Question

In figure
DE || BC.If AD=x,DB=x-2,AE=x+2 and EC =x-1,,Find the value of x.

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Answer 1

Answer:

$\fbox \pink {QUESTION}$

In figure

DE || BC.If AD=x,DB=x-2,AE=x+2 and EC =x-1,,Find the value of x.

$\fbox \pink {SOLUTION}$

GIVEN :-

ABC is a triangle,DE || BC,AD=x,DB=x-2,AE=x+2 and EC= x-1.

TO FIND :-

Value of x.

ANSWER :-

In ∆ ABC , we have

DE || BC

Therefore [By Thale's theorem]

$\bf \frac{AD}{DB} = \frac{AE }{ EC} \\ \bf AD \times EC = AE \times DB \\ \\ \bf x(x - 1) = (x - 2)(x + 2) \\ \bf = {x}^{2} - x = {x}^{2} - 4 \\ \bf \purple {➛x = 4}$

Hence the value of x is 4 .

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