.5/3.6

Or, EC= 8 cm

Thus, AC = AE + EC

= 4.5 + 8 = 12.5 cm

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

Applying Thales’ Theorem, we get :

AD/DB = AE/EC

Adding 1 to both sides, we get :

AD/DB + 1 = AE/ EC +1

⇒ AB/DB = AC/EC

⇒ 13.3/DB = 11.9/5.1

⇒ DB = 13.3×5.1/11.9 = 5.7 cm

Therefore, AD=AB-DB=13.5-5.7=7.6 cm

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

Applying Thales’ theorem, we get :

AD/DB = AE/EC

⇒ 4/7 = AE/EC

Adding 1 to both the sides, we get :

11/7 = AC/EC

⇒ EC = 6.6×7/11 = 4.2 cm

Therefore,

AE = AC – EC = 6.6 – 4.2 = 2.4 cm

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

Applying Thales’ theorem, we get:

AD/AB = AE/AC

⇒ 8/15 = AE/ AE+EC

⇒ 8/15 = AE/ AE+3.5

⇒ 8AE + 28 = 15AE

⇒ 7AE = 28

⇒ AE = 4cm