(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

From figure, D and E are the points on the sides AB and AC respectively and DE || BC

then AD/DB = AE/EC

AD = x cm, DB = (x – 2) cm, AE = (x + 2) cm and EC = (x – 1) cm.

x/(x-2) = (x+2)/(x-1)

x(x-1) = (x+2)(x-2)

Solving above equation, we get

x = 4 cm

From figure, D and E are the points on the sides AB and AC respectively and DE || BC

then AD/DB = AE/EC

AD = 4 cm, DB = (x – 4) cm, AE = 8 cm and EC = (3x – 19) cm.

AD/DB = AE/EC

4/(x-4) = 8/(3x-19)

4(3x-19) = 8(x-4)

Solving, we get x = 11 cm

Correct Answer – `(i) x=4 (ii) x=11 (ii) x=4`

Correct Answer – `(i) EC =8 cm, AC= 12.5 cm (iii) AD=7.6 cm, (iii) AE=2.4 cm, (iv) AE=4 cm`
N/A

From figure, D and E are the points on the sides AB and AC of ∆ABC.

AD = 7.2 cm, AE = 6.4 cm, AB = 12 cm and AC = 10 cm.

DB = AB – AD = 12 – 7.2 = 4.8 cm and

EC = AC – AE = 10 – 6,4 =3.6 cm

AD /DB = 7.2/4.8 = 3/2 and

AE/EC = 6.4/3.6 = 16/9

=> AD/DB ≠ AE/EC

=> DE is not parallel to BC

(i) We have: 

AD/DE = 5.7/9.5 = 0.6  

AE/EC = 4.8/8 = 0.6  

Hence, AD/DB = AE/EC 

Applying the converse of Thales’ theorem, 

We conclude that DE || BC. 

(ii) We have: 

AB = 11.7 cm, DB = 6.5 cm 

Therefore, 

AD = 11.7 -6.5 = 5.2 cm 

Similarly, 

AC = 11.2 cm, AE = 4.2 cm 

Therefore, 

EC = 11.2 – 4.2 = 7 cm 

Now, 

AD/DB = 5.2/6.5 = 4/5 

AE/EC = 4.2/7 

Thus, AD/DB ≠ AE/EC 

Applying the converse of Thales’ theorem, 

We conclude that DE is not parallel to BC. 

(iii) We have: 

AB = 10.8 cm, AD = 6.3 cm 

Therefore,

DB = 10.8 – 6.3 = 4.5 cm 

Similarly, 

AC = 9.6 cm, EC = 4cm 

Therefore, 

AE = 9.6 – 4 = 5.6 cm 

Now, 

AD/DB = 6.3/4.5 = 7/5 

AE/EC = 5.6/4 = 7/5 

⟹ AD/DB = AE/EC 

Applying the converse of Thales’ theorem, 

We conclude that DE ‖ BC. 

(iv) We have : 

AD = 7.2 cm, AB = 12 cm 

Therefore, 

DB = 12 – 7.2 = 4.8 cm 

Similarly, 

AE = 6.4 cm, AC = 10 cm 

Therefore, 

EC = 10 – 6.4 = 3.6 cm 

Now, 

AD/DB = 7.2/4.8 = 3/2 

AE/EC = 6.4/3.6 = 16/9 

This, AD/DB ≠ AE/EC 

Applying the converse of Thales’ theorem, 

We conclude that DE is not parallel to BC.

(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

From figure, D and E are the points on the sides AB and AC of ∆ABC

AD = 5.7 cm, DB = 9.5 cm, AE = 4.8 cm and EC = 8 cm

AD/DB = 5.7/9.8 = 3/5

and AE/EC = 4.8/8 = 3/5

=> AD/DB = AE/EC

=> DE || BC

From given triangle, points D and E are on the sides AB and AC respectively such that DE || BC.

AD/DB = 4/7 or AD = 4 cm, DB = 7 cm, and AC = 6.6

By Thale’s Theorem:

AD/DB = AE/EC

Add 1 on both sides

AD/DB + 1 = AE/EC + 1

(AD + DB)/DB = (AE + EC)/EC

(4+7)/7 = AC/EC = 6.6/EC

EC = (6.6 x 7)/11

= 4.2

And, AE = AC – EC

AE = 6.6 – 4.2

AE = 2.4 cm