Question

In triangle abc, DE // BC and AD/DB = 3/5. If AC= 4.8 cm, find the AE.

Answer 1

$\textbf{Given:}$

$\text{In triangle ABC, DE{\parallel}BC , \dfrac{AD}{DB}=\dfrac{3}{4} , AC=4.8 cm}$

$\textbf{To find:}$

$AE$

$\textbf{Solution:}$

$\textbf{Basic proportionality theorem(Thales theorem):}$

$\textbf{If a line is drawn parallel to one side of a}$

$\textbf{triangle, then it cuts other two sides proportionally.}$

$\text{Let}\;AE=x$

$\text{Then,}\;EC=4.8-x$

$\text{By Thales theorem, we have}$

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

$\dfrac{3}{5}=\dfrac{x}{4.8-x}$

$3(4.8-x)=5\,x$

$14.4-3\,x=5\,x$

$14.4=5\,x+3\,x$

$8\,x=14.4$

$\implies\,x=\dfrac{14.4}{8}$

$\implies\,x=1.8\,cm$

$\textbf{Answer:}$

$\bf\,AE=1.8\,cm$

Answer 2

Given :- ∆ABC.

DE//BC

& AD//DB = 3/5

AC = 4.8

To Find :- AE = ?

Proof :- In ∆ABC, DE//BC

Therefore, By BPT

$\frac{AD} { AB } = \frac{ AE }{EC}$

$\frac{3}{5} = \frac{AE}{4.8 -AE}$

$3(4.8 - AE) = 5AE \\ 14.4 - 3AE = 5AE \\ 3AE + 5AE = 14.4 \\ 8AE = 14.4 \\ AE = \frac{14.4}{8} \\ AE = 1.8cm$

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