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Home » 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.
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.
CBSE | Class 10 | Exercise 4A | Maths | RS Aggarwal | Triangles
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.

 

 

 

 

Answers:

(i)

In ∆ ABC,

DE ‖ BC

Applying Thales’ theorem,

\frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}

\begin{array}{l} \frac{x}{{x - 2}} = \frac{{x + 2}}{{x - 1}}\\ x(x - 1) = (x - 2)(x + 2)\\ {x^2} - x = {x^2} - 4\\ \therefore x = 4cm \end{array}

 

(ii)

In ∆ ABC,

DE ‖ BC

Applying Thales’ theorem,

\frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}

\begin{array}{l} \frac{4}{{x - 4}} = \frac{8}{{3x - 19}}\\ 4(3x - 19) = 8(x - 4)\\ 12x - 76 = 8x - 32\\ 4x = 44\\ \therefore x = 11cm \end{array}

 

 

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