Question

AD=(8x-7)cm,DB=(5x-3)cm,AE=(4x-3)cm and EC=(3x-1)cm,find the value of x​

Answer 1

$\tt AD = (8x - 7) cm,\; DB=(5x - 3) cm, \; AE=(4x - 3)cm \\ \tt \;and\; EC= (3x -1)cm, \;find \;the\; value\; of\; x.$

$\Large{\underbrace{\red{\sf Required \; Solution:}}}$

$\frak{\red{Given:}} \begin{cases} \orange{\sf AD = 8x - 7} \\ \orange{\sf DB = 5x - 3 }\\ \orange{\sf AER = 4x - 3} \\ \orange{ \sf EC = 3x -1} \end{cases}$

$\textsf {Required to find x}$

• By using Thαles Theorem, [ As DE ∥ BC]

       $\longmapsto \sf \dfrac{AD}{BD} = \dfrac{AE}{CE}$

       $\longmapsto \sf \dfrac{(8x-7)}{ (5x-3)} = \dfrac{ (4x-3)}{ (3x-1)}$

       $\longmapsto \sf (8x- 7)(3x - 1) = (5x - 3)(4x - 3)$

       $\longmapsto \sf 24x^2 - 29x + 7 = 20x^2 - 27x + 9$

       $\longmapsto\sf 4x^2 - 2x - 2 = 0$

       $\longmapsto \sf 2(2x^2 - x - 1) = 0$

       $\longmapsto \sf 2x^2 - x - 1 = 0$

       $\longmapsto \sf 2x^2 - 2x + x - 1 = 0$

       $\longmapsto \sf 2x(x - 1) + 1(x - 1) = 0$

       $\longmapsto \sf{ (x - 1)(2x + 1) = 0}$

       $\Longrightarrow \sf x = 1 \; or \; x = \dfrac{-1}{2}$

We know thαt the side of triαngle can never be negαtive. Therefore, we tαke the positive vαlue.

$\boxed{\orange{\sf{\therefore \; x = 1}}}$

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Hope it elps! :)