Question

In the given figure $\angle ABC={90}^{\circ }$. And $BD\perp AC$. If $AB=5.7cm,BD=3.8cm$, and $CD=4.4cm$, find length of side $BC$.

Answer 1

Given that,

$\angle ABC={90}^0\&BD\perp AC$

and

$AB=5.7cm$

$BD=3.8cm$

$CD=4.4cm$

$BC=?$

Let $AD=xcm$ and $BC=ycm$

Then,$In\Delta ABD$

By pythagoras theorem,

$A{B}^2=A{D}^2+D{B}^2$

${\left(5.7\right)}^2=y^2+{\left(3.8\right)}^2$

$32.49=y^2+13.3$

$y^2=32.49-13.3$

$y^2=19.19$

$y=4.38cm$

Now, $In\Delta ABC$

By Pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2$

${(y+4.4)}^2={\left(5.7\right)}^2+x^2$

${(4.38+4.4)}^2=32.49+x^2$

${\left(8.78\right)}^2-32.49=x^2$

$x^2=77.87-32.49$

$x^2=45.38$

$x=6.73cm$

$BC=x=6.73cm$
Hence, this is the answer.