Question

In a $\Delta ABC,\angle B={90}^o$ and $BD$ is perpendicular to $AC$. If $AD=4cm$ and $CD=5cm$, find $BD$ and $AB$.

Answer 1

R.E.F image

Given $BD\perp AC,AD=4cm,CD=5cm$

To find BD and AB

In $△$ ABC;

using Pythagoras therm $(AC{)}^2=(AB{)}^2+(BC{)}^2$

$81=(A{B}^2+(BC{)}^2)$ ___(I)

In $△BDC$

using Pythagoras therm : $(BC{)}^2=(BD{)}^2+(CD{)}^2$

$(BC{)}^2=(BD{)}^2+25$

$(BC{)}^2-(BD{)}^2=25$ ____(II)

In $△ABD$

using Pythagoras therm : $(AB{)}^2=(BD{)}^2+(AD{)}^2$

$(AB{)}^2=(BD{)}^2+16$

$(AB{)}^2-(BD{)}^2=16$ ___(III)

(II) - (III)

$(BC{)}^2-(AB{)}^2=9$___(IV)

(I)+(IV)

$2(BC{)}^2=90$

$(BC{)}^2=45$

$BC=\sqrt{45}=3\sqrt{5}cm$

$(AB{)}^2=45-9$

$(AB{)}^2=36$

$AB=6cm$

$(BD{)}^2=45-25$

$(BD{)}^2=20$

$BD=\sqrt{20}=2\sqrt{5}cm$

$\therefore BD=2\sqrt{5}cm$

$AB=6cm$