Question

In triangle ABC, DE is parallel to BC such that AD = (7x -4)cm, AE = 5 cm, EC = 3 cm and DB = (3x+4)cm, then find the value of x.

Answer 1

$\textbf{Given:}$

$\text{In \triangle ABC, DE{\parallel}BC }$

$\text{AD=(7x-4)cm, AE=5 cm, EC=3 cm and DB=(3x+4)cm}$

$\textbf{To find:}$

$\text{The value of x}$

$\textbf{Solution:}$

$\textbf{Thales theorem:}$

$\textbf{If a line is drawn parallel to one side of a triangle, then it cuts}$

$\textbf{other two sides proportionally}$

$\text{Since DE{\parallel}BC , we have}$

$\text{By Thales theorem}$

$\dfrac{AD}{DB}=\dfrac{AE}{EC}$

$\dfrac{7x-4}{3x+4}=\dfrac{5}{3}$

$3(7x-4)=5(3x+4)$

$21x-12=15x+20$

$21x-15x=20+12$

$6x=32$

$x=\dfrac{32}{6}$

$\implies\bf\,x=\dfrac{16}{3}$

$\textbf{Answer:}$

$\textbf{The value of x is \bf\,\dfrac{16}{3} }$

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