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    In the given figure, DE ∥ BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, then find the value of x.  
    Question

    In the given figure, DE ∥ BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, then find the value of x.  

    A.

    3

    B.

    4

    C.

    5

    D.

    2

    Correct option is B

    Given:

    DE ∥ BC AD = x, DB = x - 2, AE = x + 2, and EC = x - 1

    Formula Used:

    By the basic proportionality theorem, if a line is parallel to the side of a triangle and intersects the remaining two sides into two distinct points, then the lines divide the sides into proportion.

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

    Soluiton:

    By basic proportionality:

    ADDB=AEEC x(x2)=(x+2)(x1) x(x1)=(x+2)(x2) x2x=x2+2x2x4 x=4\frac{AD }{DB}= \frac{AE }{EC } \\\ \\\frac{x }{ (x - 2) }= \frac{(x + 2) }{(x - 1) } \\\ \\x(x - 1) = (x + 2)(x - 2) \\\ \\x² - x = x² + 2x - 2x - 4 \\\ \\x = 4​​

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