Question

In a quadrilateral $ABCD$ in which $AD=13$ cm, $DC=12$ cm, $BC=3$ cm and $\angle ABD=\angle BCD={90}^0$. Calculate the length of $AB$ in cm.

Answer 1

$△BCD$ is right angle triangle,

So, by using Pythagoras theorem,

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

$\Rightarrow (3{)}^2+(12{)}^2$

$\Rightarrow 9+144$

$\Rightarrow 153$

$\therefore \left(BD\right)=3\sqrt{17}$ cm.

$△ABD$ is right angle triangle,

So, by using pythagoras theorem,

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

$(13{)}^2=(AB{)}^2+(3\sqrt{17}{)}^2$

$(AB{)}^2=169-153$

$(AB{)}^2=16$

Taking square root on both sides,

$\therefore AB=4$ cm