Question

In ∆, DE ǁ BC, If AD = x + 3, AB = 2x, AE = x + 5 and AC = 2x + 3, then value of x is
​

Answer 1

In Δ ABC, DE ∥ BC

∴

DB

AD

=

EC

AE

(By basic proportionality theorem)

⇒

x−2

x

=

x−1

x+2

⇒x(x−1)=(x+2)(x−2)

⇒x

2

−x=x

2

−4

⇒x=4.

Answer 2

Given:

In ∆, DE ǁ BC, If AD = x + 3, AB = 2x, AE = x + 5 and AC = 2x + 3, then value of x is?

To find:  

The value of x  

Solution:  

We know that,  

$\boxed{\bold{Thales\:Theorem\:/\:Basic\:Proportionality \:Theorem}}:$

If a line is drawn parallel to one of the sides of a triangle intersecting the other two sides at two distinct points, then the other two sides are divided in the same proportion.

Based on the Thales Theorem for Δ ABC since DE // BC, we get  

$\frac{AD}{DB} = \frac{AE}{EC}\\\\\implies \frac{AD}{AB - AD} = \frac{AE}{AC - EC}$  

on substituting AD = x + 3, AB = 2x, AE = x + 5 and AC = 2x + 3, we get  

$\implies \frac{x + 3}{2x - (x + 3)} = \frac{x + 5}{(2x + 3) - (x + 5)}$

$\implies \frac{x + 3}{2x - x - 3} = \frac{x + 5}{2x + 3 - x - 5}$

$\implies \frac{x + 3}{ x - 3} = \frac{x + 5}{ x - 2}$

$\implies (x + 3)(x - 2)=(x + 5)(x - 3)$

$\implies x^2 - 2x + 3x- 6 = x^2 -3x + 5x - 15$  

$\implies x- 6 = 2x - 15$  

$\implies 2x - x = 15-6$

$\implies \bold{x = 9}$  

 

Thus, the value of x is → 9.

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