question_answer

In \[\Delta \,DEW,AB\parallel EW\]. If \[AD=4cm,\,\,DE\,=12cm\] and \[DW=24\text{ }cm\] , then find the of \[DB\].

Answer:

Let \[BD=x\text{ }cm\],

\[DW=24\text{ }cm\],

Then, \[BW=(24-x)\text{ }cm,\text{ }AE=12-4=8\text{ }cm\]

In \[\Delta \,DEW,\,AB\parallel EW\]

\[\therefore \frac{AD}{AE}=\frac{BD}{BW}\]                  [Thales? Theorem]

\[\Rightarrow \frac{4}{8}=\frac{x}{24-x}\]

\[\Rightarrow 8x=96-4x\]

\[\Rightarrow 12x=96\]

\[\Rightarrow x=\frac{96}{12}=8\,cm\]

\[\therefore DB=8\,cm\]