Question

In the given figure, DE || BC. If AD = 25 cm , AE=19 cm,BD=(x+3)cm, and EC=x cm. then AB is equal to​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{DE{\parallel}BC\;and\;AD=25\,cm,AE=19cm,BD=x+3\,cm,EC=x\,cm}$

$\mathsf{}$

$\underline{\textbf{To find:}}$

$\textsf{AB}$

$\underline{\textbf{Solution:}}$

$\underline{\textbf{Basic proportionality theorem:}}$

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

$\textsf{cuts other two sides proportionally}$

$\mathsf{In\;\triangle\,ABC,DE{\parallel}BC}$

$\textsf{By Basic proportionality theorem,}$

$\mathsf{\dfrac{AD}{BD}=\dfrac{AE}{EC}}}$

$\mathsf{\dfrac{25}{x+3}=\dfrac{19}{x}}$

$\mathsf{25x=19(x+3)}$

$\mathsf{25x=19x+57}$

$\mathsf{25x-19x=57}$

$\mathsf{6x=57}$

$\mathsf{x=\dfrac{57}{6}}$

$\mathsf{x=\dfrac{19}{2}}$

$\implies\mathsf{x=9.5}$

$\mathsf{Now,}$

$\mathsf{AB=AD+BD}$

$\mathsf{AB=25+(x+3)}$

$\mathsf{AB=25+(9.5+3)}$

$\implies\boxed{\mathsf{AB=37.5}}$

$\underline{\textbf{Find more:}}$

In ∆ABC, PQ||AB. If AP=10cm. PC=8cm, QC=12cm. then find BQ

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