Question

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

• A. $21.21$ cm, $84.42$ cm
• B. $2121$ cm, $84.42$ cm
• C. $21.21$ cm, $84.2$ cm
• D. $21.2$ cm, $84.42$ cm

Answer 1

Answer: (A)

$21.21$ cm, $84.42$ cm

Given, $\angle A=\angle D={90}^{\circ }$, $AB=AC$
In $△BDC$,
$B{C}^2=B{D}^2+C{D}^2$ (By Pythagoras Theorem)
$B{C}^2={18}^2+{24}^2$
$B{C}^2=900$
$BC=30$ cm
In $△ABC$,
$B{C}^2=A{B}^2+A{C}^2$ (Pythagoras Theorem)
$900=2A{B}^2$
$A{B}^2=450$
$AB=21.21$ cm
Perimeter of ABCD = $AB+BD+CD+CA$
Perimeter of ABCD = $21.21+18+24+21.21$
Perimeter of ABCD = $84.42$ cm