Question

In the given figure,
$AB=AC,\angle A=\angle D={90}^0,BD=18$ cm and $DC=24$ cm.
Calculate the length of $AB$ to the two places of decimal. Also, find the perimeter of the quadrilateral $ABDC$

Answer 1

Since, $△BDC$ is a right angled triangle, we have
${BC}^2={BD}^2+{DC}^2$
$\Rightarrow {BC}^2={18}^2+{24}^2$
$\Rightarrow {BC}^2=324+576=900$
$\Rightarrow BC=30cm$
Similarly, as $△ABC$ is a right angled triangle, we have
${BC}^2={AB}^2+{AC}^2$
$\Rightarrow {30}^2={AB}^2+{AB}^2$ (Since, $AB=AC$)
$\Rightarrow 2{AB}^2=900$
$\Rightarrow {AB}^2=450$
$\Rightarrow AB=21.21cm$
And perimeter of the quadrilateral $ABCD=21.21+21.21+18+24=84.42cm$