Question

In triangle ABC , DE parallel to BC.If AD=x, DB=x-2,AE=x+2, EC=x-1.Find the value of x..

Answer 1

In triangle ABC
DE||BC
$\frac{ad}{db} = \frac{ae}{ec} \\ \frac{x}{x - 2} = \frac{x + 2}{x - 1} \\ x(x - 1) = (x - 2)(x + 2) \\ {x}^{2} - x = {x}^{2} - {(2)}^{2} \\ {x}^{2} - {x}^{2} = x - 4 \\ x - 4 = 0 \\ x = 4$
value of x is 4

Answer 2

Concept

If ΔABC is similar to ΔPQR, then, the corresponding sides of the two triangles are in the same proportion i.e.

                                         $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}.$

Given

• A triangle ABC.

• DE is parallel to the side BC.

• AD = x

• AE = x+2

• DB = x-2

• EC = x-1

Find

The value of x.

Solution

Similarty of triangles ABC and ADE

DE is parallel to BC, so

       ∠ADE = ∠ABC,

and ∠AED = ∠ACB.

By AA similarity criterion,

ΔABC ~ ΔADE.

Relation between sides

Since triangles ABC and ADE are similar, we have,

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

Put the values of the sides in the above relation.

                       $\frac{x}{(x-2)} = \frac{(x+2)}{(x-1)}.$

By cross multiplication,

     $\;\;\;\;\;\;x(x-1) &=& (x-2)(x+2)\\\Rightarrow\;\;\;\;\;\;\; x^2 - x &=& x^2 -2^2$

     ⇒         x = 4.

The required value of x is 4.

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